INSERT INTO `exam_simulation2` (`id`, `examYear`, `questions`, `parent_que`, `child_que`, `gen_child_que_desc`, `imageDir`, `explanation`, `parent_que_expl`, `child_que_exp`, `gen_child_que_expl`, `package_id`, `subject_id`, `status`) VALUES
(87, '2016', '\n\nQuestion\n\n(a) Explain the term critical angle.\n(b) List two factors which determine the deviation of a ray of light by a triangular glass prism.\n(c) The angle of refraction (r) of a ray of white light from air through a triangular glass prism of refractive index 1.5 is 29.0°. Calculate the angle through which the ray is least deviated.\n(d) Study the ray diagram below and use it to answer the questions that follow.\n\n\n\nCalculate the:\n(i) values of angles P,Q and R;\n(ii) refractive index n of the glass prism;\n(iii) value of e;\n(iv) total deviation D.\n\n\n(a) Explain the term critical angle.\n(b) List two factors which determine the deviation of a ray of light by a triangular glass prism.\n(c) The angle of refraction (r) of a ray of white light from air through a triangular glass prism of refractive index 1.5 is 29.0°. Calculate the angle through which the ray is least deviated.\n(d) Study the ray diagram below and use it to answer the questions that follow.\n\n\n', '', '(a) Explain the term critical angle.\n(b) List two factors which determine the deviation of a ray of light by a triangular glass prism.\n(c) The angle of refraction (r) of a ray of white light from air through a triangular glass prism of refractive index 1.5 is 29.0°. Calculate the angle through which the ray is least deviated.\n(d) Study the ray diagram below and use it to answer the questions that follow.\n\n\n\nCalculate the:\n(i) values of angles P,Q and R;\n(ii) refractive index n of the glass prism;\n(iii) value of e;\n(iv) total deviation D.\n\n\n(a) Explain the term critical angle.\n(b) List two factors which determine the deviation of a ray of light by a triangular glass prism.\n(c) The angle of refraction (r) of a ray of white light from air through a triangular glass prism of refractive index 1.5 is 29.0°. Calculate the angle through which the ray is least deviated.\n(d) Study the ray diagram below and use it to answer the questions that follow.\n', '(a) Explain the term critical angle.\n(b) List two factors which determine the deviation of a ray of light by a triangular glass prism.\n(c) The angle of refraction (r) of a ray of white light from air through a triangular glass prism of refractive index 1.5 is 29.0°. Calculate the angle through which the ray is least deviated.\n(d) Study the ray diagram below and use it to answer the questions that follow.\n\n\n\nCalculate the:\n(i) values of angles P,Q and R;\n(ii) refractive index n of the glass prism;\n(iii) value of e;\n(iv) total deviation D.\n\n\n(a) Explain the term critical angle.~(b) List two factors which determine the deviation of a ray of light by a triangular glass prism.\n(c) The angle of refraction (r) of a ray of white light from air through a triangular glass prism of refractive index 1.5 is 29.0°. Calculate the angle through which the ray is least deviated.\n(d) Study the ray diagram below and use it to answer the questions that follow.\n\n\n\nCalculate the:\n(i) values of angles P,Q and R;\n(ii) refractive index n of the glass prism;\n(iii) value of e;\n(iv) total deviation D.\n\n\n(a) Explain the term critical angle.\n(b) List two factors which determine the deviation of a ray of light by a triangular glass prism.\n(c) The angle of refraction (r) of a ray of white light from air through a triangular glass prism of refractive index 1.5 is 29.0°. Calculate the angle through which the ray is least deviated.\n(d) Study the ray diagram below and use it to answer the questions that follow.\n~(c) The angle of refraction (r) of a ray of white light from air through a triangular glass prism of refractive index 1.5 is 29.0°. Calculate the angle through which the ray is least deviated.\n(d) Study the ray diagram below and use it to answer the questions that follow.\n\n\n\nCalculate the:\n(i) values of angles P,Q and R;\n(ii) refractive index n of the glass prism;\n(iii) value of e;\n(iv) total deviation D.\n\n\n(a) Explain the term critical angle.\n(b) List two factors which determine the deviation of a ray of light by a triangular glass prism.\n(c) The angle of refraction (r) of a ray of white light from air through a triangular glass prism of refractive index 1.5 is 29.0°. Calculate the angle through which the ray is least deviated.\n(d) Study the ray diagram below and use it to answer the questions that follow.\n~(d) Study the ray diagram below and use it to answer the questions that follow.\n\n\n\nCalculate the:\n(i) values of angles P,Q and R;\n(ii) refractive index n of the glass prism;\n(iii) value of e;\n(iv) total deviation D.\n\n\n(a) Explain the term critical angle.\n(b) List two factors which determine the deviation of a ray of light by a triangular glass prism.\n(c) The angle of refraction (r) of a ray of white light from air through a triangular glass prism of refractive index 1.5 is 29.0°. Calculate the angle through which the ray is least deviated.\n(d) Study the ray diagram below and use it to answer the questions that follow.\n', NULL, '\n\nExplanation\nCritical angle is the angle of incidence in the denser medium when the angle of refraction in the less dense medium is 90°.\n\n(b) Deviation of a ray in a triangular prism is determined by the factors:\n- Angle of incidence\n- Refracting angle of the prism, and\n- Refractive index of the prism material.\n(c) A = 2r = 2 x 29 = 58\\(^o\\)\nBut n = \\(\\frac{sin \\frac{1}{2} (A + Dm)}{sin \\frac{1}{2}A} \\)\nsin \\(\\frac{1}{2}\\)(A + Dm) = n sin A \\(\\frac{1}{2}\\)A\n58 + Dm = 93.32 - 58 = 35.32\\(^o\\)\n(d) Angle P = 45° - 20° = 25°\nBut Angle Q = 180°- 60° = 120°\nP +Q + R = 180°\nHence R = 180 - (P + Q) =180 - (25 + 120) = 35°\n(ii) n = \\(\\frac{Sini}{sinP} = \\frac{sin45^o}{sin45^o} = 1.673\\)\n(iii) n = \\(\\frac{sini}{sine} = \\frac{sine}{sinR} = \\frac{sinee^o}{sin35^o}\\)\n= 1.673 = \\(\\frac{sine}{sin35^o}\\)\ne = sin\\(^{-1}\\) (1.673 x sin 35\\(^o\\)) = 73.7\\(^o\\)\n(iv) D = (i + e) - A = (45 + 73.3) - 60\\(^o\\)\n= 58.3\\(^o\\)\n\n\n', '', 'Explanation\nCritical angle is the angle of incidence in the denser medium when the angle of refraction in the less dense medium is 90°.\n\n(b) Deviation of a ray in a triangular prism is determined by the factors:\n- Angle of incidence\n- Refracting angle of the prism, and\n- Refractive index of the prism material.\n(c) A = 2r = 2 x 29 = 58\\(^o\\)\nBut n = \\(\\frac{sin \\frac{1}{2} (A + Dm)}{sin \\frac{1}{2}A} \\)\nsin \\(\\frac{1}{2}\\)(A + Dm) = n sin A \\(\\frac{1}{2}\\)A\n58 + Dm = 93.32 - 58 = 35.32\\(^o\\)\n(d) Angle P = 45° - 20° = 25°\nBut Angle Q = 180°- 60° = 120°\nP +Q + R = 180°\nHence R = 180 - (P + Q) =180 - (25 + 120) = 35°\n(ii) n = \\(\\frac{Sini}{sinP} = \\frac{sin45^o}{sin45^o} = 1.673\\)\n(iii) n = \\(\\frac{sini}{sine} = \\frac{sine}{sinR} = \\frac{sinee^o}{sin35^o}\\)\n= 1.673 = \\(\\frac{sine}{sin35^o}\\)\ne = sin\\(^{-1}\\) (1.673 x sin 35\\(^o\\)) = 73.7\\(^o\\)\n(iv) D = (i + e) - A = (45 + 73.3) - 60\\(^o\\)\n= 58.3\\(^o\\)\n\n\n', '~(b) Deviation of a ray in a triangular prism is determined by the factors:\n- Angle of incidence\n- Refracting angle of the prism, and\n- Refractive index of the prism material.\n(c) A = 2r = 2 x 29 = 58\\(^o\\)\nBut n = \\(\\frac{sin \\frac{1}{2} (A + Dm)}{sin \\frac{1}{2}A} \\)\nsin \\(\\frac{1}{2}\\)(A + Dm) = n sin A \\(\\frac{1}{2}\\)A\n58 + Dm = 93.32 - 58 = 35.32\\(^o\\)\n(d) Angle P = 45° - 20° = 25°\nBut Angle Q = 180°- 60° = 120°\nP +Q + R = 180°\nHence R = 180 - (P + Q) =180 - (25 + 120) = 35°\n(ii) n = \\(\\frac{Sini}{sinP} = \\frac{sin45^o}{sin45^o} = 1.673\\)\n(iii) n = \\(\\frac{sini}{sine} = \\frac{sine}{sinR} = \\frac{sinee^o}{sin35^o}\\)\n= 1.673 = \\(\\frac{sine}{sin35^o}\\)\ne = sin\\(^{-1}\\) (1.673 x sin 35\\(^o\\)) = 73.7\\(^o\\)\n(iv) D = (i + e) - A = (45 + 73.3) - 60\\(^o\\)\n= 58.3\\(^o\\)\n\n~(c) A = 2r = 2 x 29 = 58\\(^o\\)\nBut n = \\(\\frac{sin \\frac{1}{2} (A + Dm)}{sin \\frac{1}{2}A} \\)\nsin \\(\\frac{1}{2}\\)(A + Dm) = n sin A \\(\\frac{1}{2}\\)A\n58 + Dm = 93.32 - 58 = 35.32\\(^o\\)\n(d) Angle P = 45° - 20° = 25°\nBut Angle Q = 180°- 60° = 120°\nP +Q + R = 180°\nHence R = 180 - (P + Q) =180 - (25 + 120) = 35°\n(ii) n = \\(\\frac{Sini}{sinP} = \\frac{sin45^o}{sin45^o} = 1.673\\)\n(iii) n = \\(\\frac{sini}{sine} = \\frac{sine}{sinR} = \\frac{sinee^o}{sin35^o}\\)\n= 1.673 = \\(\\frac{sine}{sin35^o}\\)\ne = sin\\(^{-1}\\) (1.673 x sin 35\\(^o\\)) = 73.7\\(^o\\)\n(iv) D = (i + e) - A = (45 + 73.3) - 60\\(^o\\)\n= 58.3\\(^o\\)\n\n~(d) Angle P = 45° - 20° = 25°\nBut Angle Q = 180°- 60° = 120°\nP +Q + R = 180°\nHence R = 180 - (P + Q) =180 - (25 + 120) = 35°\n(ii) n = \\(\\frac{Sini}{sinP} = \\frac{sin45^o}{sin45^o} = 1.673\\)\n(iii) n = \\(\\frac{sini}{sine} = \\frac{sine}{sinR} = \\frac{sinee^o}{sin35^o}\\)\n= 1.673 = \\(\\frac{sine}{sin35^o}\\)\ne = sin\\(^{-1}\\) (1.673 x sin 35\\(^o\\)) = 73.7\\(^o\\)\n(iv) D = (i + e) - A = (45 + 73.3) - 60\\(^o\\)\n= 58.3\\(^o\\)\n\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(88, '2016', '\n\nQuestion\n\n(a) Define magnetic line of force.\n(b) A wire of length 10 cm carrying a current of 4.0 A is placed between the poles of a powerful electromagnet of magnetic flux density 2.0 T. Calculate the:\n(i) force on the wire when it is parallel to the field;\n(ii) maximum force on the wire;\n(iii) force on the wire when it makes an angle of 60° with the field.\n(c) Describe how keepers can be used to preserve the magnetic strength of permanent bar magnets.\n(d) A sailor observes that his mariners\' compass reads N 10° W at a place where the angle of declination is N15° W. Calculate the true bearing of the place.\n\n', '', '(a) Define magnetic line of force.\n(b) A wire of length 10 cm carrying a current of 4.0 A is placed between the poles of a powerful electromagnet of magnetic flux density 2.0 T. Calculate the:\n(i) force on the wire when it is parallel to the field;\n(ii) maximum force on the wire;\n(iii) force on the wire when it makes an angle of 60° with the field.\n(c) Describe how keepers can be used to preserve the magnetic strength of permanent bar magnets.\n(d) A sailor observes that his mariners\' compass reads N 10° W at a place where the angle of declination is N15° W. Calculate the true bearing of the place.', '(a) Define magnetic line of force.~(b) A wire of length 10 cm carrying a current of 4.0 A is placed between the poles of a powerful electromagnet of magnetic flux density 2.0 T. Calculate the:\n(i) force on the wire when it is parallel to the field;\n(ii) maximum force on the wire;\n(iii) force on the wire when it makes an angle of 60° with the field.\n(c) Describe how keepers can be used to preserve the magnetic strength of permanent bar magnets.\n(d) A sailor observes that his mariners\' compass reads N 10° W at a place where the angle of declination is N15° W. Calculate the true bearing of the place.~(c) Describe how keepers can be used to preserve the magnetic strength of permanent bar magnets.\n(d) A sailor observes that his mariners\' compass reads N 10° W at a place where the angle of declination is N15° W. Calculate the true bearing of the place.~(d) A sailor observes that his mariners\' compass reads N 10° W at a place where the angle of declination is N15° W. Calculate the true bearing of the place.', NULL, '\n\nExplanation\n(a) Magnetic line of force is defined as an imaginary line in a mangetic field along which a free or isolated north pole would move. OR, it is an imaginary line in a magnetic field such that the targent to it at any point gives the direction of the field at that point.\n\n(b) F = 1BL Sin\\(\\theta\\)\n(i) When wire is parallel to the field F = IBL = 4x 2 x 0.1 =0N\n(ii) When force is maximum F = IBL Sin90° = 4 x 2 x 0.1 x Sin90°= 0.8N\n(iii) When the wire is at 60° to the field\nF = IBL Sin\\(\\theta\\) = 4 x 2 x 0.1 x Sin60° = 0.693N\n\n(c) Magnets are stored in pairs with their poles adjacent, placing soft iron pieces called keepers across their ends. The keepers become magnets by induction decreasing the reluctance of the air gap; hence both magnets and keepers form closed loops with no free poles preventing demagnetization\n.\n\n(d) True bearing = 15 + 10 = N25°W.\n\n\n', '', 'Explanation\n(a) Magnetic line of force is defined as an imaginary line in a mangetic field along which a free or isolated north pole would move. OR, it is an imaginary line in a magnetic field such that the targent to it at any point gives the direction of the field at that point.\n\n(b) F = 1BL Sin\\(\\theta\\)\n(i) When wire is parallel to the field F = IBL = 4x 2 x 0.1 =0N\n(ii) When force is maximum F = IBL Sin90° = 4 x 2 x 0.1 x Sin90°= 0.8N\n(iii) When the wire is at 60° to the field\nF = IBL Sin\\(\\theta\\) = 4 x 2 x 0.1 x Sin60° = 0.693N\n\n(c) Magnets are stored in pairs with their poles adjacent, placing soft iron pieces called keepers across their ends. The keepers become magnets by induction decreasing the reluctance of the air gap; hence both magnets and keepers form closed loops with no free poles preventing demagnetization\n.\n\n(d) True bearing = 15 + 10 = N25°W.\n\n\n', '(a) Magnetic line of force is defined as an imaginary line in a mangetic field along which a free or isolated north pole would move. OR, it is an imaginary line in a magnetic field such that the targent to it at any point gives the direction of the field at that point.\n~(b) F = 1BL Sin\\(\\theta\\)\n(i) When wire is parallel to the field F = IBL = 4x 2 x 0.1 =0N\n(ii) When force is maximum F = IBL Sin90° = 4 x 2 x 0.1 x Sin90°= 0.8N\n(iii) When the wire is at 60° to the field\nF = IBL Sin\\(\\theta\\) = 4 x 2 x 0.1 x Sin60° = 0.693N\n\n(c) Magnets are stored in pairs with their poles adjacent, placing soft iron pieces called keepers across their ends. The keepers become magnets by induction decreasing the reluctance of the air gap; hence both magnets and keepers form closed loops with no free poles preventing demagnetization\n.\n\n(d) True bearing = 15 + 10 = N25°W.\n\n~(c) Magnets are stored in pairs with their poles adjacent, placing soft iron pieces called keepers across their ends. The keepers become magnets by induction decreasing the reluctance of the air gap; hence both magnets and keepers form closed loops with no free poles preventing demagnetization\n.\n\n(d) True bearing = 15 + 10 = N25°W.\n\n~(d) True bearing = 15 + 10 = N25°W.\n\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(89, '2016', '\n\nQuestion\n\n(a) (i) Explain why x-rays can be used to produce photographs of fractures in bones.\n(ii) List four uses of x-rays other than in medicine.\n(b) State the energy transformations which takes place during the operation of an x-ray tube.\n(c) (i) Explain three named dangers to which human beings may be exposed when subjected to large doses of x-rays.\n(ii) State two precautions that must be taken by persons working with x-rays.\n(d) In an x-ray tube, an electron is accelerated from rest towards a tungsten target biased at a potential of 33 kV. Calculate, for the electron, the\n(i) kinetic energy;\n(ii) velocity. [h = 6.6 x 10\\(^{-14}\\) Js; Me = 9.1 x10\\(^{-31}\\) kg; c = 3.0 x 10\\(^4\\) ms\\(^{-1}\\); e = 1.6 x 10\\(^{-19}\\) C.]\n\n', '', '(a) (i) Explain why x-rays can be used to produce photographs of fractures in bones.\n(ii) List four uses of x-rays other than in medicine.\n(b) State the energy transformations which takes place during the operation of an x-ray tube.\n(c) (i) Explain three named dangers to which human beings may be exposed when subjected to large doses of x-rays.\n(ii) State two precautions that must be taken by persons working with x-rays.\n(d) In an x-ray tube, an electron is accelerated from rest towards a tungsten target biased at a potential of 33 kV. Calculate, for the electron, the\n(i) kinetic energy;\n(ii) velocity. [h = 6.6 x 10\\(^{-14}\\) Js; Me = 9.1 x10\\(^{-31}\\) kg; c = 3.0 x 10\\(^4\\) ms\\(^{-1}\\); e = 1.6 x 10\\(^{-19}\\) C.]', '(a) (i) Explain why x-rays can be used to produce photographs of fractures in bones.\n(ii) List four uses of x-rays other than in medicine.~(b) State the energy transformations which takes place during the operation of an x-ray tube.\n(c) (i) Explain three named dangers to which human beings may be exposed when subjected to large doses of x-rays.\n(ii) State two precautions that must be taken by persons working with x-rays.\n(d) In an x-ray tube, an electron is accelerated from rest towards a tungsten target biased at a potential of 33 kV. Calculate, for the electron, the\n(i) kinetic energy;\n(ii) velocity. [h = 6.6 x 10\\(^{-14}\\) Js; Me = 9.1 x10\\(^{-31}\\) kg; c = 3.0 x 10\\(^4\\) ms\\(^{-1}\\); e = 1.6 x 10\\(^{-19}\\) C.]~(c) (i) Explain three named dangers to which human beings may be exposed when subjected to large doses of x-rays.\n(ii) State two precautions that must be taken by persons working with x-rays.\n(d) In an x-ray tube, an electron is accelerated from rest towards a tungsten target biased at a potential of 33 kV. Calculate, for the electron, the\n(i) kinetic energy;\n(ii) velocity. [h = 6.6 x 10\\(^{-14}\\) Js; Me = 9.1 x10\\(^{-31}\\) kg; c = 3.0 x 10\\(^4\\) ms\\(^{-1}\\); e = 1.6 x 10\\(^{-19}\\) C.]~(d) In an x-ray tube, an electron is accelerated from rest towards a tungsten target biased at a potential of 33 kV. Calculate, for the electron, the\n(i) kinetic energy;\n(ii) velocity. [h = 6.6 x 10\\(^{-14}\\) Js; Me = 9.1 x10\\(^{-31}\\) kg; c = 3.0 x 10\\(^4\\) ms\\(^{-1}\\); e = 1.6 x 10\\(^{-19}\\) C.]', NULL, '\n\nExplanation\n(a)(i) X-rays can produce photographs of fractures in bones; it can pass through fractures but cannot penetrate bones.\n(ii) Uses of x-ray other than in medicine:\n—detection of cracks in metallic products\n—detection of alternations in art works\n—detection of fake gems\n—examination of bags and luggage\n—identification of chemical elements and their isotopes.\n(b) The energy transformations: Electrical energy\n—> Heat energy\n—> Kinetic energy\n—> Electromagnetic energy\n—> X-rays and Heat energy.\n\n(c)(i) Dangers of over exposure to x-rays are:\n— Cell mutation, i.e. change in cell genetic codes.\n— Leukaemia: the destruction of living cells, tissues and organs.\n— Skins harm, causes skin burn and tanning.\n\n(ii) A person working on x-rays should take the following precautions:\n— wear lead protective coats\n— wear dosimeter\n— wear protective goggles\n— go for routine regular medical check-up.\n\n(d)(i) K.E. = eV = 1.6x10\\(^{-19}\\) x 33 x 10\\(^3\\)\n= 5.28 x 10\\(^{-15}\\)J\nV = \\(\\sqrt{\\frac{2K.E}{m}} = \\sqrt{\\frac{2 \\times 5.28 \\times 10^{-15}}{9.1 \\times 10^{-31}}}\\)\n\n', '', 'Explanation\n(a)(i) X-rays can produce photographs of fractures in bones; it can pass through fractures but cannot penetrate bones.\n(ii) Uses of x-ray other than in medicine:\n—detection of cracks in metallic products\n—detection of alternations in art works\n—detection of fake gems\n—examination of bags and luggage\n—identification of chemical elements and their isotopes.\n(b) The energy transformations: Electrical energy\n—> Heat energy\n—> Kinetic energy\n—> Electromagnetic energy\n—> X-rays and Heat energy.\n\n(c)(i) Dangers of over exposure to x-rays are:\n— Cell mutation, i.e. change in cell genetic codes.\n— Leukaemia: the destruction of living cells, tissues and organs.\n— Skins harm, causes skin burn and tanning.\n\n(ii) A person working on x-rays should take the following precautions:\n— wear lead protective coats\n— wear dosimeter\n— wear protective goggles\n— go for routine regular medical check-up.\n\n(d)(i) K.E. = eV = 1.6x10\\(^{-19}\\) x 33 x 10\\(^3\\)\n= 5.28 x 10\\(^{-15}\\)J\nV = \\(\\sqrt{\\frac{2K.E}{m}} = \\sqrt{\\frac{2 \\times 5.28 \\times 10^{-15}}{9.1 \\times 10^{-31}}}\\)\n\n', '(a)(i) X-rays can produce photographs of fractures in bones; it can pass through fractures but cannot penetrate bones.\n(ii) Uses of x-ray other than in medicine:\n—detection of cracks in metallic products\n—detection of alternations in art works\n—detection of fake gems\n—examination of bags and luggage\n—identification of chemical elements and their isotopes.~(b) The energy transformations: Electrical energy\n—> Heat energy\n—> Kinetic energy\n—> Electromagnetic energy\n—> X-rays and Heat energy.\n\n(c)(i) Dangers of over exposure to x-rays are:\n— Cell mutation, i.e. change in cell genetic codes.\n— Leukaemia: the destruction of living cells, tissues and organs.\n— Skins harm, causes skin burn and tanning.\n\n(ii) A person working on x-rays should take the following precautions:\n— wear lead protective coats\n— wear dosimeter\n— wear protective goggles\n— go for routine regular medical check-up.\n\n(d)(i) K.E. = eV = 1.6x10\\(^{-19}\\) x 33 x 10\\(^3\\)\n= 5.28 x 10\\(^{-15}\\)J\nV = \\(\\sqrt{\\frac{2K.E}{m}} = \\sqrt{\\frac{2 \\times 5.28 \\times 10^{-15}}{9.1 \\times 10^{-31}}}\\)\n~(c)(i) Dangers of over exposure to x-rays are:\n— Cell mutation, i.e. change in cell genetic codes.\n— Leukaemia: the destruction of living cells, tissues and organs.\n— Skins harm, causes skin burn and tanning.\n\n(ii) A person working on x-rays should take the following precautions:\n— wear lead protective coats\n— wear dosimeter\n— wear protective goggles\n— go for routine regular medical check-up.\n\n(d)(i) K.E. = eV = 1.6x10\\(^{-19}\\) x 33 x 10\\(^3\\)\n= 5.28 x 10\\(^{-15}\\)J\nV = \\(\\sqrt{\\frac{2K.E}{m}} = \\sqrt{\\frac{2 \\times 5.28 \\times 10^{-15}}{9.1 \\times 10^{-31}}}\\)\n~(d)(i) K.E. = eV = 1.6x10\\(^{-19}\\) x 33 x 10\\(^3\\)\n= 5.28 x 10\\(^{-15}\\)J\nV = \\(\\sqrt{\\frac{2K.E}{m}} = \\sqrt{\\frac{2 \\times 5.28 \\times 10^{-15}}{9.1 \\times 10^{-31}}}\\)\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(90, '2016', '\n\nQuestion\n\nA spiral spring with a metal extends by 10.5 cm in air. When the metal is fully submerged in water, the spring extends by 6.8 cm. Calculate the relative density of the metal. (Assume Hooke\'s law is obeyed)\n\n', 'A spiral spring with a metal extends by 10.5 cm in air. When the metal is fully submerged in water, the spring extends by 6.8 cm. Calculate the relative density of the metal. (Assume Hooke\'s law is obeyed)', '', '', NULL, '\n\nExplanation\nR.d = \\(\\frac{\\text{weight of object in air}}{\\text{upthrust in water}}\\)\n= \\(\\frac{10.5}{10.5 - 6.8} = \\frac{10.5}{3.7} = 2.84\\)\n\n\n', 'R.d = \\(\\frac{\\text{weight of object in air}}{\\text{upthrust in water}}\\)\n= \\(\\frac{10.5}{10.5 - 6.8} = \\frac{10.5}{3.7} = 2.84\\)\n\n\n', '', '', 'WAEC-PK-1', 'WAEC-PHY', 1),
(91, '2016', '\n\nQuestion\n\nA stone projected horizontally from the top of a tower with a speed of 4 ms\\(^{-1}\\) lands on the level ground at a horizontal distance of 25 m from the foot of the tower. Calculate the height of the tower. = 10 ms\\(^{-2}\\)]\n\n', 'A stone projected horizontally from the top of a tower with a speed of 4 ms\\(^{-1}\\) lands on the level ground at a horizontal distance of 25 m from the foot of the tower. Calculate the height of the tower. = 10 ms\\(^{-2}\\)]', '', '', NULL, '\n\nExplanation\nR = U\\(\\sqrt{\\frac{2h}{g}}\\)\nh = \\(\\frac{R^2g}{2u^2}\\)\n= \\(\\frac{25^2 \\times 10}{2 \\times 4^2}\\)\n= 195.3m\n\n', 'R = U\\(\\sqrt{\\frac{2h}{g}}\\)\nh = \\(\\frac{R^2g}{2u^2}\\)\n= \\(\\frac{25^2 \\times 10}{2 \\times 4^2}\\)\n= 195.3m\n\n', '', '', 'WAEC-PK-1', 'WAEC-PHY', 1),
(92, '2016', '\n\nQuestion\n\n(a) Explain diffusion.\n(b) Give one reason why the rate of diffusion is higher in gases than in liquids at the same temperature.\n\n', '', '(a) Explain diffusion.\n(b) Give one reason why the rate of diffusion is higher in gases than in liquids at the same temperature.', '(a) Explain diffusion.~(b) Give one reason why the rate of diffusion is higher in gases than in liquids at the same temperature.', NULL, '\n\nExplanation\n(a) Diffusion is a process whereby fluids mix intimately with one another due to random molecular motion.\n(b) Diffusion rate in gases is higher than that in liquids at the same temperature because molecular motion in gases are faster, the intermolecular forces is less and the density is also less than those in liquids.\n\n\n', '', 'Explanation\n(a) Diffusion is a process whereby fluids mix intimately with one another due to random molecular motion.\n(b) Diffusion rate in gases is higher than that in liquids at the same temperature because molecular motion in gases are faster, the intermolecular forces is less and the density is also less than those in liquids.\n\n\n', '(a) Diffusion is a process whereby fluids mix intimately with one another due to random molecular motion.~(b) Diffusion rate in gases is higher than that in liquids at the same temperature because molecular motion in gases are faster, the intermolecular forces is less and the density is also less than those in liquids.\n\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(93, '2016', '\n\nQuestion\n\nState: (a) the difference between plane polarized light and ordinary light;\n(b) two uses of polaroids.\n\n', '', 'State: (a) the difference between plane polarized light and ordinary light;\n(b) two uses of polaroids.', '(a) the difference between plane polarized light and ordinary light;~(b) two uses of polaroids.', NULL, '\n\nExplanation\n(a) In plane polarized light the vibrations of light waves are perpendicular in one plane only while in ordinary light the vibrations of the particles are in all planes.\n(b) Two uses of polaroids are:\n- in sunglass to reduce sunlight intensity\n- to eliminate reflected light or glare\n- in the measurement of sugar solution concentration or saccharimetry; and\n- in photo elasticity or stress analysis.\n\n', '', 'Explanation\n(a) In plane polarized light the vibrations of light waves are perpendicular in one plane only while in ordinary light the vibrations of the particles are in all planes.\n(b) Two uses of polaroids are:\n- in sunglass to reduce sunlight intensity\n- to eliminate reflected light or glare\n- in the measurement of sugar solution concentration or saccharimetry; and\n- in photo elasticity or stress analysis.\n\n', '(a) In plane polarized light the vibrations of light waves are perpendicular in one plane only while in ordinary light the vibrations of the particles are in all planes.~(b) Two uses of polaroids are:\n- in sunglass to reduce sunlight intensity\n- to eliminate reflected light or glare\n- in the measurement of sugar solution concentration or saccharimetry; and\n- in photo elasticity or stress analysis.\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(94, '2016', '\n\nQuestion\n\n(a). State: (a) two applications of electrolysis in an industry\n(b) one application of electrolysis in a school laboratory.\n\n', '', '(a). State: (a) two applications of electrolysis in an industry\n(b) one application of electrolysis in a school laboratory.', '(a). State: (a) two applications of electrolysis in an industry~(b) one application of electrolysis in a school laboratory.', NULL, '\n\nExplanation\na) Industrial application of electrolysis are\n- purification of some metals\n- electroplating\n- production of gases, e.g. Hydrogen from water\n- extraction of some metals from their ores, e.g. Aluminium from bauxite.\n(b) Applications of electrolysis in school laboratory\n- electrolysis of acidified water\n- calibration of ammeter.\n\n', '', 'Explanation\na) Industrial application of electrolysis are\n- purification of some metals\n- electroplating\n- production of gases, e.g. Hydrogen from water\n- extraction of some metals from their ores, e.g. Aluminium from bauxite.\n(b) Applications of electrolysis in school laboratory\n- electrolysis of acidified water\n- calibration of ammeter.\n\n', 'a) Industrial application of electrolysis are\n- purification of some metals\n- electroplating\n- production of gases, e.g. Hydrogen from water\n- extraction of some metals from their ores, e.g. Aluminium from bauxite.~(b) Applications of electrolysis in school laboratory\n- electrolysis of acidified water\n- calibration of ammeter.\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(95, '2016', '\n\nQuestion\n\nIn fig. la and fig.1 b above, Ida and Wb represent the respective loads on a spring placedtiear a 30 cm rule, when in air and when in water:\n(a) Identify the force causing a shrink in the spring in fig.(b).\n(b) Given that the force constant of the spring is 2.0 x 10\\(^{11}\\) Nm\\(^{-1}\\), calculate the work done by the force in causing the shrink.\n\n\n\n', '', 'In fig. la and fig.1 b above, Ida and Wb represent the respective loads on a spring placedtiear a 30 cm rule, when in air and when in water:\n(a) Identify the force causing a shrink in the spring in fig.(b).\n(b) Given that the force constant of the spring is 2.0 x 10\\(^{11}\\) Nm\\(^{-1}\\), calculate the work done by the force in causing the shrink.\n\n', '(a) Identify the force causing a shrink in the spring in fig.(b).~(b).\n(b) Given that the force constant of the spring is 2.0 x 10\\(^{11}\\) Nm\\(^{-1}\\), calculate the work done by the force in causing the shrink.\n\n', NULL, '\n\nExplanation\n(a) The force causing a shrink is the upthrust of the water.\n(b) Work done = \\(\\frac{1}{2}ke^2\\)\n= \\(\\frac{1}{2}(2 \\times 10^{11}) \\times (2 \\times 10^{-2})^2\\) = 4.0 x 10\\(^7\\)\n\n', '', 'Explanation\n(a) The force causing a shrink is the upthrust of the water.\n(b) Work done = \\(\\frac{1}{2}ke^2\\)\n= \\(\\frac{1}{2}(2 \\times 10^{11}) \\times (2 \\times 10^{-2})^2\\) = 4.0 x 10\\(^7\\)\n\n', '(a) The force causing a shrink is the upthrust of the water.~(b) Work done = \\(\\frac{1}{2}ke^2\\)\n= \\(\\frac{1}{2}(2 \\times 10^{11}) \\times (2 \\times 10^{-2})^2\\) = 4.0 x 10\\(^7\\)\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(96, '2016', '\n\nQuestion\n\nExplain why water in a narrow glass tube has a concave meniscus while mercury, in the same tube, has a convex meniscus.\n\n\n', 'Explain why water in a narrow glass tube has a concave meniscus while mercury, in the same tube, has a convex meniscus.\n', '', '', NULL, '\n\nExplanation\nIn the case of water in axarrow glass tube, the force of adhesion between the water and glass molecules is greater than the cohesive force among water molecule, hence, water clings to the glass tube forming a concave meniscus. In the case of mercury, the cohesive force among its molecules is greater than the adhesive force, between mercury molecules and water molecules hence mercury molecules cling to each other forming a convex meniscus in the capillary glass tube.\n\n\n', 'In the case of water in axarrow glass tube, the force of adhesion between the water and glass molecules is greater than the cohesive force among water molecule, hence, water clings to the glass tube forming a concave meniscus. In the case of mercury, the cohesive force among its molecules is greater than the adhesive force, between mercury molecules and water molecules hence mercury molecules cling to each other forming a convex meniscus in the capillary glass tube.\n\n\n', '', '', 'WAEC-PK-1', 'WAEC-PHY', 1),
(97, '2016', '\n\nQuestion\n\nState three methods of polarizing an unpolarized light\n\n', 'State three methods of polarizing an unpolarized light', '', '', NULL, '\n\nExplanation\nThe methods of polarizing an unpolarized light are:\n- reflection from plane, smooth surfaces.\n- double refraction through crystals.\n- selective absorption\n- polaroid.\n- Scattering by small particles, etc.\n\n', 'The methods of polarizing an unpolarized light are:\n- reflection from plane, smooth surfaces.\n- double refraction through crystals.\n- selective absorption\n- polaroid.\n- Scattering by small particles, etc.\n\n', '', '', 'WAEC-PK-1', 'WAEC-PHY', 1),
(98, '2016', '\n\nQuestion\n\nAn electron of charge 1.60 x 10\\(^{-19}\\) C is accelerated under a potential difference of 1.0 x 10\\(^5\\) V. Calculate the energy of the electron in joules.\n\n', 'An electron of charge 1.60 x 10\\(^{-19}\\) C is accelerated under a potential difference of 1.0 x 10\\(^5\\) V. Calculate the energy of the electron in joules.', '', '', NULL, '\n\nExplanation\nE = eV= 1.6 x 10\\(^{-19}\\) x 1.0 x 10\\(^5\\)\n=1.6 x 10\\(^{-14}\\)J\n\n', 'E = eV= 1.6 x 10\\(^{-19}\\) x 1.0 x 10\\(^5\\)\n=1.6 x 10\\(^{-14}\\)J\n\n', '', '', 'WAEC-PK-1', 'WAEC-PHY', 1),
(99, '2016', '\n\nQuestion\n\nThe uncertainty in determining the duration during which an electron remains in a particular energy level before returning to the ground state is 2.0 x 10\\(^{-9}\\)s. Calculate the uncertainty in determining its energy at that level. (\\(\\frac{h}{2 \\pi}\\) = h = 1.054 x 10\\(^{-34}\\)Js)\n\n', 'The uncertainty in determining the duration during which an electron remains in a particular energy level before returning to the ground state is 2.0 x 10\\(^{-9}\\)s. Calculate the uncertainty in determining its energy at that level. (\\(\\frac{h}{2 \\pi}\\) = h = 1.054 x 10\\(^{-34}\\)Js)', '', '', NULL, '\n\nExplanation\n\\(\\Delta E \\geq \\frac{h}{2 \\pi} \\times \\frac{1}{\\Delta t}\\)\n\\(\\geq 1.054 \\times 10^{-37} \\times \\frac{1}{2 \\times 10^{-9}}\\)\n\\(\\geq 5.27 \\times 10^{-28}\\)J\n\n', '\\(\\Delta E \\geq \\frac{h}{2 \\pi} \\times \\frac{1}{\\Delta t}\\)\n\\(\\geq 1.054 \\times 10^{-37} \\times \\frac{1}{2 \\times 10^{-9}}\\)\n\\(\\geq 5.27 \\times 10^{-28}\\)J\n\n', '', '', 'WAEC-PK-1', 'WAEC-PHY', 1),
(100, '2016', '\n\nQuestion\n\n(a) What is a vector quantity?\n(b) Three vectors 3 ms\\(^{-1}\\) N 45° W, 12 ms\\(^{-1}\\)W and 5 ms\\(^{-1}\\)S act at a point.\n(i) Sketch a vector diagram to illustrate the given information.\n(ii) Calculate the resultant of the vectors.\n(c) In a laboratory experiment to determine the force constant of a spiral spring, the mass or, the spring was varied and the corresponding extensions were measured and recorded as shown in the table below.\n\n\n\nMass M/g\nWeight W/N\nExtension e/cm\n\n\n\n50\n100\n150\n200\n250\n\n\n\n6.5\n11.0\n15.0\n20.0\n25.0\n\n\n\n\n(i) Copy and complete the table. (Take g = 10 ms\\(^{-2}\\))\n(ii) Plot a graph with weight, W, on the vertical axis and extension, e, on the horizontal axis.\n(iii) Using the graph, determine the force constant of the spring.\n(iv) Determine the natural length of the spring if its length was 38.0 cm when loaded with 250 g mass.\n\n\n(a) What is a vector quantity?\n(b) Three vectors 3 ms\\(^{-1}\\) N 45° W, 12 ms\\(^{-1}\\)W and 5 ms\\(^{-1}\\)S act at a point.\n(i) Sketch a vector diagram to illustrate the given information.\n(ii) Calculate the resultant of the vectors.\n(c) In a laboratory experiment to determine the force constant of a spiral spring, the mass or, the spring was varied and the corresponding extensions were measured and recorded as shown in the table below.\n\n\n\nMass M/g\nWeight W/N\nExtension e/cm\n\n\n\n50\n100\n150\n200\n250\n\n\n\n6.5\n11.0\n15.0\n20.0\n25.0\n\n\n\n', '', '(a) What is a vector quantity?\n(b) Three vectors 3 ms\\(^{-1}\\) N 45° W, 12 ms\\(^{-1}\\)W and 5 ms\\(^{-1}\\)S act at a point.\n(i) Sketch a vector diagram to illustrate the given information.\n(ii) Calculate the resultant of the vectors.\n(c) In a laboratory experiment to determine the force constant of a spiral spring, the mass or, the spring was varied and the corresponding extensions were measured and recorded as shown in the table below.\n\n\n\nMass M/g\nWeight W/N\nExtension e/cm\n\n\n\n50\n100\n150\n200\n250\n\n\n\n6.5\n11.0\n15.0\n20.0\n25.0\n\n\n\n\n(i) Copy and complete the table. (Take g = 10 ms\\(^{-2}\\))\n(ii) Plot a graph with weight, W, on the vertical axis and extension, e, on the horizontal axis.\n(iii) Using the graph, determine the force constant of the spring.\n(iv) Determine the natural length of the spring if its length was 38.0 cm when loaded with 250 g mass.\n\n\n(a) What is a vector quantity?\n(b) Three vectors 3 ms\\(^{-1}\\) N 45° W, 12 ms\\(^{-1}\\)W and 5 ms\\(^{-1}\\)S act at a point.\n(i) Sketch a vector diagram to illustrate the given information.\n(ii) Calculate the resultant of the vectors.\n(c) In a laboratory experiment to determine the force constant of a spiral spring, the mass or, the spring was varied and the corresponding extensions were measured and recorded as shown in the table below.\n\n\n\nMass M/g\nWeight W/N\nExtension e/cm\n\n\n\n50\n100\n150\n200\n250\n\n\n\n6.5\n11.0\n15.0\n20.0\n25.0\n\n', '(a) What is a vector quantity?\n(b) Three vectors 3 ms\\(^{-1}\\) N 45° W, 12 ms\\(^{-1}\\)W and 5 ms\\(^{-1}\\)S act at a point.\n(i) Sketch a vector diagram to illustrate the given information.\n(ii) Calculate the resultant of the vectors.\n(c) In a laboratory experiment to determine the force constant of a spiral spring, the mass or, the spring was varied and the corresponding extensions were measured and recorded as shown in the table below.\n\n\n\nMass M/g\nWeight W/N\nExtension e/cm\n\n\n\n50\n100\n150\n200\n250\n\n\n\n6.5\n11.0\n15.0\n20.0\n25.0\n\n\n\n\n(i) Copy and complete the table. (Take g = 10 ms\\(^{-2}\\))\n(ii) Plot a graph with weight, W, on the vertical axis and extension, e, on the horizontal axis.\n(iii) Using the graph, determine the force constant of the spring.\n(iv) Determine the natural length of the spring if its length was 38.0 cm when loaded with 250 g mass.\n\n\n(a) What is a vector quantity?~(b) Three vectors 3 ms\\(^{-1}\\) N 45° W, 12 ms\\(^{-1}\\)W and 5 ms\\(^{-1}\\)S act at a point.\n(i) Sketch a vector diagram to illustrate the given information.\n(ii) Calculate the resultant of the vectors.\n(c) In a laboratory experiment to determine the force constant of a spiral spring, the mass or, the spring was varied and the corresponding extensions were measured and recorded as shown in the table below.\n\n\n\nMass M/g\nWeight W/N\nExtension e/cm\n\n\n\n50\n100\n150\n200\n250\n\n\n\n6.5\n11.0\n15.0\n20.0\n25.0\n\n\n\n\n(i) Copy and complete the table. (Take g = 10 ms\\(^{-2}\\))\n(ii) Plot a graph with weight, W, on the vertical axis and extension, e, on the horizontal axis.\n(iii) Using the graph, determine the force constant of the spring.\n(iv) Determine the natural length of the spring if its length was 38.0 cm when loaded with 250 g mass.\n\n\n(a) What is a vector quantity?\n(b) Three vectors 3 ms\\(^{-1}\\) N 45° W, 12 ms\\(^{-1}\\)W and 5 ms\\(^{-1}\\)S act at a point.\n(i) Sketch a vector diagram to illustrate the given information.\n(ii) Calculate the resultant of the vectors.\n(c) In a laboratory experiment to determine the force constant of a spiral spring, the mass or, the spring was varied and the corresponding extensions were measured and recorded as shown in the table below.\n\n\n\nMass M/g\nWeight W/N\nExtension e/cm\n\n\n\n50\n100\n150\n200\n250\n\n\n\n6.5\n11.0\n15.0\n20.0\n25.0\n\n~(c) In a laboratory experiment to determine the force constant of a spiral spring, the mass or, the spring was varied and the corresponding extensions were measured and recorded as shown in the table below.\n\n\n\nMass M/g\nWeight W/N\nExtension e/cm\n\n\n\n50\n100\n150\n200\n250\n\n\n\n6.5\n11.0\n15.0\n20.0\n25.0\n\n\n\n\n(i) Copy and complete the table. (Take g = 10 ms\\(^{-2}\\))\n(ii) Plot a graph with weight, W, on the vertical axis and extension, e, on the horizontal axis.\n(iii) Using the graph, determine the force constant of the spring.\n(iv) Determine the natural length of the spring if its length was 38.0 cm when loaded with 250 g mass.\n\n\n(a) What is a vector quantity?\n(b) Three vectors 3 ms\\(^{-1}\\) N 45° W, 12 ms\\(^{-1}\\)W and 5 ms\\(^{-1}\\)S act at a point.\n(i) Sketch a vector diagram to illustrate the given information.\n(ii) Calculate the resultant of the vectors.\n(c) In a laboratory experiment to determine the force constant of a spiral spring, the mass or, the spring was varied and the corresponding extensions were measured and recorded as shown in the table below.\n\n\n\nMass M/g\nWeight W/N\nExtension e/cm\n\n\n\n50\n100\n150\n200\n250\n\n\n\n6.5\n11.0\n15.0\n20.0\n25.0\n\n', NULL, '\n\nExplanation\n(a) Vector quantity is any quantity that has both magnitude and direction, e.g. velocity.\n\n(b)(i)\n\n\n(ii)\n\n\n\nx\ny\n\n\n\n-3 cos 45 = -2.121\n -12.000\n 0.000 \n V_x = -14.121 \n\n\n+ 3 sin 45 = 2.121\n 0000\n -5000 \n V_y = -2.879\n\n\n\n\n\n\n\nR = \\(\\sqrt{V_x^2 + Vy^2}\\)\n= \\(\\sqrt{-14.121^2} + -2.879^2\\) = 14.4m/s\nTa n \\(\\theta = \\frac{2.879}{14.121} = 0.2039\\)\n\\(\\theta = tan^{-1} 0.2039 = 11.5^o\\)\nR = 14.4m/s S78.5\\(^o\\)W\n\n(c)(i)\n\n\n\nMass/g\nWeight W/N\nExt.e/cm\n\n\n\n50\n100\n150\n200\n250\n\n\n0.5\n1.0\n1.5\n2.0\n2.5\n\n\n6,5\n11.0\n15.0\n20.0\n25.0\n\n\n\n\n\n\n(iii) Force constant = slope\n\\(\\frac{\\Delta W}{\\Delta e} = \\frac{3.1 - 1.0}{30 - 11} = \\frac{2.1}{9} = 0.1105\\)\n(iv) Nature length = 38.0 - 25 = 13cm\n\n\n', '', 'Explanation\n(a) Vector quantity is any quantity that has both magnitude and direction, e.g. velocity.\n\n(b)(i)\n\n\n(ii)\n\n\n\nx\ny\n\n\n\n-3 cos 45 = -2.121\n -12.000\n 0.000 \n V_x = -14.121 \n\n\n+ 3 sin 45 = 2.121\n 0000\n -5000 \n V_y = -2.879\n\n\n\n\n\n\n\nR = \\(\\sqrt{V_x^2 + Vy^2}\\)\n= \\(\\sqrt{-14.121^2} + -2.879^2\\) = 14.4m/s\nTa n \\(\\theta = \\frac{2.879}{14.121} = 0.2039\\)\n\\(\\theta = tan^{-1} 0.2039 = 11.5^o\\)\nR = 14.4m/s S78.5\\(^o\\)W\n\n(c)(i)\n\n\n\nMass/g\nWeight W/N\nExt.e/cm\n\n\n\n50\n100\n150\n200\n250\n\n\n0.5\n1.0\n1.5\n2.0\n2.5\n\n\n6,5\n11.0\n15.0\n20.0\n25.0\n\n\n\n\n\n\n(iii) Force constant = slope\n\\(\\frac{\\Delta W}{\\Delta e} = \\frac{3.1 - 1.0}{30 - 11} = \\frac{2.1}{9} = 0.1105\\)\n(iv) Nature length = 38.0 - 25 = 13cm\n\n\n', '(a) Vector quantity is any quantity that has both magnitude and direction, e.g. velocity.\n~(b)(i)\n\n\n(ii)\n\n\n\nx\ny\n\n\n\n-3 cos 45 = -2.121\n -12.000\n 0.000 \n V_x = -14.121 \n\n\n+ 3 sin 45 = 2.121\n 0000\n -5000 \n V_y = -2.879\n\n\n\n\n\n\n\nR = \\(\\sqrt{V_x^2 + Vy^2}\\)\n= \\(\\sqrt{-14.121^2} + -2.879^2\\) = 14.4m/s\nTa n \\(\\theta = \\frac{2.879}{14.121} = 0.2039\\)\n\\(\\theta = tan^{-1} 0.2039 = 11.5^o\\)\nR = 14.4m/s S78.5\\(^o\\)W\n\n(c)(i)\n\n\n\nMass/g\nWeight W/N\nExt.e/cm\n\n\n\n50\n100\n150\n200\n250\n\n\n0.5\n1.0\n1.5\n2.0\n2.5\n\n\n6,5\n11.0\n15.0\n20.0\n25.0\n\n\n\n\n\n\n(iii) Force constant = slope\n\\(\\frac{\\Delta W}{\\Delta e} = \\frac{3.1 - 1.0}{30 - 11} = \\frac{2.1}{9} = 0.1105\\)\n(iv) Nature length = 38.0 - 25 = 13cm\n\n~(c)(i)\n\n\n\nMass/g\nWeight W/N\nExt.e/cm\n\n\n\n50\n100\n150\n200\n250\n\n\n0.5\n1.0\n1.5\n2.0\n2.5\n\n\n6,5\n11.0\n15.0\n20.0\n25.0\n\n\n\n\n\n\n(iii) Force constant = slope\n\\(\\frac{\\Delta W}{\\Delta e} = \\frac{3.1 - 1.0}{30 - 11} = \\frac{2.1}{9} = 0.1105\\)\n(iv) Nature length = 38.0 - 25 = 13cm\n\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(101, '2016', '\n\nQuestion\n\n(a)(i) State two advantages of alcohol over mercury as a thermometric liquid.\n(ii) When the bulb of a thermometer is placed in a beaker of hot water, the level of the mercury first falls and then rises gradually. Explain this observation.\n(b) List two factors, other than temperature, that affect the rate of evaporation of a liquid.\n(b) A block of lead of mass 100 kg in a crucible and at a temperature of 40 °C was placed in an electric furnace rated 10 kW. If the melting point of lead is 320 °C, calculate the:\n(i) quantity of heat required to heat the lead to its melting point;\n(ii) additional heat energy required to melt the lead;\n(iii) time taken to supply this additional energy. (Specific heat capacity of lead = 120 Jkg\\(^{-1}\\) K\\(^{-1}\\) (Specific Idtent heat of fusion lead = 2.5 x 10\\(^{4} JK^{-1}\\))\n(d) State two precautions necessary in an experiment to determine the specific latent heat of vaporization of water.\n\n', '', '(a)(i) State two advantages of alcohol over mercury as a thermometric liquid.\n(ii) When the bulb of a thermometer is placed in a beaker of hot water, the level of the mercury first falls and then rises gradually. Explain this observation.\n(b) List two factors, other than temperature, that affect the rate of evaporation of a liquid.\n(b) A block of lead of mass 100 kg in a crucible and at a temperature of 40 °C was placed in an electric furnace rated 10 kW. If the melting point of lead is 320 °C, calculate the:\n(i) quantity of heat required to heat the lead to its melting point;\n(ii) additional heat energy required to melt the lead;\n(iii) time taken to supply this additional energy. (Specific heat capacity of lead = 120 Jkg\\(^{-1}\\) K\\(^{-1}\\) (Specific Idtent heat of fusion lead = 2.5 x 10\\(^{4} JK^{-1}\\))\n(d) State two precautions necessary in an experiment to determine the specific latent heat of vaporization of water.', '(a)(i) State two advantages of alcohol over mercury as a thermometric liquid.\n(ii) When the bulb of a thermometer is placed in a beaker of hot water, the level of the mercury first falls and then rises gradually. Explain this observation.\n(b) List two factors, other than temperature, that affect the rate of evaporation of a liquid.~(b) List two factors, other than temperature, that affect the rate of evaporation of a liquid.\n(b) A block of lead of mass 100 kg in a crucible and at a temperature of 40 °C was placed in an electric furnace rated 10 kW. If the melting point of lead is 320 °C, calculate the:\n(i) quantity of heat required to heat the lead to its melting point;\n(ii) additional heat energy required to melt the lead;\n(iii) time taken to supply this additional energy. (Specific heat capacity of lead = 120 Jkg\\(^{-1}\\) K\\(^{-1}\\) (Specific Idtent heat of fusion lead = 2.5 x 10\\(^{4} JK^{-1}\\))\n(d) State two precautions necessary in an experiment to determine the specific latent heat of vaporization of water.~(d) State two precautions necessary in an experiment to determine the specific latent heat of vaporization of water.', NULL, '\n\nExplanation\na)(i) Advantages of alcohol over mercury as a thermometric liquid are:\n- Alcohol is cheaper\n-Alcohol is easily available\n-Alcohol has higher expansivity\n- Alcohol has lower freezing point.\n(ii) The glass bulb of the thermometer first got heated and expands, therefore the mercury level in it falls. Then the mercury expands and increases more than that of the glass and then rises.\n(b) Factors that affect the rate of evaporation of a liquid, temperature exclusively are:\n- Pressure, area of liquid surface exposed, humidity, nature of the liquid, wind and dryness of air, etc.\n(c)(i) Q = mc\n\\(\\Delta \\theta = 100 \\times 120 \\times (320 - 40)\\)\nii) Q = Ml = 100 x 0.25 x 10\\(^5)\\) = 2500000J\n= 2.5MJ\nii) Pt = ml\nt = \\(\\frac{ml}{p} = \\frac{2500000}{10000} = 250s\\)\n(d) Precautions to be considered are:\n- Calorimeter should be well lagged.\n- Use only dry steam.\n- Stir the mixture gently and continually.\n- Shield the calorimeter from heat sources.\n\n', '', 'Explanation\na)(i) Advantages of alcohol over mercury as a thermometric liquid are:\n- Alcohol is cheaper\n-Alcohol is easily available\n-Alcohol has higher expansivity\n- Alcohol has lower freezing point.\n(ii) The glass bulb of the thermometer first got heated and expands, therefore the mercury level in it falls. Then the mercury expands and increases more than that of the glass and then rises.\n(b) Factors that affect the rate of evaporation of a liquid, temperature exclusively are:\n- Pressure, area of liquid surface exposed, humidity, nature of the liquid, wind and dryness of air, etc.\n(c)(i) Q = mc\n\\(\\Delta \\theta = 100 \\times 120 \\times (320 - 40)\\)\nii) Q = Ml = 100 x 0.25 x 10\\(^5)\\) = 2500000J\n= 2.5MJ\nii) Pt = ml\nt = \\(\\frac{ml}{p} = \\frac{2500000}{10000} = 250s\\)\n(d) Precautions to be considered are:\n- Calorimeter should be well lagged.\n- Use only dry steam.\n- Stir the mixture gently and continually.\n- Shield the calorimeter from heat sources.\n\n', 'a)(i) Advantages of alcohol over mercury as a thermometric liquid are:\n- Alcohol is cheaper\n-Alcohol is easily available\n-Alcohol has higher expansivity\n- Alcohol has lower freezing point.\n(ii) The glass bulb of the thermometer first got heated and expands, therefore the mercury level in it falls. Then the mercury expands and increases more than that of the glass and then rises.~(b) Factors that affect the rate of evaporation of a liquid, temperature exclusively are:\n- Pressure, area of liquid surface exposed, humidity, nature of the liquid, wind and dryness of air, etc.\n(c)(i) Q = mc\n\\(\\Delta \\theta = 100 \\times 120 \\times (320 - 40)\\)\nii) Q = Ml = 100 x 0.25 x 10\\(^5)\\) = 2500000J\n= 2.5MJ\nii) Pt = ml\nt = \\(\\frac{ml}{p} = \\frac{2500000}{10000} = 250s\\)\n(d) Precautions to be considered are:\n- Calorimeter should be well lagged.\n- Use only dry steam.\n- Stir the mixture gently and continually.\n- Shield the calorimeter from heat sources.\n~(c)(i) Q = mc\n\\(\\Delta \\theta = 100 \\times 120 \\times (320 - 40)\\)\nii) Q = Ml = 100 x 0.25 x 10\\(^5)\\) = 2500000J\n= 2.5MJ\nii) Pt = ml\nt = \\(\\frac{ml}{p} = \\frac{2500000}{10000} = 250s\\)\n(d) Precautions to be considered are:\n- Calorimeter should be well lagged.\n- Use only dry steam.\n- Stir the mixture gently and continually.\n- Shield the calorimeter from heat sources.\n~(d) Precautions to be considered are:\n- Calorimeter should be well lagged.\n- Use only dry steam.\n- Stir the mixture gently and continually.\n- Shield the calorimeter from heat sources.\n', 'WAEC-PK-1', 'WAEC-PHY', 1);
INSERT INTO `exam_simulation2` (`id`, `examYear`, `questions`, `parent_que`, `child_que`, `gen_child_que_desc`, `imageDir`, `explanation`, `parent_que_expl`, `child_que_exp`, `gen_child_que_expl`, `package_id`, `subject_id`, `status`) VALUES
(102, '2016', '\n\nQuestion\n\n(a)(i) What is an eclipse?\n(ii) List the three types of eclipse.\n\n(b) A student in a lecture theatre can read from the board clearly but requires a pair of spectacles to read from a book.\n(i) What eye defect has this student?\n(ii) What type of lens is needed to correct the eye defect?\n(iii) The focal length of the lens used to correct this defect is 10cm. Calculate the power of the lens.\n\n(c) A car B moves towards a stationary car A. If B produces an ultrasonic sound at a point and it takes 5.6 x 10\\(^{-3}\\)s for a beep to be heard in B, calculate the distance between the two cars at that instant. (Speed. sound in air = 340 ms\\(^-\\))\n(d) The image of an object is located 9 cm behind a convex mirror. If magnification produced is 0.6, calculate the focal length of the mirror.\n\n', '', '(a)(i) What is an eclipse?\n(ii) List the three types of eclipse.\n\n(b) A student in a lecture theatre can read from the board clearly but requires a pair of spectacles to read from a book.\n(i) What eye defect has this student?\n(ii) What type of lens is needed to correct the eye defect?\n(iii) The focal length of the lens used to correct this defect is 10cm. Calculate the power of the lens.\n\n(c) A car B moves towards a stationary car A. If B produces an ultrasonic sound at a point and it takes 5.6 x 10\\(^{-3}\\)s for a beep to be heard in B, calculate the distance between the two cars at that instant. (Speed. sound in air = 340 ms\\(^-\\))\n(d) The image of an object is located 9 cm behind a convex mirror. If magnification produced is 0.6, calculate the focal length of the mirror.', '(a)(i) What is an eclipse?\n(ii) List the three types of eclipse.\n~(b) A student in a lecture theatre can read from the board clearly but requires a pair of spectacles to read from a book.\n(i) What eye defect has this student?\n(ii) What type of lens is needed to correct the eye defect?\n(iii) The focal length of the lens used to correct this defect is 10cm. Calculate the power of the lens.\n\n(c) A car B moves towards a stationary car A. If B produces an ultrasonic sound at a point and it takes 5.6 x 10\\(^{-3}\\)s for a beep to be heard in B, calculate the distance between the two cars at that instant. (Speed. sound in air = 340 ms\\(^-\\))\n(d) The image of an object is located 9 cm behind a convex mirror. If magnification produced is 0.6, calculate the focal length of the mirror.~(c) A car B moves towards a stationary car A. If B produces an ultrasonic sound at a point and it takes 5.6 x 10\\(^{-3}\\)s for a beep to be heard in B, calculate the distance between the two cars at that instant. (Speed. sound in air = 340 ms\\(^-\\))\n(d) The image of an object is located 9 cm behind a convex mirror. If magnification produced is 0.6, calculate the focal length of the mirror.~(d) The image of an object is located 9 cm behind a convex mirror. If magnification produced is 0.6, calculate the focal length of the mirror.', NULL, '\n\nExplanation\n(i) An eclipse is a shadow that results from the obstruction of light from the sun by the moon or the earth.\n(ii) Types of eclipse:\n-Eclipse of the sun or solar eclipse.\n-Annular eclipse\n- Eclipse of the moon or lunar eclipse.\n(b i) The eye defect is long sight or hypermetropia.\n(ii) The type of lens to correct the defect is convex or converging lens.\n(iii) Power of lens = \\(\\frac{1}{f} = \\frac{1}{0.1} = dioptre(D)\\)\n(c) Velocity of an echo = \\(\\frac{2d}{t}\\)\nd = \\(\\frac{vt}{2} = \\frac{340 \\times 5 \\times 10^{-3}}{2} = 0.952 m\\)\n(d) M = \\(\\frac{v}{u}\\)\nU = \\(\\frac{V}{M} = \\frac{9}{0.6} = 15cm\\)\n\\(\\frac{1}{f} = \\frac{1}{u} + \\frac{1}{v} = \\frac{1}{15} + \\frac{1}{-9}\\)\nf = -22.5cm\n\n', '', 'Explanation\n(i) An eclipse is a shadow that results from the obstruction of light from the sun by the moon or the earth.\n(ii) Types of eclipse:\n-Eclipse of the sun or solar eclipse.\n-Annular eclipse\n- Eclipse of the moon or lunar eclipse.\n(b i) The eye defect is long sight or hypermetropia.\n(ii) The type of lens to correct the defect is convex or converging lens.\n(iii) Power of lens = \\(\\frac{1}{f} = \\frac{1}{0.1} = dioptre(D)\\)\n(c) Velocity of an echo = \\(\\frac{2d}{t}\\)\nd = \\(\\frac{vt}{2} = \\frac{340 \\times 5 \\times 10^{-3}}{2} = 0.952 m\\)\n(d) M = \\(\\frac{v}{u}\\)\nU = \\(\\frac{V}{M} = \\frac{9}{0.6} = 15cm\\)\n\\(\\frac{1}{f} = \\frac{1}{u} + \\frac{1}{v} = \\frac{1}{15} + \\frac{1}{-9}\\)\nf = -22.5cm\n\n', '~(c) Velocity of an echo = \\(\\frac{2d}{t}\\)\nd = \\(\\frac{vt}{2} = \\frac{340 \\times 5 \\times 10^{-3}}{2} = 0.952 m\\)\n(d) M = \\(\\frac{v}{u}\\)\nU = \\(\\frac{V}{M} = \\frac{9}{0.6} = 15cm\\)\n\\(\\frac{1}{f} = \\frac{1}{u} + \\frac{1}{v} = \\frac{1}{15} + \\frac{1}{-9}\\)\nf = -22.5cm\n~(d) M = \\(\\frac{v}{u}\\)\nU = \\(\\frac{V}{M} = \\frac{9}{0.6} = 15cm\\)\n\\(\\frac{1}{f} = \\frac{1}{u} + \\frac{1}{v} = \\frac{1}{15} + \\frac{1}{-9}\\)\nf = -22.5cm\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(103, '2016', '\n\nQuestion\n\n(a) Explain mutual induction\n(b) State four use of electromagnets\n\n\nThe diagram above illustrates two coils X an arranged so that their axes are collinear. X is connected to an a.c. supply and has an amme in series with it while Y is connected to a lamp Explain the following observations. The (i) lamp is lit when the a.c. supply is switched on;\n(ii) brightness of the light from the lamp increases when distance between X. and Y is decreased;\n(iii) filament of the lamp glows brighter whet bundle of insulatttriron wires is placed along ccmmon axis of the coils.\n(d) Two cells, one have an emf of 2.0 V and an internal resistance of 0.4 and the other having an emf of 2.0 V and an internal resistance of 0.1 \\(\\Omega\\), are connected in parallel. The combination is then connected in series with a resistor.\n(i) Draw a circuit diagram of the arrangement.\n(ii) Calculate the current through 5\\(\\Omega\\) resistor.\n\n\n\n\n(a) Explain mutual induction\n(b) State four use of electromagnets\n\n', '', '(a) Explain mutual induction\n(b) State four use of electromagnets\n\n\nThe diagram above illustrates two coils X an arranged so that their axes are collinear. X is connected to an a.c. supply and has an amme in series with it while Y is connected to a lamp Explain the following observations. The (i) lamp is lit when the a.c. supply is switched on;\n(ii) brightness of the light from the lamp increases when distance between X. and Y is decreased;\n(iii) filament of the lamp glows brighter whet bundle of insulatttriron wires is placed along ccmmon axis of the coils.\n(d) Two cells, one have an emf of 2.0 V and an internal resistance of 0.4 and the other having an emf of 2.0 V and an internal resistance of 0.1 \\(\\Omega\\), are connected in parallel. The combination is then connected in series with a resistor.\n(i) Draw a circuit diagram of the arrangement.\n(ii) Calculate the current through 5\\(\\Omega\\) resistor.\n\n\n\n\n(a) Explain mutual induction\n(b) State four use of electromagnets', '(a) Explain mutual induction\n(b) State four use of electromagnets\n\n\nThe diagram above illustrates two coils X an arranged so that their axes are collinear. X is connected to an a.c. supply and has an amme in series with it while Y is connected to a lamp Explain the following observations. The (i) lamp is lit when the a.c. supply is switched on;\n(ii) brightness of the light from the lamp increases when distance between X. and Y is decreased;\n(iii) filament of the lamp glows brighter whet bundle of insulatttriron wires is placed along ccmmon axis of the coils.\n(d) Two cells, one have an emf of 2.0 V and an internal resistance of 0.4 and the other having an emf of 2.0 V and an internal resistance of 0.1 \\(\\Omega\\), are connected in parallel. The combination is then connected in series with a resistor.\n(i) Draw a circuit diagram of the arrangement.\n(ii) Calculate the current through 5\\(\\Omega\\) resistor.\n\n\n\n\n(a) Explain mutual induction~(b) State four use of electromagnets\n\n\nThe diagram above illustrates two coils X an arranged so that their axes are collinear. X is connected to an a.c. supply and has an amme in series with it while Y is connected to a lamp Explain the following observations. The (i) lamp is lit when the a.c. supply is switched on;\n(ii) brightness of the light from the lamp increases when distance between X. and Y is decreased;\n(iii) filament of the lamp glows brighter whet bundle of insulatttriron wires is placed along ccmmon axis of the coils.\n(d) Two cells, one have an emf of 2.0 V and an internal resistance of 0.4 and the other having an emf of 2.0 V and an internal resistance of 0.1 \\(\\Omega\\), are connected in parallel. The combination is then connected in series with a resistor.\n(i) Draw a circuit diagram of the arrangement.\n(ii) Calculate the current through 5\\(\\Omega\\) resistor.\n\n\n\n\n(a) Explain mutual induction\n(b) State four use of electromagnets~(d) Two cells, one have an emf of 2.0 V and an internal resistance of 0.4 and the other having an emf of 2.0 V and an internal resistance of 0.1 \\(\\Omega\\), are connected in parallel. The combination is then connected in series with a resistor.\n(i) Draw a circuit diagram of the arrangement.\n(ii) Calculate the current through 5\\(\\Omega\\) resistor.\n\n\n\n\n(a) Explain mutual induction\n(b) State four use of electromagnets', NULL, '\n\nExplanation\n(a) Mutual Induction: Mutual Induction is a phenomenon which occurs when the current in a coil is changing, an e.m.f will be induced in a tear-by circuit due to some of the magnetic flux produced by the first circuit linking the second circuit.\n(b) The uses of electromagnets are in:\n- electric bells or telephone earpieces\n- generators or electric motors\n- industrial robots\n- separation of iron scraps from debris\n- electromagnetic switches or magnetic relays\n- cranes fbr lifting or transporting heavy magnetic materials\n- magnetic resonance scanners, i.e. in medical diagnosis.\n(i) The changing magnetic flux in X induces changing magnetic flux in Y which induces the current that light up the bulb.\n(ii) As the distance between X and Y decreases, the magnetic flux density linking coil Y increases, hence the induced e.m.f. increase hence the brightners of the lamp.\n(iii) The insulated soft iron wires increases the magnetic flux through coil Y, hence increase in the induced current in Y, therefore, the filament of the lamp glows brighter.\n(d)\n\nTotal e,m,f. E = 2V\nTotal internal resistance r\n\\(\\frac{1}{r} + \\frac{1}{r_2} = \\frac{1}{0.4} + \\frac{1}{0.1}\\)\ne = 0.08\\(\\Omega\\)\nE = I(R + r)\n2 = I(5 + 0.08)\nI = \\(\\frac{2}{5.08}\\) = 0.394A\n\n', '', 'Explanation\n(a) Mutual Induction: Mutual Induction is a phenomenon which occurs when the current in a coil is changing, an e.m.f will be induced in a tear-by circuit due to some of the magnetic flux produced by the first circuit linking the second circuit.\n(b) The uses of electromagnets are in:\n- electric bells or telephone earpieces\n- generators or electric motors\n- industrial robots\n- separation of iron scraps from debris\n- electromagnetic switches or magnetic relays\n- cranes fbr lifting or transporting heavy magnetic materials\n- magnetic resonance scanners, i.e. in medical diagnosis.\n(i) The changing magnetic flux in X induces changing magnetic flux in Y which induces the current that light up the bulb.\n(ii) As the distance between X and Y decreases, the magnetic flux density linking coil Y increases, hence the induced e.m.f. increase hence the brightners of the lamp.\n(iii) The insulated soft iron wires increases the magnetic flux through coil Y, hence increase in the induced current in Y, therefore, the filament of the lamp glows brighter.\n(d)\n\nTotal e,m,f. E = 2V\nTotal internal resistance r\n\\(\\frac{1}{r} + \\frac{1}{r_2} = \\frac{1}{0.4} + \\frac{1}{0.1}\\)\ne = 0.08\\(\\Omega\\)\nE = I(R + r)\n2 = I(5 + 0.08)\nI = \\(\\frac{2}{5.08}\\) = 0.394A\n\n', '(a) Mutual Induction: Mutual Induction is a phenomenon which occurs when the current in a coil is changing, an e.m.f will be induced in a tear-by circuit due to some of the magnetic flux produced by the first circuit linking the second circuit.~(b) The uses of electromagnets are in:\n- electric bells or telephone earpieces\n- generators or electric motors\n- industrial robots\n- separation of iron scraps from debris\n- electromagnetic switches or magnetic relays\n- cranes fbr lifting or transporting heavy magnetic materials\n- magnetic resonance scanners, i.e. in medical diagnosis.\n(i) The changing magnetic flux in X induces changing magnetic flux in Y which induces the current that light up the bulb.\n(ii) As the distance between X and Y decreases, the magnetic flux density linking coil Y increases, hence the induced e.m.f. increase hence the brightners of the lamp.\n(iii) The insulated soft iron wires increases the magnetic flux through coil Y, hence increase in the induced current in Y, therefore, the filament of the lamp glows brighter.\n(d)\n\nTotal e,m,f. E = 2V\nTotal internal resistance r\n\\(\\frac{1}{r} + \\frac{1}{r_2} = \\frac{1}{0.4} + \\frac{1}{0.1}\\)\ne = 0.08\\(\\Omega\\)\nE = I(R + r)\n2 = I(5 + 0.08)\nI = \\(\\frac{2}{5.08}\\) = 0.394A\n~(d)\n\nTotal e,m,f. E = 2V\nTotal internal resistance r\n\\(\\frac{1}{r} + \\frac{1}{r_2} = \\frac{1}{0.4} + \\frac{1}{0.1}\\)\ne = 0.08\\(\\Omega\\)\nE = I(R + r)\n2 = I(5 + 0.08)\nI = \\(\\frac{2}{5.08}\\) = 0.394A\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(104, '2016', '\n\nQuestion\n\n(a) Explain the terms:\n(i) transmutation as it relates to radioactivity; (ii) stopping potential.\n(b) \\(^{23}_{11}A + ^2_1B\\) —> \\(^p_qC\\) + proton\n\\(^p_qC\\) —> \\(^r_sE\\) + beta\n\nA nucleus C, formed artificially from A and B radioactive and quickly decays to another nucli E as indicated in the nuclear equations abc Datermine the values of p, q, r and s.\n(c) A certain metal of work function 1.6 eV is irradiated with ultra-violet light of wavelength 3.6 x 10\\(^{-7}\\) Calculate the maximum\n(i) kinetic energy of ejected electron in joules;\n(ii) speed of an emitted electron. (1eV = 1.6 x 10\\(^{-18}\\) J; C = 3.0 x 10\\(^{8}\\) ms\\(^{-1}\\); m, = 9.1 x 10\\(^{-31}\\) kg; h = 6.6 x 10\\(^{-34}\\) Js)\n(d) If source of the ultra-violet light in (c) above is mo away from the surface of the metal, state the of on the maximum speed of the ejected electron\n\n', '', '(a) Explain the terms:\n(i) transmutation as it relates to radioactivity; (ii) stopping potential.\n(b) \\(^{23}_{11}A + ^2_1B\\) —> \\(^p_qC\\) + proton\n\\(^p_qC\\) —> \\(^r_sE\\) + beta\n\nA nucleus C, formed artificially from A and B radioactive and quickly decays to another nucli E as indicated in the nuclear equations abc Datermine the values of p, q, r and s.\n(c) A certain metal of work function 1.6 eV is irradiated with ultra-violet light of wavelength 3.6 x 10\\(^{-7}\\) Calculate the maximum\n(i) kinetic energy of ejected electron in joules;\n(ii) speed of an emitted electron. (1eV = 1.6 x 10\\(^{-18}\\) J; C = 3.0 x 10\\(^{8}\\) ms\\(^{-1}\\); m, = 9.1 x 10\\(^{-31}\\) kg; h = 6.6 x 10\\(^{-34}\\) Js)\n(d) If source of the ultra-violet light in (c) above is mo away from the surface of the metal, state the of on the maximum speed of the ejected electron', '(a) Explain the terms:\n(i) transmutation as it relates to radioactivity; (ii) stopping potential.~(b) \\(^{23}_{11}A + ^2_1B\\) —> \\(^p_qC\\) + proton\n\\(^p_qC\\) —> \\(^r_sE\\) + beta\n\nA nucleus C, formed artificially from A and B radioactive and quickly decays to another nucli E as indicated in the nuclear equations abc Datermine the values of p, q, r and s.\n(c) A certain metal of work function 1.6 eV is irradiated with ultra-violet light of wavelength 3.6 x 10\\(^{-7}\\) Calculate the maximum\n(i) kinetic energy of ejected electron in joules;\n(ii) speed of an emitted electron. (1eV = 1.6 x 10\\(^{-18}\\) J; C = 3.0 x 10\\(^{8}\\) ms\\(^{-1}\\); m, = 9.1 x 10\\(^{-31}\\) kg; h = 6.6 x 10\\(^{-34}\\) Js)\n(d) If source of the ultra-violet light in (c) above is mo away from the surface of the metal, state the of on the maximum speed of the ejected electron~(c) A certain metal of work function 1.6 eV is irradiated with ultra-violet light of wavelength 3.6 x 10\\(^{-7}\\) Calculate the maximum\n(i) kinetic energy of ejected electron in joules;\n(ii) speed of an emitted electron. (1eV = 1.6 x 10\\(^{-18}\\) J; C = 3.0 x 10\\(^{8}\\) ms\\(^{-1}\\); m, = 9.1 x 10\\(^{-31}\\) kg; h = 6.6 x 10\\(^{-34}\\) Js)\n(d) If source of the ultra-violet light in (c) above is mo away from the surface of the metal, state the of on the maximum speed of the ejected electron~(d) If source of the ultra-violet light in (c) above is mo away from the surface of the metal, state the of on the maximum speed of the ejected electron', NULL, '\n\nExplanation\n(a)(i) Transmutation is the change of one chemical element into another by radioactive disintegration or nuclear bombardment with particles. Example is the conversion of mercury to gold.\n(ii) Stopping potential is the p.d of the anode of a photocell with respect to the cathode at which no electrons will reach the anode stopping potential Vs\n= \\(\\frac{ht}{e} - \\frac{Wo}{e}\\)\n\n(b) \\(^{23}_{11}A + ^2_1B + ^1_1P\\)\nP + 1 = 23 + 2\nP = 24\nq + 1 = 11 + 1\nq = 11\n\\(^{24}_{11}C \\to ^r_sE + ^o_{-1}B\\)\nr + 0 = 24; r = 24\nS - 1 = 11; S = 12\n\n(c) (i) W = 1.6eV = 1.6 x 1.6 x 10\\(^{-19}\\)\n= 2.56 x 10\\(^{-19}\\)J\nE = \\(\\frac{hc}{\\lambda} = \\frac{6.6 \\times 10{-34} \\times 3 \\times 10^{8}}{3.6 \\times 10^{-7}}\\)\nE = W + K.E\\(_{max}\\)\n\nii) K.E = \\(\\frac{1}{2}mv^2\\)\n2.94 x 10\\(^{-19}\\) = \\(\\frac{1}{2} \\times 9 \\times 10^{-3} \\times V^2_{max}\\)\nV\\(_{max} = 8.04 \\times 10^5 m/s\\)\n\n(d) No effect.\n\n\n', '', 'Explanation\n(a)(i) Transmutation is the change of one chemical element into another by radioactive disintegration or nuclear bombardment with particles. Example is the conversion of mercury to gold.\n(ii) Stopping potential is the p.d of the anode of a photocell with respect to the cathode at which no electrons will reach the anode stopping potential Vs\n= \\(\\frac{ht}{e} - \\frac{Wo}{e}\\)\n\n(b) \\(^{23}_{11}A + ^2_1B + ^1_1P\\)\nP + 1 = 23 + 2\nP = 24\nq + 1 = 11 + 1\nq = 11\n\\(^{24}_{11}C \\to ^r_sE + ^o_{-1}B\\)\nr + 0 = 24; r = 24\nS - 1 = 11; S = 12\n\n(c) (i) W = 1.6eV = 1.6 x 1.6 x 10\\(^{-19}\\)\n= 2.56 x 10\\(^{-19}\\)J\nE = \\(\\frac{hc}{\\lambda} = \\frac{6.6 \\times 10{-34} \\times 3 \\times 10^{8}}{3.6 \\times 10^{-7}}\\)\nE = W + K.E\\(_{max}\\)\n\nii) K.E = \\(\\frac{1}{2}mv^2\\)\n2.94 x 10\\(^{-19}\\) = \\(\\frac{1}{2} \\times 9 \\times 10^{-3} \\times V^2_{max}\\)\nV\\(_{max} = 8.04 \\times 10^5 m/s\\)\n\n(d) No effect.\n\n\n', '(a)(i) Transmutation is the change of one chemical element into another by radioactive disintegration or nuclear bombardment with particles. Example is the conversion of mercury to gold.\n(ii) Stopping potential is the p.d of the anode of a photocell with respect to the cathode at which no electrons will reach the anode stopping potential Vs\n= \\(\\frac{ht}{e} - \\frac{Wo}{e}\\)\n~(b) \\(^{23}_{11}A + ^2_1B + ^1_1P\\)\nP + 1 = 23 + 2\nP = 24\nq + 1 = 11 + 1\nq = 11\n\\(^{24}_{11}C \\to ^r_sE + ^o_{-1}B\\)\nr + 0 = 24; r = 24\nS - 1 = 11; S = 12\n\n(c) (i) W = 1.6eV = 1.6 x 1.6 x 10\\(^{-19}\\)\n= 2.56 x 10\\(^{-19}\\)J\nE = \\(\\frac{hc}{\\lambda} = \\frac{6.6 \\times 10{-34} \\times 3 \\times 10^{8}}{3.6 \\times 10^{-7}}\\)\nE = W + K.E\\(_{max}\\)\n\nii) K.E = \\(\\frac{1}{2}mv^2\\)\n2.94 x 10\\(^{-19}\\) = \\(\\frac{1}{2} \\times 9 \\times 10^{-3} \\times V^2_{max}\\)\nV\\(_{max} = 8.04 \\times 10^5 m/s\\)\n\n(d) No effect.\n\n~(c) (i) W = 1.6eV = 1.6 x 1.6 x 10\\(^{-19}\\)\n= 2.56 x 10\\(^{-19}\\)J\nE = \\(\\frac{hc}{\\lambda} = \\frac{6.6 \\times 10{-34} \\times 3 \\times 10^{8}}{3.6 \\times 10^{-7}}\\)\nE = W + K.E\\(_{max}\\)\n\nii) K.E = \\(\\frac{1}{2}mv^2\\)\n2.94 x 10\\(^{-19}\\) = \\(\\frac{1}{2} \\times 9 \\times 10^{-3} \\times V^2_{max}\\)\nV\\(_{max} = 8.04 \\times 10^5 m/s\\)\n\n(d) No effect.\n\n~(d) No effect.\n\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(105, '2016', '\n\nQuestion\n\nState three properties of cathode rays which suggest the particle nature of matter\n\n', 'State three properties of cathode rays which suggest the particle nature of matter', '', '', NULL, '\n\nExplanation\nProperties of cathode rays that show particle nature of matter:\n- Casting of shadows of objects along their path.\n- They have momentum\n- Heat up objects that stop thorn, that is they obey energy conservation law.\n- They are deflected in electric or magnetic fields, showing they are tiny particles of matter, etc.\n\n', 'Properties of cathode rays that show particle nature of matter:\n- Casting of shadows of objects along their path.\n- They have momentum\n- Heat up objects that stop thorn, that is they obey energy conservation law.\n- They are deflected in electric or magnetic fields, showing they are tiny particles of matter, etc.\n\n', '', '', 'WAEC-PK-1', 'WAEC-PHY', 1),
(106, '2016', '\n\nQuestion\n\nA stone is projected vertically upward with a speed of 30ms\\(^{1}\\) from the top of a tower of height 50 m. Neglecting air resistance, determine the maximum height it reached from the ground. [g = 10 ms\\(^-2\\)]\n\n', 'A stone is projected vertically upward with a speed of 30ms\\(^{1}\\) from the top of a tower of height 50 m. Neglecting air resistance, determine the maximum height it reached from the ground. [g = 10 ms\\(^-2\\)]', '', '', NULL, '\n\nExplanation\nV\\(^2 = U^2 - 2gs\\)\nS = \\(\\frac{V^2 - U^2}{-2g} = \\frac{O^2 - 30^2}{-2 \\times 20}\\) = 45m\nHeight from ground = 50 + 45 = 95m\nHeight from ground = 50 + 45 = 95m\n\n', 'V\\(^2 = U^2 - 2gs\\)\nS = \\(\\frac{V^2 - U^2}{-2g} = \\frac{O^2 - 30^2}{-2 \\times 20}\\) = 45m\nHeight from ground = 50 + 45 = 95m\nHeight from ground = 50 + 45 = 95m\n\n', '', '', 'WAEC-PK-1', 'WAEC-PHY', 1),
(107, '2016', '\n\nQuestion\n\nA force of 40 N is applied at the free end of a wire fixed at one end to produce an extension of 0.24 mm. If the original length and diameter of the wire art., 3 m and 2.0 mm respectively, calculate the: (a) stress on the wire; (b) strain in the wire.\n\n', 'A force of 40 N is applied at the free end of a wire fixed at one end to produce an extension of 0.24 mm. If the original length and diameter of the wire art., 3 m and 2.0 mm respectively, calculate the: (a) stress on the wire; (b) strain in the wire.', '', '', NULL, '\n\nExplanation\n(a) Stree = \\(\\frac{F}{A}\\) = \\(\\frac{40}{3.14(1.0 \\times 10^{-3})}^2\\)\n= \\(1.27 \\times 10^{7} \\times 10^7 NM ^{-2}\\)\n(b) Strain = \\(\\frac{e}{L} = \\frac{0.24 \\times 10^{-3}}{3} = 8.0 \\times 10^{-5}\\)\n\n', '', 'Explanation\n(a) Stree = \\(\\frac{F}{A}\\) = \\(\\frac{40}{3.14(1.0 \\times 10^{-3})}^2\\)\n= \\(1.27 \\times 10^{7} \\times 10^7 NM ^{-2}\\)\n(b) Strain = \\(\\frac{e}{L} = \\frac{0.24 \\times 10^{-3}}{3} = 8.0 \\times 10^{-5}\\)\n\n', '(a) Stree = \\(\\frac{F}{A}\\) = \\(\\frac{40}{3.14(1.0 \\times 10^{-3})}^2\\)\n= \\(1.27 \\times 10^{7} \\times 10^7 NM ^{-2}\\)~(b) Strain = \\(\\frac{e}{L} = \\frac{0.24 \\times 10^{-3}}{3} = 8.0 \\times 10^{-5}\\)\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(108, '2016', '\n\nQuestion\n\nWhen a lead-acid accumulator is fully charged, evolution of gases occurs at the electrodes. Name these gases and the respective electrodes at which thcy are given off.\n\n', 'When a lead-acid accumulator is fully charged, evolution of gases occurs at the electrodes. Name these gases and the respective electrodes at which thcy are given off.', '', '', NULL, '\n\nExplanation\nGases are:\n(i) Hydrogen at Anode or Lead peroxide.\n(ii) Oxygen at cathode or lead plate.\n\n', '', 'Explanation\nGases are:\n(i) Hydrogen at Anode or Lead peroxide.\n(ii) Oxygen at cathode or lead plate.\n\n', '', 'WAEC-PK-1', 'WAEC-PHY', 1),
(109, '2016', '\n\nQuestion\n\n(a) State two factors which affect the mass of elements deposited during electrolysis.\n(b) List two non-electrolysis.\n\n\n', '', '(a) State two factors which affect the mass of elements deposited during electrolysis.\n(b) List two non-electrolysis.\n', '(a) State two factors which affect the mass of elements deposited during electrolysis.~(b) List two non-electrolysis.\n', NULL, '\n\nExplanation\nFactors are:\n- Current, time and electrochemical equivalent of the element\n(b) Non-electrolytes are:\n- Petrol, water, ethanol, kerosine, etc.\n\n', '', 'Explanation\nFactors are:\n- Current, time and electrochemical equivalent of the element\n(b) Non-electrolytes are:\n- Petrol, water, ethanol, kerosine, etc.\n\n', '~(b) Non-electrolytes are:\n- Petrol, water, ethanol, kerosine, etc.\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(110, '2016', '\n\nQuestion\n\n(a) Differentiate between plane polarization and interference as applied to waves.\n(b) List two uses of polaroids.\n\n', '', '(a) Differentiate between plane polarization and interference as applied to waves.\n(b) List two uses of polaroids.', '(a) Differentiate between plane polarization and interference as applied to waves.~(b) List two uses of polaroids.', NULL, '\n\nExplanation\n(a)(i) In polarization vibration of wave is in one plane while interference is an effect produced when two-waves of the same wave length or amplified or frequency travelling in the same direction in a medium are super imposed.\n(ii) Only transverse waves can be polarised while in interference both transverse and longitudinal waves are involved.\n(b) Uses of polaroids are in:\n- sunglasses\n- Polaroid cameras\n- Products on the 3-dimensional films, and\n- determination of sugar solution concentration.\n\n\n', '', 'Explanation\n(a)(i) In polarization vibration of wave is in one plane while interference is an effect produced when two-waves of the same wave length or amplified or frequency travelling in the same direction in a medium are super imposed.\n(ii) Only transverse waves can be polarised while in interference both transverse and longitudinal waves are involved.\n(b) Uses of polaroids are in:\n- sunglasses\n- Polaroid cameras\n- Products on the 3-dimensional films, and\n- determination of sugar solution concentration.\n\n\n', '(a)(i) In polarization vibration of wave is in one plane while interference is an effect produced when two-waves of the same wave length or amplified or frequency travelling in the same direction in a medium are super imposed.\n(ii) Only transverse waves can be polarised while in interference both transverse and longitudinal waves are involved.~(b) Uses of polaroids are in:\n- sunglasses\n- Polaroid cameras\n- Products on the 3-dimensional films, and\n- determination of sugar solution concentration.\n\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(111, '2016', '\n\nQuestion\n\n(a) List two examples each of substances with:\n(i) low viscosity;\n(ii) high viscosity.\n(b) When is a liquid said to be viscostatic?\n\n', '', '(a) List two examples each of substances with:\n(i) low viscosity;\n(ii) high viscosity.\n(b) When is a liquid said to be viscostatic?', '(a) List two examples each of substances with:\n(i) low viscosity;\n(ii) high viscosity.~(b) When is a liquid said to be viscostatic?', NULL, '\n\nExplanation\n(a)(i) Substances with ions viscosity are: water, ethanol, salt solution, kerosine, methylated spirit, etc.\n(ii) Substances with high viscosity: engine oil, glycerine, palm oil, etc.\n(b) A liquid is viscostatic when its viscosity does not change appreciably with change in temperature.\n\n', '', 'Explanation\n(a)(i) Substances with ions viscosity are: water, ethanol, salt solution, kerosine, methylated spirit, etc.\n(ii) Substances with high viscosity: engine oil, glycerine, palm oil, etc.\n(b) A liquid is viscostatic when its viscosity does not change appreciably with change in temperature.\n\n', '(a)(i) Substances with ions viscosity are: water, ethanol, salt solution, kerosine, methylated spirit, etc.\n(ii) Substances with high viscosity: engine oil, glycerine, palm oil, etc.~(b) A liquid is viscostatic when its viscosity does not change appreciably with change in temperature.\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(112, '2016', '\n\nQuestion\n\nState one reason each why cathode rays:\n(a) are not electromagnetic waves;\n(b) cast sharp shadows of objects in their path;\n(c) can rotate a light paddle wheel inside a discharge tube.\n\n', '', 'State one reason each why cathode rays:\n(a) are not electromagnetic waves;\n(b) cast sharp shadows of objects in their path;\n(c) can rotate a light paddle wheel inside a discharge tube.', '(a) are not electromagnetic waves;~(b) cast sharp shadows of objects in their path;\n(c) can rotate a light paddle wheel inside a discharge tube.~(c) can rotate a light paddle wheel inside a discharge tube.', NULL, '\n\nExplanation\nThe reasons are:\n(a) deflected by electric or magnetic fields.\n(b) travel in straight lines or paths.\n(c) possess momentum.\n\n', '', 'Explanation\nThe reasons are:\n(a) deflected by electric or magnetic fields.\n(b) travel in straight lines or paths.\n(c) possess momentum.\n\n', '(a) deflected by electric or magnetic fields.~(b) travel in straight lines or paths.\n(c) possess momentum.\n~(c) possess momentum.\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(113, '2016', '\n\nQuestion\n\nAn X-ray tube operates at a potential of 2500 V. If the power of the tube is 750 W, calculate the speed of the electron striking the target. [e = 1.6 x 10\\(^{-19}\\) C; mass of electron = 9.1 x 10\\(^{-3}\\) kg]\n\n', 'An X-ray tube operates at a potential of 2500 V. If the power of the tube is 750 W, calculate the speed of the electron striking the target. [e = 1.6 x 10\\(^{-19}\\) C; mass of electron = 9.1 x 10\\(^{-3}\\) kg]', '', '', NULL, '\n\nExplanation\n(a) eV \\(\\frac{1}{2}MV^2\\)\nV\\(^2 = \\frac{2eV}{m}\\)\n= \\(\\frac{2 \\times 1.6 \\times 10^{-19} \\times 2500}{9.1 \\times 10^{-31}}\\)\nV = \\(\\sqrt{8.79 \\times 10^{14}}\\)\n= 2.96 x 10\\(^{7}\\)\n\n', '(a) eV \\(\\frac{1}{2}MV^2\\)\nV\\(^2 = \\frac{2eV}{m}\\)\n= \\(\\frac{2 \\times 1.6 \\times 10^{-19} \\times 2500}{9.1 \\times 10^{-31}}\\)\nV = \\(\\sqrt{8.79 \\times 10^{14}}\\)\n= 2.96 x 10\\(^{7}\\)\n\n', '', '', 'WAEC-PK-1', 'WAEC-PHY', 1),
(114, '2016', '\n\nQuestion\n\n(a) Write down the names of two particles used in explaining the wave nature of matter.\n(b) State the wave characteristics which are exhibited by the particles named in (a) above.\n\n\n', '', '(a) Write down the names of two particles used in explaining the wave nature of matter.\n(b) State the wave characteristics which are exhibited by the particles named in (a) above.\n', '(a) Write down the names of two particles used in explaining the wave nature of matter.~(b) State the wave characteristics which are exhibited by the particles named in (a) above.\n', NULL, '\n\nExplanation\n(a) Particle used are: Electron, Helium atom, Neutron, Hydrogen atom or Proton, Helium nucleus or alpha particle\n(b) Exhibited wave characteristics include: diffraction and interference.\n\n', '', 'Explanation\n(a) Particle used are: Electron, Helium atom, Neutron, Hydrogen atom or Proton, Helium nucleus or alpha particle\n(b) Exhibited wave characteristics include: diffraction and interference.\n\n', '(a) Particle used are: Electron, Helium atom, Neutron, Hydrogen atom or Proton, Helium nucleus or alpha particle~(b) Exhibited wave characteristics include: diffraction and interference.\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(115, '2016', '\n\nQuestion\n\n(s) State the Heisenberg Uncertainty Principle.\n(b) The product of the uncertainties in Heisenberg Uncertainty Principle is equal to or greater than a constant. State the mathematical expression for this constant.\n\n', '', '(s) State the Heisenberg Uncertainty Principle.\n(b) The product of the uncertainties in Heisenberg Uncertainty Principle is equal to or greater than a constant. State the mathematical expression for this constant.', '~(b) The product of the uncertainties in Heisenberg Uncertainty Principle is equal to or greater than a constant. State the mathematical expression for this constant.', NULL, '\n\nExplanation\n(a) In the simultaneous determination of the position x and momentum p of a particle, there is an uncertainty in its position \\(\\Delta\\) x and an uncertainty in its momentum \\(\\Delta\\) p.\n(b) The constant = h or \\(\\frac{h}{2 \\pi}\\)\n\n', '', 'Explanation\n(a) In the simultaneous determination of the position x and momentum p of a particle, there is an uncertainty in its position \\(\\Delta\\) x and an uncertainty in its momentum \\(\\Delta\\) p.\n(b) The constant = h or \\(\\frac{h}{2 \\pi}\\)\n\n', '(a) In the simultaneous determination of the position x and momentum p of a particle, there is an uncertainty in its position \\(\\Delta\\) x and an uncertainty in its momentum \\(\\Delta\\) p.~(b) The constant = h or \\(\\frac{h}{2 \\pi}\\)\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(116, '2016', '\n\nQuestion\n\n(a) Distinguish between perfectly elastic collision and perfectly inelastic collision.\n(b) Sketch a distance — time graph for a particle moving in a straight line with:\n(i) uniform speed;\n(ii) variable speed.\n(c) A body starts from rest and travels distances of 120, 300 and 180m in successive equal time intervals of 12 s. During each interval the body is uniformly accelerated. (i) Calculate the velocity of the body at the end of each successive time interval.\n(ii) Sketch a velocity-time graph for the motion.\n\n', '', '(a) Distinguish between perfectly elastic collision and perfectly inelastic collision.\n(b) Sketch a distance — time graph for a particle moving in a straight line with:\n(i) uniform speed;\n(ii) variable speed.\n(c) A body starts from rest and travels distances of 120, 300 and 180m in successive equal time intervals of 12 s. During each interval the body is uniformly accelerated. (i) Calculate the velocity of the body at the end of each successive time interval.\n(ii) Sketch a velocity-time graph for the motion.', '(a) Distinguish between perfectly elastic collision and perfectly inelastic collision.~(b) Sketch a distance — time graph for a particle moving in a straight line with:\n(i) uniform speed;\n(ii) variable speed.\n(c) A body starts from rest and travels distances of 120, 300 and 180m in successive equal time intervals of 12 s. During each interval the body is uniformly accelerated. (i) Calculate the velocity of the body at the end of each successive time interval.\n(ii) Sketch a velocity-time graph for the motion.~(c) A body starts from rest and travels distances of 120, 300 and 180m in successive equal time intervals of 12 s. During each interval the body is uniformly accelerated. (i) Calculate the velocity of the body at the end of each successive time interval.\n(ii) Sketch a velocity-time graph for the motion.', NULL, '\n\nExplanation\nPerfectly Elastic\n\n\nPerfectly Inelastic\n\n\n\n\nK.E. is conserved\nColliding objects do not stick together after collision\nIn case of head-on collision directions of the bodies are reversed\n\n\nK.E. decreases\nColliding objects stick and move together after collision\nDirection of both bodies may or may not be reserved\n\n\n\n\n\n(b)\n\n(c)(i) Stage 1; S = ut + \\(\\frac{1}{2}at^2\\)\n= 120 = 0 x 12 + \\(\\frac{1}{2}\\) x a x 12\\(^2\\)\na = \\(\\frac{5}{3}\\)\n\\(V_1 = u + at\\)\n= 20 + \\(\\frac{5}{6} \\times 12 = 30m/s\\)\nStage 3; S = ut + \\(\\frac{1}{2}\\) at\\(^2\\)\n-180 = 30 x 12v + \\(\\frac{1}{2}\\) x a x 12\\(^2\\)\na = \\(\\frac{-5}{2}\\)\nV\\(_3\\) = u + at = 30 + \\(\\frac{-5}{2}\\) x 12 = 0m/s\n(ii) v/m\\(^{s - 1}\\)\n\n\n\n\n\n', '', 'Explanation\nPerfectly Elastic\n\n\nPerfectly Inelastic\n\n\n\n\nK.E. is conserved\nColliding objects do not stick together after collision\nIn case of head-on collision directions of the bodies are reversed\n\n\nK.E. decreases\nColliding objects stick and move together after collision\nDirection of both bodies may or may not be reserved\n\n\n\n\n\n(b)\n\n(c)(i) Stage 1; S = ut + \\(\\frac{1}{2}at^2\\)\n= 120 = 0 x 12 + \\(\\frac{1}{2}\\) x a x 12\\(^2\\)\na = \\(\\frac{5}{3}\\)\n\\(V_1 = u + at\\)\n= 20 + \\(\\frac{5}{6} \\times 12 = 30m/s\\)\nStage 3; S = ut + \\(\\frac{1}{2}\\) at\\(^2\\)\n-180 = 30 x 12v + \\(\\frac{1}{2}\\) x a x 12\\(^2\\)\na = \\(\\frac{-5}{2}\\)\nV\\(_3\\) = u + at = 30 + \\(\\frac{-5}{2}\\) x 12 = 0m/s\n(ii) v/m\\(^{s - 1}\\)\n\n\n\n\n\n', '~(b)\n\n(c)(i) Stage 1; S = ut + \\(\\frac{1}{2}at^2\\)\n= 120 = 0 x 12 + \\(\\frac{1}{2}\\) x a x 12\\(^2\\)\na = \\(\\frac{5}{3}\\)\n\\(V_1 = u + at\\)\n= 20 + \\(\\frac{5}{6} \\times 12 = 30m/s\\)\nStage 3; S = ut + \\(\\frac{1}{2}\\) at\\(^2\\)\n-180 = 30 x 12v + \\(\\frac{1}{2}\\) x a x 12\\(^2\\)\na = \\(\\frac{-5}{2}\\)\nV\\(_3\\) = u + at = 30 + \\(\\frac{-5}{2}\\) x 12 = 0m/s\n(ii) v/m\\(^{s - 1}\\)\n\n\n\n\n~(c)(i) Stage 1; S = ut + \\(\\frac{1}{2}at^2\\)\n= 120 = 0 x 12 + \\(\\frac{1}{2}\\) x a x 12\\(^2\\)\na = \\(\\frac{5}{3}\\)\n\\(V_1 = u + at\\)\n= 20 + \\(\\frac{5}{6} \\times 12 = 30m/s\\)\nStage 3; S = ut + \\(\\frac{1}{2}\\) at\\(^2\\)\n-180 = 30 x 12v + \\(\\frac{1}{2}\\) x a x 12\\(^2\\)\na = \\(\\frac{-5}{2}\\)\nV\\(_3\\) = u + at = 30 + \\(\\frac{-5}{2}\\) x 12 = 0m/s\n(ii) v/m\\(^{s - 1}\\)\n\n\n\n\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(117, '2016', '\n\nQuestion\n\n(a) Explain the terms: (i) inertia; (ii) inertial mass.\n(b) List three factors which affect the rate of evaporation of water in a pond.\n(c) Two ice cubes pressed together for some time were found to stick together when the pressure was removed. Explain this observation.\n(d) Two vertical capillary tubes of the same diameter are lowered into beakers situated at the same level, containing liquids A and B of densities 9.2 x 10\\(^2\\) kgm\\(^{-3}\\) and 1.30 x 10\\(^3\\) kgm\\(^{-3}\\) respectively. A suction pump is used to withdraw air from the top of the liquid columns in the tubes by means of a T-piece arrangement until the liquid in A rises to a height of 26.0 cm. Calculate the height of the liquid in tube B.\n\n\n', '', '(a) Explain the terms: (i) inertia; (ii) inertial mass.\n(b) List three factors which affect the rate of evaporation of water in a pond.\n(c) Two ice cubes pressed together for some time were found to stick together when the pressure was removed. Explain this observation.\n(d) Two vertical capillary tubes of the same diameter are lowered into beakers situated at the same level, containing liquids A and B of densities 9.2 x 10\\(^2\\) kgm\\(^{-3}\\) and 1.30 x 10\\(^3\\) kgm\\(^{-3}\\) respectively. A suction pump is used to withdraw air from the top of the liquid columns in the tubes by means of a T-piece arrangement until the liquid in A rises to a height of 26.0 cm. Calculate the height of the liquid in tube B.\n', '(a) Explain the terms: (i) inertia; (ii) inertial mass.~(b) List three factors which affect the rate of evaporation of water in a pond.\n(c) Two ice cubes pressed together for some time were found to stick together when the pressure was removed. Explain this observation.\n(d) Two vertical capillary tubes of the same diameter are lowered into beakers situated at the same level, containing liquids A and B of densities 9.2 x 10\\(^2\\) kgm\\(^{-3}\\) and 1.30 x 10\\(^3\\) kgm\\(^{-3}\\) respectively. A suction pump is used to withdraw air from the top of the liquid columns in the tubes by means of a T-piece arrangement until the liquid in A rises to a height of 26.0 cm. Calculate the height of the liquid in tube B.\n~(c) Two ice cubes pressed together for some time were found to stick together when the pressure was removed. Explain this observation.\n(d) Two vertical capillary tubes of the same diameter are lowered into beakers situated at the same level, containing liquids A and B of densities 9.2 x 10\\(^2\\) kgm\\(^{-3}\\) and 1.30 x 10\\(^3\\) kgm\\(^{-3}\\) respectively. A suction pump is used to withdraw air from the top of the liquid columns in the tubes by means of a T-piece arrangement until the liquid in A rises to a height of 26.0 cm. Calculate the height of the liquid in tube B.\n~(d) Two vertical capillary tubes of the same diameter are lowered into beakers situated at the same level, containing liquids A and B of densities 9.2 x 10\\(^2\\) kgm\\(^{-3}\\) and 1.30 x 10\\(^3\\) kgm\\(^{-3}\\) respectively. A suction pump is used to withdraw air from the top of the liquid columns in the tubes by means of a T-piece arrangement until the liquid in A rises to a height of 26.0 cm. Calculate the height of the liquid in tube B.\n', NULL, '\n\nExplanation\n(a)(i) Inertia is the reluctance of a body to move if it is at rest or to stop if it is in motion. The more the mass of a body has the greater the inertia.\n(ii) Inertial mass represents resistance to any type of force. The more mass a body has the more force required to give it an acceleration. Force = mass x acceleration.\n(b)Factors that affect the rate of evaporation are: Wind, pressure, temperature, area of surface exposed, humidity or dryness of air.\n(c) Ice melts when sufficient pressure is applied to it, because increase in pressure lowers the melting point of ice. When the two ice cubes are pressed together the melting point is lowered, hence they melt to form thin layer of water between them. On releasing the pressure the water formed freezes because decrease in pressure raises the freezing point. Therefore the two cubes stick together.\n(d)h\\(_A\\)e\\(_A\\) = \\(h_Be_B\\)\n\\(h_B = \\frac{h_Ae_A}{e_B} = \\frac{26 \\times 0.92 \\times 10^3}{1.30 \\times 10^3}\\)\n= 18.4 cm\n\n', '', 'Explanation\n(a)(i) Inertia is the reluctance of a body to move if it is at rest or to stop if it is in motion. The more the mass of a body has the greater the inertia.\n(ii) Inertial mass represents resistance to any type of force. The more mass a body has the more force required to give it an acceleration. Force = mass x acceleration.\n(b)Factors that affect the rate of evaporation are: Wind, pressure, temperature, area of surface exposed, humidity or dryness of air.\n(c) Ice melts when sufficient pressure is applied to it, because increase in pressure lowers the melting point of ice. When the two ice cubes are pressed together the melting point is lowered, hence they melt to form thin layer of water between them. On releasing the pressure the water formed freezes because decrease in pressure raises the freezing point. Therefore the two cubes stick together.\n(d)h\\(_A\\)e\\(_A\\) = \\(h_Be_B\\)\n\\(h_B = \\frac{h_Ae_A}{e_B} = \\frac{26 \\times 0.92 \\times 10^3}{1.30 \\times 10^3}\\)\n= 18.4 cm\n\n', '(a)(i) Inertia is the reluctance of a body to move if it is at rest or to stop if it is in motion. The more the mass of a body has the greater the inertia.\n(ii) Inertial mass represents resistance to any type of force. The more mass a body has the more force required to give it an acceleration. Force = mass x acceleration.~(b)Factors that affect the rate of evaporation are: Wind, pressure, temperature, area of surface exposed, humidity or dryness of air.\n(c) Ice melts when sufficient pressure is applied to it, because increase in pressure lowers the melting point of ice. When the two ice cubes are pressed together the melting point is lowered, hence they melt to form thin layer of water between them. On releasing the pressure the water formed freezes because decrease in pressure raises the freezing point. Therefore the two cubes stick together.\n(d)h\\(_A\\)e\\(_A\\) = \\(h_Be_B\\)\n\\(h_B = \\frac{h_Ae_A}{e_B} = \\frac{26 \\times 0.92 \\times 10^3}{1.30 \\times 10^3}\\)\n= 18.4 cm\n~(c) Ice melts when sufficient pressure is applied to it, because increase in pressure lowers the melting point of ice. When the two ice cubes are pressed together the melting point is lowered, hence they melt to form thin layer of water between them. On releasing the pressure the water formed freezes because decrease in pressure raises the freezing point. Therefore the two cubes stick together.\n(d)h\\(_A\\)e\\(_A\\) = \\(h_Be_B\\)\n\\(h_B = \\frac{h_Ae_A}{e_B} = \\frac{26 \\times 0.92 \\times 10^3}{1.30 \\times 10^3}\\)\n= 18.4 cm\n~(d)h\\(_A\\)e\\(_A\\) = \\(h_Be_B\\)\n\\(h_B = \\frac{h_Ae_A}{e_B} = \\frac{26 \\times 0.92 \\times 10^3}{1.30 \\times 10^3}\\)\n= 18.4 cm\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(118, '2016', '\n\nQuestion\n\n(a) State two factors which affect the angle of deviation of a ray of light through a triangular glass prism.\n(b) Seven virtual images of an object are formed when two plane mirrors are inclined at an angle 0 to each other. Calculate the value of 0.\n(c) By means of a ripple tank, a student was able to generate series of transverse waves by varying the frequency of the dipper and all the waves so generated covered a distance of 0.80 m in 0.2s.\n(i) Determine the speed, v, of the waves.\n\nCopy and complete the table given in your answer booklet.\n(iii) Plot a graph with f on the vertical axis and \\(\\lambda ^{-1}\\) on the horizontal axis.\n(iv) What does the slope of the graph represent?\n\n', '', '(a) State two factors which affect the angle of deviation of a ray of light through a triangular glass prism.\n(b) Seven virtual images of an object are formed when two plane mirrors are inclined at an angle 0 to each other. Calculate the value of 0.\n(c) By means of a ripple tank, a student was able to generate series of transverse waves by varying the frequency of the dipper and all the waves so generated covered a distance of 0.80 m in 0.2s.\n(i) Determine the speed, v, of the waves.\n\nCopy and complete the table given in your answer booklet.\n(iii) Plot a graph with f on the vertical axis and \\(\\lambda ^{-1}\\) on the horizontal axis.\n(iv) What does the slope of the graph represent?', '(a) State two factors which affect the angle of deviation of a ray of light through a triangular glass prism.~(b) Seven virtual images of an object are formed when two plane mirrors are inclined at an angle 0 to each other. Calculate the value of 0.\n(c) By means of a ripple tank, a student was able to generate series of transverse waves by varying the frequency of the dipper and all the waves so generated covered a distance of 0.80 m in 0.2s.\n(i) Determine the speed, v, of the waves.\n\nCopy and complete the table given in your answer booklet.\n(iii) Plot a graph with f on the vertical axis and \\(\\lambda ^{-1}\\) on the horizontal axis.\n(iv) What does the slope of the graph represent?~(c) By means of a ripple tank, a student was able to generate series of transverse waves by varying the frequency of the dipper and all the waves so generated covered a distance of 0.80 m in 0.2s.\n(i) Determine the speed, v, of the waves.\n\nCopy and complete the table given in your answer booklet.\n(iii) Plot a graph with f on the vertical axis and \\(\\lambda ^{-1}\\) on the horizontal axis.\n(iv) What does the slope of the graph represent?', NULL, '\n\nExplanation\n(a) Factors that affect angle of deviation:\n- angle of incidence\n- refracting angle of the prism\n- refractive index of the prism material.\n(b) N = \\(\\frac{360}{\\theta}\\) = 1\n7 = \\(\\frac{360}{\\theta}\\) = 1\n\\(\\theta = 45^o\\)\n(c) V = s/t = \\(\\frac{0.8}{0.2}\\)\n= 4 m/s\n(ii)\n\n(ii)\n\n(iv) The slope represents speed or velocity\n\n\n', '', 'Explanation\n(a) Factors that affect angle of deviation:\n- angle of incidence\n- refracting angle of the prism\n- refractive index of the prism material.\n(b) N = \\(\\frac{360}{\\theta}\\) = 1\n7 = \\(\\frac{360}{\\theta}\\) = 1\n\\(\\theta = 45^o\\)\n(c) V = s/t = \\(\\frac{0.8}{0.2}\\)\n= 4 m/s\n(ii)\n\n(ii)\n\n(iv) The slope represents speed or velocity\n\n\n', '(a) Factors that affect angle of deviation:\n- angle of incidence\n- refracting angle of the prism\n- refractive index of the prism material.~(b) N = \\(\\frac{360}{\\theta}\\) = 1\n7 = \\(\\frac{360}{\\theta}\\) = 1\n\\(\\theta = 45^o\\)\n(c) V = s/t = \\(\\frac{0.8}{0.2}\\)\n= 4 m/s\n(ii)\n\n(ii)\n\n(iv) The slope represents speed or velocity\n\n~(c) V = s/t = \\(\\frac{0.8}{0.2}\\)\n= 4 m/s\n(ii)\n\n(ii)\n\n(iv) The slope represents speed or velocity\n\n', 'WAEC-PK-1', 'WAEC-PHY', 1);
INSERT INTO `exam_simulation2` (`id`, `examYear`, `questions`, `parent_que`, `child_que`, `gen_child_que_desc`, `imageDir`, `explanation`, `parent_que_expl`, `child_que_exp`, `gen_child_que_expl`, `package_id`, `subject_id`, `status`) VALUES
(119, '2016', '\n\nQuestion\n\n(a)\n\nWhen a positively charged conductor is placed near a candle flame, the flame spreads out as shown in the diagram above. Explain this observation.\n(b) A proton moving with a speed of 5.0 x 10\\(^{5}\\) ms\\(^{-1}\\) enters a magnetic field of flux density 0.2 T at an angle of 30° to the field. Calculate the magnitude of the magnetic fcrce exerted on the proton. [Proton charge = 1.6 x 10\\(^{-19}\\) C]\n(c)\n\nThe diagram above illustrates a 9.0 V battery of internal resistance 0.5 \\(\\Omega\\) connected to two resistors of values 2.0 \\(\\Omega\\) and R \\(\\Omega\\). A\\(_1\\) A\\(_2\\) and A\\(_3\\) are ammeters of negligible internal resistances. If Al reads 4.0 A, calculate the:\n(i) equivalent resistance of the combined resistors 2.0 \\(\\Omega\\) and R \\(\\Omega\\);\n(ii) currents through A\\(_1\\) and A\\(_3\\) ; (iii) value of R.\n\n', '', '(a)\n\nWhen a positively charged conductor is placed near a candle flame, the flame spreads out as shown in the diagram above. Explain this observation.\n(b) A proton moving with a speed of 5.0 x 10\\(^{5}\\) ms\\(^{-1}\\) enters a magnetic field of flux density 0.2 T at an angle of 30° to the field. Calculate the magnitude of the magnetic fcrce exerted on the proton. [Proton charge = 1.6 x 10\\(^{-19}\\) C]\n(c)\n\nThe diagram above illustrates a 9.0 V battery of internal resistance 0.5 \\(\\Omega\\) connected to two resistors of values 2.0 \\(\\Omega\\) and R \\(\\Omega\\). A\\(_1\\) A\\(_2\\) and A\\(_3\\) are ammeters of negligible internal resistances. If Al reads 4.0 A, calculate the:\n(i) equivalent resistance of the combined resistors 2.0 \\(\\Omega\\) and R \\(\\Omega\\);\n(ii) currents through A\\(_1\\) and A\\(_3\\) ; (iii) value of R.', '(a)\n\nWhen a positively charged conductor is placed near a candle flame, the flame spreads out as shown in the diagram above. Explain this observation.~(b) A proton moving with a speed of 5.0 x 10\\(^{5}\\) ms\\(^{-1}\\) enters a magnetic field of flux density 0.2 T at an angle of 30° to the field. Calculate the magnitude of the magnetic fcrce exerted on the proton. [Proton charge = 1.6 x 10\\(^{-19}\\) C]\n(c)\n\nThe diagram above illustrates a 9.0 V battery of internal resistance 0.5 \\(\\Omega\\) connected to two resistors of values 2.0 \\(\\Omega\\) and R \\(\\Omega\\). A\\(_1\\) A\\(_2\\) and A\\(_3\\) are ammeters of negligible internal resistances. If Al reads 4.0 A, calculate the:\n(i) equivalent resistance of the combined resistors 2.0 \\(\\Omega\\) and R \\(\\Omega\\);\n(ii) currents through A\\(_1\\) and A\\(_3\\) ; (iii) value of R.~(c)\n\nThe diagram above illustrates a 9.0 V battery of internal resistance 0.5 \\(\\Omega\\) connected to two resistors of values 2.0 \\(\\Omega\\) and R \\(\\Omega\\). A\\(_1\\) A\\(_2\\) and A\\(_3\\) are ammeters of negligible internal resistances. If Al reads 4.0 A, calculate the:\n(i) equivalent resistance of the combined resistors 2.0 \\(\\Omega\\) and R \\(\\Omega\\);\n(ii) currents through A\\(_1\\) and A\\(_3\\) ; (iii) value of R.', NULL, '\n\nExplanation\n(a) The positively charged conductor attracts the negative charges in the air and repels the positive charges i.e. the candle flame ionizes the air around it.\n= 1.6 x 10\\(^{-19}\\) X 5.0 X 10\\(^5\\) x 0.2 x Sin 30°\n= 0.8 x 10\\(^{-1}\\)\n(c)(i)E = I (Rc + r)\n9 = 4(Rc + 0.5)\nR: = 9/4-0.5 \n= 1.75Ω\n\n(ii) Lost volt = Ir = 4.0 x 0.5 =2.0 V\nV oltage across 2Ω and RΩ\nV = 9.0 - 2.0 = 7.0V\nCurrent in A\\(_2\\) = I\\(_2\\) = \\(\\frac{2}{VR}\\) = 7/2 = 3.5A\nCurrent in A\\(_3\\) = I\\(_3\\) = V/R = 7/14 = 0.5A\nOR\nA\\(_3\\) = A\\(_1\\) - A\\(_2\\)\n= 4.0 - 3.5 = 0.5A\n(iii) 1/RC = \\(\\frac{1}{R}\\) + \\(\\frac{1}{R_2}\\)\n1/1.75 = \\(\\frac{1}{R}\\) + \\(\\frac{1}{2}\\)\n\\(\\frac{1}{R}\\) = 1/1.75 - \\(\\frac{1}{2}\\)\nR = 14Ω\n\n\n', '', 'Explanation\n(a) The positively charged conductor attracts the negative charges in the air and repels the positive charges i.e. the candle flame ionizes the air around it.\n= 1.6 x 10\\(^{-19}\\) X 5.0 X 10\\(^5\\) x 0.2 x Sin 30°\n= 0.8 x 10\\(^{-1}\\)\n(c)(i)E = I (Rc + r)\n9 = 4(Rc + 0.5)\nR: = 9/4-0.5 \n= 1.75Ω\n\n(ii) Lost volt = Ir = 4.0 x 0.5 =2.0 V\nV oltage across 2Ω and RΩ\nV = 9.0 - 2.0 = 7.0V\nCurrent in A\\(_2\\) = I\\(_2\\) = \\(\\frac{2}{VR}\\) = 7/2 = 3.5A\nCurrent in A\\(_3\\) = I\\(_3\\) = V/R = 7/14 = 0.5A\nOR\nA\\(_3\\) = A\\(_1\\) - A\\(_2\\)\n= 4.0 - 3.5 = 0.5A\n(iii) 1/RC = \\(\\frac{1}{R}\\) + \\(\\frac{1}{R_2}\\)\n1/1.75 = \\(\\frac{1}{R}\\) + \\(\\frac{1}{2}\\)\n\\(\\frac{1}{R}\\) = 1/1.75 - \\(\\frac{1}{2}\\)\nR = 14Ω\n\n\n', '~(c)(i)E = I (Rc + r)\n9 = 4(Rc + 0.5)\nR: = 9/4-0.5 \n= 1.75Ω\n\n(ii) Lost volt = Ir = 4.0 x 0.5 =2.0 V\nV oltage across 2Ω and RΩ\nV = 9.0 - 2.0 = 7.0V\nCurrent in A\\(_2\\) = I\\(_2\\) = \\(\\frac{2}{VR}\\) = 7/2 = 3.5A\nCurrent in A\\(_3\\) = I\\(_3\\) = V/R = 7/14 = 0.5A\nOR\nA\\(_3\\) = A\\(_1\\) - A\\(_2\\)\n= 4.0 - 3.5 = 0.5A\n(iii) 1/RC = \\(\\frac{1}{R}\\) + \\(\\frac{1}{R_2}\\)\n1/1.75 = \\(\\frac{1}{R}\\) + \\(\\frac{1}{2}\\)\n\\(\\frac{1}{R}\\) = 1/1.75 - \\(\\frac{1}{2}\\)\nR = 14Ω\n\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(120, '2016', '\n\nQuestion\n\n(a) State three conclusions that can be drawn from Rutherford\'s experiment on the scattering of alpha particles by a thin metal foil in relation to the structure of the atom\n\nThe diagram above illustrates th3 energy levels of an electron in an atom. If an excited electron moves from n\\(_2\\) to n\\(_\\theta\\), calculate the:\n(i) frequency;\n(ii) wavelength of the emitted radiation. [ h = 6.6 x 10\\(^{-34}\\) Js; le V = .6 x 10\\(^{-19}\\) J; C = 3.0 x 10\\(^8\\) ms\\(^{-1}\\)]\n(c) The following nuclear equations represent two types of radioactivity.\n\\(^{226}_{88}R_a \\to ^{222}_{86}R_n + ^a_2a\\) (Equation A)\n\\(^{14}_7N + ^4_2a \\to ^{17}_8O + ^1_1p\\) (Equation B)\nIdentify each type and explain briefly the difference between them\n\n', '', '(a) State three conclusions that can be drawn from Rutherford\'s experiment on the scattering of alpha particles by a thin metal foil in relation to the structure of the atom\n\nThe diagram above illustrates th3 energy levels of an electron in an atom. If an excited electron moves from n\\(_2\\) to n\\(_\\theta\\), calculate the:\n(i) frequency;\n(ii) wavelength of the emitted radiation. [ h = 6.6 x 10\\(^{-34}\\) Js; le V = .6 x 10\\(^{-19}\\) J; C = 3.0 x 10\\(^8\\) ms\\(^{-1}\\)]\n(c) The following nuclear equations represent two types of radioactivity.\n\\(^{226}_{88}R_a \\to ^{222}_{86}R_n + ^a_2a\\) (Equation A)\n\\(^{14}_7N + ^4_2a \\to ^{17}_8O + ^1_1p\\) (Equation B)\nIdentify each type and explain briefly the difference between them', '~(c) The following nuclear equations represent two types of radioactivity.\n\\(^{226}_{88}R_a \\to ^{222}_{86}R_n + ^a_2a\\) (Equation A)\n\\(^{14}_7N + ^4_2a \\to ^{17}_8O + ^1_1p\\) (Equation B)\nIdentify each type and explain briefly the difference between them', NULL, '\n\nExplanation\n(a) Conclusion from Rutherford\'s experiment;\n- The nucleus is massive\n- Most of the atom is empty space\n- The nucleus is positively charged\n- Electrons revolve round the nucleus\n- It explains the periodicity of elements in the periodic table\n\nb )(i) E\\(_2\\) = E\\(_2\\) - E\\(_o\\)\n= - 2.0 - (-12.0)\n= 10.0 e V \n= 10 x 1.6 x 1O\\(^{-19}\\)J\n= 1.6 x 10\\(^{-18}\\)J\nE = hf\n\nOr\nf = E/h\n= 1.6 x 10\\(^{18}\\)\n= 6.6 x10\\(^{-34}\\)\n= 2.424 x 10\\(^{15}\\) Hz\n\n(ii) c = fλ\n λ= c/f\n= 3 x 10\\(^8\\)\n2.424 x 10\\(^{15}\\) Hz\n\n(c) A: Natural radioactivity\nB: Artificial radioactivity\nAny correct difference\n\n\n\n', '', 'Explanation\n(a) Conclusion from Rutherford\'s experiment;\n- The nucleus is massive\n- Most of the atom is empty space\n- The nucleus is positively charged\n- Electrons revolve round the nucleus\n- It explains the periodicity of elements in the periodic table\n\nb )(i) E\\(_2\\) = E\\(_2\\) - E\\(_o\\)\n= - 2.0 - (-12.0)\n= 10.0 e V \n= 10 x 1.6 x 1O\\(^{-19}\\)J\n= 1.6 x 10\\(^{-18}\\)J\nE = hf\n\nOr\nf = E/h\n= 1.6 x 10\\(^{18}\\)\n= 6.6 x10\\(^{-34}\\)\n= 2.424 x 10\\(^{15}\\) Hz\n\n(ii) c = fλ\n λ= c/f\n= 3 x 10\\(^8\\)\n2.424 x 10\\(^{15}\\) Hz\n\n(c) A: Natural radioactivity\nB: Artificial radioactivity\nAny correct difference\n\n\n\n', '~(c) A: Natural radioactivity\nB: Artificial radioactivity\nAny correct difference\n\n\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(121, '2017', '\n\nQuestion\n\nA particle is dropped from a vertical height h and falls freely for a time t. With the aid of a sketch, explain how h varies with\n (a) t;\n (b) t\\(^2\\).\n\n', 'A particle is dropped from a vertical height h and falls freely for a time t. With the aid of a sketch, explain how h varies with\n (a) t;\n (b) t\\(^2\\).', '', '', NULL, '\n\nExplanation\na) Diagram\n\n- Correct shape in relation to h and t\n- h varies parabolically with t  \n- correct shape in relation to h and t\\(^{2 }\\)\n- h varies linearly with t\\(^2\\)\n\n', 'a) Diagram\n\n- Correct shape in relation to h and t\n- h varies parabolically with t  \n- correct shape in relation to h and t\\(^{2 }\\)\n- h varies linearly with t\\(^2\\)\n\n', '', '', 'WAEC-PK-1', 'WAEC-PHY', 1),
(122, '2017', '\n\nQuestion\n\nA particle is projected horizontally at 15ms\\(^{-1}\\) from a height of 20m.\nCalculate the horizontal distance covered by the particle just before hitting the ground.\n[g = 10 ms\\(^{-2}\\)]\n\n', 'A particle is projected horizontally at 15ms\\(^{-1}\\) from a height of 20m.\nCalculate the horizontal distance covered by the particle just before hitting the ground.\n[g = 10 ms\\(^{-2}\\)]', '', '', NULL, '\n\nExplanation\nThe expected answer is:\nLet R represent the horizontal distance covered at time, t.\n  h = ½ gt\\(^2\\) \n  20 = ½ x 10 x t\\(^2\\) \n \n  t = 2 s\n R = ut\n = 15 x 2\n\n  = 30m\n\n\n', 'The expected answer is:\nLet R represent the horizontal distance covered at time, t.\n  h = ½ gt\\(^2\\) \n  20 = ½ x 10 x t\\(^2\\) \n \n  t = 2 s\n R = ut\n = 15 x 2\n\n  = 30m\n\n\n', '', '', 'WAEC-PK-1', 'WAEC-PHY', 1),
(123, '2017', '\n\nQuestion\n\nExplain why mercury does not wet glass while water does.\n\n\n', 'Explain why mercury does not wet glass while water does.\n', '', '', NULL, '\n\nExplanation\nMercury does not wet glass because (force of) cohesion/attraction of mercury molecules is greater than (force of) adhesion/attraction between glass and mercury molecules.\nWater wets glass because (force of) adhesion/attraction between glass and water molecules is greater than (force of) cohesion/attraction of water molecules.\n\n', 'Mercury does not wet glass because (force of) cohesion/attraction of mercury molecules is greater than (force of) adhesion/attraction between glass and mercury molecules.\nWater wets glass because (force of) adhesion/attraction between glass and water molecules is greater than (force of) cohesion/attraction of water molecules.\n\n', '', '', 'WAEC-PK-1', 'WAEC-PHY', 1),
(124, '2017', '\n\nQuestion\n\n(a) Explain what is meant by cations\n(b) Draw and label an electrolytic cell\n\n\n', '', '(a) Explain what is meant by cations\n(b) Draw and label an electrolytic cell\n', '(a) Explain what is meant by cations~(b) Draw and label an electrolytic cell\n', NULL, '\n\nExplanation\n(a)  Cations – are positive ions that are attracted to the cathode (during electrolysis)\n\n(b) Diagram\n\nCorrect diagram/sketch: labeling cathode, anode and electrolyte. \n\n', '', 'Explanation\n(a)  Cations – are positive ions that are attracted to the cathode (during electrolysis)\n\n(b) Diagram\n\nCorrect diagram/sketch: labeling cathode, anode and electrolyte. \n\n', '(a)  Cations – are positive ions that are attracted to the cathode (during electrolysis)\n~(b) Diagram\n\nCorrect diagram/sketch: labeling cathode, anode and electrolyte. \n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(125, '2017', '\n\nQuestion\n\n(a) State Faraday’s second law of electrolysis.\n(b) An electric charge of 9.6 x 10\\(^4\\) C liberates 1 mole of substance containing 6.0 x 10\\(^{23}\\) atoms. Determine the value of the electronic charge\n\n', '', '(a) State Faraday’s second law of electrolysis.\n(b) An electric charge of 9.6 x 10\\(^4\\) C liberates 1 mole of substance containing 6.0 x 10\\(^{23}\\) atoms. Determine the value of the electronic charge', '(a) State Faraday’s second law of electrolysis.~(b) An electric charge of 9.6 x 10\\(^4\\) C liberates 1 mole of substance containing 6.0 x 10\\(^{23}\\) atoms. Determine the value of the electronic charge', NULL, '\n\nExplanation\n(a) If the same quantity of electricity is passed through different voltameters/electrolytes connected in series the masses of the substances liberated/deposited during electrolysis is (directly) proportional to their chemical equivalents.  \nOR The mass of an element deposited/liberated during electrolysis is (directly) proportional to the chemical equivalent of the element.\n\n\n(b) Let e represent the electronic charge.\n\n e = Faraday’s constant\n  Avogadro’s number\n = 9.6 x 10\\(^4\\)  \n 6. 0. x 10\\(^{23}\\)\n= 1.60 x 10\\(^{-19}\\)C\n\n', '', 'Explanation\n(a) If the same quantity of electricity is passed through different voltameters/electrolytes connected in series the masses of the substances liberated/deposited during electrolysis is (directly) proportional to their chemical equivalents.  \nOR The mass of an element deposited/liberated during electrolysis is (directly) proportional to the chemical equivalent of the element.\n\n\n(b) Let e represent the electronic charge.\n\n e = Faraday’s constant\n  Avogadro’s number\n = 9.6 x 10\\(^4\\)  \n 6. 0. x 10\\(^{23}\\)\n= 1.60 x 10\\(^{-19}\\)C\n\n', '(a) If the same quantity of electricity is passed through different voltameters/electrolytes connected in series the masses of the substances liberated/deposited during electrolysis is (directly) proportional to their chemical equivalents.  \nOR The mass of an element deposited/liberated during electrolysis is (directly) proportional to the chemical equivalent of the element.\n\n~(b) Let e represent the electronic charge.\n\n e = Faraday’s constant\n  Avogadro’s number\n = 9.6 x 10\\(^4\\)  \n 6. 0. x 10\\(^{23}\\)\n= 1.60 x 10\\(^{-19}\\)C\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(126, '2017', '\n\nQuestion\n\nExplain the following terms:\n(a) tensile stress;\n(b)Young’s modulus\n\n', '', 'Explain the following terms:\n(a) tensile stress;\n(b)Young’s modulus', '(a) tensile stress;~(b)Young’s modulus', NULL, '\n\nExplanation\n(a) Tensile stress is the ratio of force to cross-sectional area of a wire/rod. It is expressed in Newton per metre squared (Nm\\(^{-2}\\))\n\n(b) Young’s modulus is the ratio of tensile stress to tensile strain. It is expressed in Newton per metre squared. (Nm\\(^{-2}\\))\n\n', '', 'Explanation\n(a) Tensile stress is the ratio of force to cross-sectional area of a wire/rod. It is expressed in Newton per metre squared (Nm\\(^{-2}\\))\n\n(b) Young’s modulus is the ratio of tensile stress to tensile strain. It is expressed in Newton per metre squared. (Nm\\(^{-2}\\))\n\n', '(a) Tensile stress is the ratio of force to cross-sectional area of a wire/rod. It is expressed in Newton per metre squared (Nm\\(^{-2}\\))\n~(b) Young’s modulus is the ratio of tensile stress to tensile strain. It is expressed in Newton per metre squared. (Nm\\(^{-2}\\))\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(127, '2017', '\n\nQuestion\n\na) Define diffusion.\n(b) State two applications of electrical conduction through gases.\n\n', '', 'a) Define diffusion.\n(b) State two applications of electrical conduction through gases.', '~(b) State two applications of electrical conduction through gases.', NULL, '\n\nExplanation\n(a) Diffusion is the process by which substances mix (intimately) with one another due to the random motion of their molecules.\n\n(b) Applications of electrical conduction through gases include:.\n- In advertising industry/Neon signs\n- In lighting/fluorescent tubes\n- Identification of gases\n- Cathode ray oscilloscope/T.V. tubes\n\n', '', 'Explanation\n(a) Diffusion is the process by which substances mix (intimately) with one another due to the random motion of their molecules.\n\n(b) Applications of electrical conduction through gases include:.\n- In advertising industry/Neon signs\n- In lighting/fluorescent tubes\n- Identification of gases\n- Cathode ray oscilloscope/T.V. tubes\n\n', '(a) Diffusion is the process by which substances mix (intimately) with one another due to the random motion of their molecules.\n~(b) Applications of electrical conduction through gases include:.\n- In advertising industry/Neon signs\n- In lighting/fluorescent tubes\n- Identification of gases\n- Cathode ray oscilloscope/T.V. tubes\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(128, '2017', '\n\nQuestion\n\na) List two properties of cathode rays.\n(b) Explain how the intensity and energy of cathode rays may be increased\n\n', '', 'a) List two properties of cathode rays.\n(b) Explain how the intensity and energy of cathode rays may be increased', '~(b) Explain how the intensity and energy of cathode rays may be increased', NULL, '\n\nExplanation\n(a) Properties of cathode rays:\nCathode rays\n\n\n\nare negatively charged.\ntravel in straight line in field free space.\nare deflected by electric/magnetic field.\npossess(kinetic) energy.\npossess momentum.\n\n\n\n(b)(i) The intensity of cathode rays may be increased by raising the temperature of the cathode/increasing the current through the heater.\n(ii) The energy of cathode rays may be increased by raising the potential difference between the anode and the cathode/the anode potential.\n\n', '', 'Explanation\n(a) Properties of cathode rays:\nCathode rays\n\n\n\nare negatively charged.\ntravel in straight line in field free space.\nare deflected by electric/magnetic field.\npossess(kinetic) energy.\npossess momentum.\n\n\n\n(b)(i) The intensity of cathode rays may be increased by raising the temperature of the cathode/increasing the current through the heater.\n(ii) The energy of cathode rays may be increased by raising the potential difference between the anode and the cathode/the anode potential.\n\n', '(a) Properties of cathode rays:\nCathode rays\n\n\n\nare negatively charged.\ntravel in straight line in field free space.\nare deflected by electric/magnetic field.\npossess(kinetic) energy.\npossess momentum.\n\n\n~(b)(i) The intensity of cathode rays may be increased by raising the temperature of the cathode/increasing the current through the heater.\n(ii) The energy of cathode rays may be increased by raising the potential difference between the anode and the cathode/the anode potential.\n~c) energy.\npossess momentum.\n\n\n\n(b)(i) The intensity of cathode rays may be increased by raising the temperature of the cathode/increasing the current through the heater.\n(ii) The energy of cathode rays may be increased by raising the potential difference between the anode and the cathode/the anode potential.\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(129, '2017', '\n\nQuestion\n\na) Given a retort stand and clamp, a stout pin, a simple pendulum and a pencil, describe how you would use these apparatus to determine the centre of gravity of an irregularly shaped piece of cardboard of a moderate size.\n(b) Using a suitable diagram, explain how the following can be obtained from a velocity-time graph:\n(i) acceleration;\n(ii) total distance covered.\n(c ) A body at rest is given an initial uniform acceleration of 6.0 ms\\(^{-2}\\) for 20s after which the acceleration is reduced to 4.0 ms\\(^{-2}\\) for the next 10s.\nThe body maintains the speed attained for 30s.\n\nDraw the velocity-time graph of the motion using the information given above. From the graph, calculate the:\n\nmaximum speed attained during the motion;\ntotal distance traveled during the first 30 s;\naverage speed during the same time interval as in (ii) above.\n\n\n\n\na) Given a retort stand and clamp, a stout pin, a simple pendulum and a pencil, describe how you would use these apparatus to determine the centre of gravity of an irregularly shaped piece of cardboard of a moderate size.\n(b) Using a suitable diagram, explain how the following can be obtained from a velocity-time graph:\n(i) acceleration;\n(ii) total distance covered.\n(c ) A body at rest is given an initial uniform acceleration of 6.0 ms\\(^{-2}\\) for 20s after which the acceleration is reduced to 4.0 ms\\(^{-2}\\) for the next 10s.\nThe body maintains the speed attained for 30s.\n\nDraw the velocity-time graph of the motion using the information given above. From the graph, calculate the:\n\nmaximum speed attained during the motion;\ntotal distance traveled during the first 30 s;\naverage speed during the same time interval as in (ii) above.\n\n', '', 'a) Given a retort stand and clamp, a stout pin, a simple pendulum and a pencil, describe how you would use these apparatus to determine the centre of gravity of an irregularly shaped piece of cardboard of a moderate size.\n(b) Using a suitable diagram, explain how the following can be obtained from a velocity-time graph:\n(i) acceleration;\n(ii) total distance covered.\n(c ) A body at rest is given an initial uniform acceleration of 6.0 ms\\(^{-2}\\) for 20s after which the acceleration is reduced to 4.0 ms\\(^{-2}\\) for the next 10s.\nThe body maintains the speed attained for 30s.\n\nDraw the velocity-time graph of the motion using the information given above. From the graph, calculate the:\n\nmaximum speed attained during the motion;\ntotal distance traveled during the first 30 s;\naverage speed during the same time interval as in (ii) above.\n\n\n\n\na) Given a retort stand and clamp, a stout pin, a simple pendulum and a pencil, describe how you would use these apparatus to determine the centre of gravity of an irregularly shaped piece of cardboard of a moderate size.\n(b) Using a suitable diagram, explain how the following can be obtained from a velocity-time graph:\n(i) acceleration;\n(ii) total distance covered.\n(c ) A body at rest is given an initial uniform acceleration of 6.0 ms\\(^{-2}\\) for 20s after which the acceleration is reduced to 4.0 ms\\(^{-2}\\) for the next 10s.\nThe body maintains the speed attained for 30s.\n\nDraw the velocity-time graph of the motion using the information given above. From the graph, calculate the:\n\nmaximum speed attained during the motion;\ntotal distance traveled during the first 30 s;\naverage speed during the same time interval as in (ii) above.', '(i) acceleration;\n(ii) total distance covered.\n(c ) A body at rest is given an initial uniform acceleration of 6.0 ms\\(^{-2}\\) for 20s after which the acceleration is reduced to 4.0 ms\\(^{-2}\\) for the next 10s.\nThe body maintains the speed attained for 30s.\n\nDraw the velocity-time graph of the motion using the information given above. From the graph, calculate the:\n\nmaximum speed attained during the motion;\ntotal distance traveled during the first 30 s;\naverage speed during the same time interval as in (ii) above.\n\n\n\n\na) Given a retort stand and clamp, a stout pin, a simple pendulum and a pencil, describe how you would use these apparatus to determine the centre of gravity of an irregularly shaped piece of cardboard of a moderate size.~(b) Using a suitable diagram, explain how the following can be obtained from a velocity-time graph:\n(i) acceleration;\n(ii) total distance covered.\n(c ) A body at rest is given an initial uniform acceleration of 6.0 ms\\(^{-2}\\) for 20s after which the acceleration is reduced to 4.0 ms\\(^{-2}\\) for the next 10s.\nThe body maintains the speed attained for 30s.\n\nDraw the velocity-time graph of the motion using the information given above. From the graph, calculate the:\n\nmaximum speed attained during the motion;\ntotal distance traveled during the first 30 s;\naverage speed during the same time interval as in (ii) above.\n\n\n\n\na) Given a retort stand and clamp, a stout pin, a simple pendulum and a pencil, describe how you would use these apparatus to determine the centre of gravity of an irregularly shaped piece of cardboard of a moderate size.\n(b) Using a suitable diagram, explain how the following can be obtained from a velocity-time graph:\n(i) acceleration;\n(ii) total distance covered.\n(c ) A body at rest is given an initial uniform acceleration of 6.0 ms\\(^{-2}\\) for 20s after which the acceleration is reduced to 4.0 ms\\(^{-2}\\) for the next 10s.\nThe body maintains the speed attained for 30s.\n\nDraw the velocity-time graph of the motion using the information given above. From the graph, calculate the:\n\nmaximum speed attained during the motion;\ntotal distance traveled during the first 30 s;\naverage speed during the same time interval as in (ii) above.', NULL, '\n\nExplanation\n(a) PROCEDURE\nMake at least 3 well-spaced pin holes round the edge of the cardboard. Clamp the pin horizontally and suspend the cardboard on it through one of the pin-holes such that the cardboard can swing freely\nHang the simple pendulum on the same pin and let its string be very close to the cardboard.  \nWhen the whole system is at rest (or in equilibrium) trace the plumbline on the cardboard. Repeat the procedure for each of the two other pin holes. \n CONCLUSION\nThe point at which the (three) traced lines intersect is the centre of gravity of the cardboard. \n PRECAUTIONS \n-  Repeat procedure\n- Pin rigidly and firmly held by retort stand and clamp\n- Allow the simple pendulum to rest before tracing the shadow of\n the plumbline on the cardboard.\n-  The string to be close to the cardboard\nb)  Diagram: \n\n  Both axes correctly labelled  \n  Any correct shape of graph showing acceleration segment \n\nAcceleration = gradient of AB  \nTotal distance covered = area under the graph  \n\n(c) Diagram\n\n- At least one axis labelled \n- correct shape of graph  \nLet V\\(_1\\) = maximum velocity after 20 sec.\n V\\(_2\\) = maximum velocity after 30 sec.\n(i) Then  \\(\\frac{V_1}{20}\\)  = 6\n \n ', '', 'Explanation\n(a) PROCEDURE\nMake at least 3 well-spaced pin holes round the edge of the cardboard. Clamp the pin horizontally and suspend the cardboard on it through one of the pin-holes such that the cardboard can swing freely\nHang the simple pendulum on the same pin and let its string be very close to the cardboard.  \nWhen the whole system is at rest (or in equilibrium) trace the plumbline on the cardboard. Repeat the procedure for each of the two other pin holes. \n CONCLUSION\nThe point at which the (three) traced lines intersect is the centre of gravity of the cardboard. \n PRECAUTIONS \n-  Repeat procedure\n- Pin rigidly and firmly held by retort stand and clamp\n- Allow the simple pendulum to rest before tracing the shadow of\n the plumbline on the cardboard.\n-  The string to be close to the cardboard\nb)  Diagram: \n\n  Both axes correctly labelled  \n  Any correct shape of graph showing acceleration segment \n\nAcceleration = gradient of AB  \nTotal distance covered = area under the graph  \n\n(c) Diagram\n\n- At least one axis labelled \n- correct shape of graph  \nLet V\\(_1\\) = maximum velocity after 20 sec.\n V\\(_2\\) = maximum velocity after 30 sec.\n(i) Then  \\(\\frac{V_1}{20}\\)  = 6\n \n ', '~b)  Diagram: \n\n  Both axes correctly labelled  \n  Any correct shape of graph showing acceleration segment \n\nAcceleration = gradient of AB  \nTotal distance covered = area under the graph  \n\n(c) Diagram\n\n- At least one axis labelled \n- correct shape of graph  \nLet V\\(_1\\) = maximum velocity after 20 sec.\n V\\(_2\\) = maximum velocity after 30 sec.\n(i) Then  \\(\\frac{V_1}{20}\\)  = 6\n ~(c) Diagram\n\n- At least one axis labelled \n- correct shape of graph  \nLet V\\(_1\\) = maximum velocity after 20 sec.\n V\\(_2\\) = maximum velocity after 30 sec.\n(i) Then  \\(\\frac{V_1}{20}\\)  = 6\n ', 'WAEC-PK-1', 'WAEC-PHY', 1),
(130, '2017', '\n\nQuestion\n\n(a) Explain why it is not advisable to sterilize a clinical thermometer in boiling water at normal atmospheric pressure.\n(b) State the effect of an increase in pressure on the\n (i) boiling point; and\n (ii) melting point of water.\n(c) Diagram:\n\nThe graph shown above is that of the saturated vapour pressure (s.v.p.) of water against temperature.\nPure water is known to boil at 100\\(^o\\)C and at an atmospheric pressure of 760 mmHg. What general conclusion can be drawn from the information given above?\nd) A thread of mercury of length 20 cm is used to trap some air in a capillary tube with uniform cross-sectional area and closed at one end. With the tube vertical and the open end uppermost, the length of the trapped air column is 15cm. Calculate the length of the air column when the tube is held:\ni) horizontally;\nii) vertically with the open end underneath. [Atmospheric pressure = 76 cmHg ]\n\n', '', '(a) Explain why it is not advisable to sterilize a clinical thermometer in boiling water at normal atmospheric pressure.\n(b) State the effect of an increase in pressure on the\n (i) boiling point; and\n (ii) melting point of water.\n(c) Diagram:\n\nThe graph shown above is that of the saturated vapour pressure (s.v.p.) of water against temperature.\nPure water is known to boil at 100\\(^o\\)C and at an atmospheric pressure of 760 mmHg. What general conclusion can be drawn from the information given above?\nd) A thread of mercury of length 20 cm is used to trap some air in a capillary tube with uniform cross-sectional area and closed at one end. With the tube vertical and the open end uppermost, the length of the trapped air column is 15cm. Calculate the length of the air column when the tube is held:\ni) horizontally;\nii) vertically with the open end underneath. [Atmospheric pressure = 76 cmHg ]', '(a) Explain why it is not advisable to sterilize a clinical thermometer in boiling water at normal atmospheric pressure.~(b) State the effect of an increase in pressure on the\n (i) boiling point; and\n (ii) melting point of water.\n(c) Diagram:\n\nThe graph shown above is that of the saturated vapour pressure (s.v.p.) of water against temperature.\nPure water is known to boil at 100\\(^o\\)C and at an atmospheric pressure of 760 mmHg. What general conclusion can be drawn from the information given above?\nd) A thread of mercury of length 20 cm is used to trap some air in a capillary tube with uniform cross-sectional area and closed at one end. With the tube vertical and the open end uppermost, the length of the trapped air column is 15cm. Calculate the length of the air column when the tube is held:\ni) horizontally;\nii) vertically with the open end underneath. [Atmospheric pressure = 76 cmHg ]~(c) Diagram:\n\nThe graph shown above is that of the saturated vapour pressure (s.v.p.) of water against temperature.\nPure water is known to boil at 100\\(^o\\)C and at an atmospheric pressure of 760 mmHg. What general conclusion can be drawn from the information given above?\nd) A thread of mercury of length 20 cm is used to trap some air in a capillary tube with uniform cross-sectional area and closed at one end. With the tube vertical and the open end uppermost, the length of the trapped air column is 15cm. Calculate the length of the air column when the tube is held:\ni) horizontally;\nii) vertically with the open end underneath. [Atmospheric pressure = 76 cmHg ]', NULL, '\n\nExplanation\n (a) A clinical thermometer has small temperature range. The glass will crack/burst due to excessive pressure created by expansion of mercury.\n(b) Increase in pressure\n (i) increases the boiling point; \n (ii) decreases the melting point. \n\n(c) At the boiling point of pure water, the saturated vapour pressure (s.v.p) of the water is equal to the external atmospheric pressure.  \n(d) Diagram\n\nVolume V is proportional to length Ɩ stated or implied \n(i)  P\\(_1\\)V\\(_1\\) = P\\(_2\\)V\\(_2\\) implies P\\(_1\\)P\\(_1\\) = P\\(_2\\)P\\(_2\\) \n(76 + 20) 15 = 76 x L\\(_2\\) \n Ɩ\\(_2\\) = 96 x 15\n    76\n Ɩ\\(_2\\) = 18.95 cm\n(ii) P\\(_1\\)L\\(_1\\) = P\\(_3\\) L\\(_3\\) \n= (76 + 20) 15 = (76 – 20) x Ɩ\\(_3\\)\n Ɩ\\(_3\\) = 96 x 15 \n              56\nƖ\\(_3\\) = 25.7l cm\n\n', '', 'Explanation\n (a) A clinical thermometer has small temperature range. The glass will crack/burst due to excessive pressure created by expansion of mercury.\n(b) Increase in pressure\n (i) increases the boiling point; \n (ii) decreases the melting point. \n\n(c) At the boiling point of pure water, the saturated vapour pressure (s.v.p) of the water is equal to the external atmospheric pressure.  \n(d) Diagram\n\nVolume V is proportional to length Ɩ stated or implied \n(i)  P\\(_1\\)V\\(_1\\) = P\\(_2\\)V\\(_2\\) implies P\\(_1\\)P\\(_1\\) = P\\(_2\\)P\\(_2\\) \n(76 + 20) 15 = 76 x L\\(_2\\) \n Ɩ\\(_2\\) = 96 x 15\n    76\n Ɩ\\(_2\\) = 18.95 cm\n(ii) P\\(_1\\)L\\(_1\\) = P\\(_3\\) L\\(_3\\) \n= (76 + 20) 15 = (76 – 20) x Ɩ\\(_3\\)\n Ɩ\\(_3\\) = 96 x 15 \n              56\nƖ\\(_3\\) = 25.7l cm\n\n', '(a) A clinical thermometer has small temperature range. The glass will crack/burst due to excessive pressure created by expansion of mercury.~(b) Increase in pressure\n (i) increases the boiling point; \n (ii) decreases the melting point. \n\n(c) At the boiling point of pure water, the saturated vapour pressure (s.v.p) of the water is equal to the external atmospheric pressure.  \n(d) Diagram\n\nVolume V is proportional to length Ɩ stated or implied \n(i)  P\\(_1\\)V\\(_1\\) = P\\(_2\\)V\\(_2\\) implies P\\(_1\\)P\\(_1\\) = P\\(_2\\)P\\(_2\\) \n(76 + 20) 15 = 76 x L\\(_2\\) \n Ɩ\\(_2\\) = 96 x 15\n    76\n Ɩ\\(_2\\) = 18.95 cm\n(ii) P\\(_1\\)L\\(_1\\) = P\\(_3\\) L\\(_3\\) \n= (76 + 20) 15 = (76 – 20) x Ɩ\\(_3\\)\n Ɩ\\(_3\\) = 96 x 15 \n              56\nƖ\\(_3\\) = 25.7l cm\n~(c) At the boiling point of pure water, the saturated vapour pressure (s.v.p) of the water is equal to the external atmospheric pressure.  \n(d) Diagram\n\nVolume V is proportional to length Ɩ stated or implied \n(i)  P\\(_1\\)V\\(_1\\) = P\\(_2\\)V\\(_2\\) implies P\\(_1\\)P\\(_1\\) = P\\(_2\\)P\\(_2\\) \n(76 + 20) 15 = 76 x L\\(_2\\) \n Ɩ\\(_2\\) = 96 x 15\n    76\n Ɩ\\(_2\\) = 18.95 cm\n(ii) P\\(_1\\)L\\(_1\\) = P\\(_3\\) L\\(_3\\) \n= (76 + 20) 15 = (76 – 20) x Ɩ\\(_3\\)\n Ɩ\\(_3\\) = 96 x 15 \n              56\nƖ\\(_3\\) = 25.7l cm\n~(d) Diagram\n\nVolume V is proportional to length Ɩ stated or implied \n(i)  P\\(_1\\)V\\(_1\\) = P\\(_2\\)V\\(_2\\) implies P\\(_1\\)P\\(_1\\) = P\\(_2\\)P\\(_2\\) \n(76 + 20) 15 = 76 x L\\(_2\\) \n Ɩ\\(_2\\) = 96 x 15\n    76\n Ɩ\\(_2\\) = 18.95 cm\n(ii) P\\(_1\\)L\\(_1\\) = P\\(_3\\) L\\(_3\\) \n= (76 + 20) 15 = (76 – 20) x Ɩ\\(_3\\)\n Ɩ\\(_3\\) = 96 x 15 \n              56\nƖ\\(_3\\) = 25.7l cm\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(131, '2017', '\n\nQuestion\n\n(a) State two differences between a sound wave and a radio wave.\n(b) Explain why a vibrating tuning fork sounds louder when its stem is pressed against a table top than when held in air.\n(c)State two conditions necessary for the:\n\nproduction of stationary wave in a medium;\nformation of interference wave patterns;\noccurrence of total internal reflection of a wave.\n\n(d) A ray of light is incident on one face of an equilateral glass prism.\n\nDraw a ray diagram to show the path of the ray through the prism.\nCalculate the refractive index of the glass if the angle of minimum deviation is 41\\(^o\\).\n\n\n\n\n(a) State two differences between a sound wave and a radio wave.\n(b) Explain why a vibrating tuning fork sounds louder when its stem is pressed against a table top than when held in air.\n(c)State two conditions necessary for the:\n\nproduction of stationary wave in a medium;\nformation of interference wave patterns;\noccurrence of total internal reflection of a wave.\n\n(d) A ray of light is incident on one face of an equilateral glass prism.\n\nDraw a ray diagram to show the path of the ray through the prism.\nCalculate the refractive index of the glass if the angle of minimum deviation is 41\\(^o\\).\n\n', '', '(a) State two differences between a sound wave and a radio wave.\n(b) Explain why a vibrating tuning fork sounds louder when its stem is pressed against a table top than when held in air.\n(c)State two conditions necessary for the:\n\nproduction of stationary wave in a medium;\nformation of interference wave patterns;\noccurrence of total internal reflection of a wave.\n\n(d) A ray of light is incident on one face of an equilateral glass prism.\n\nDraw a ray diagram to show the path of the ray through the prism.\nCalculate the refractive index of the glass if the angle of minimum deviation is 41\\(^o\\).\n\n\n\n\n(a) State two differences between a sound wave and a radio wave.\n(b) Explain why a vibrating tuning fork sounds louder when its stem is pressed against a table top than when held in air.\n(c)State two conditions necessary for the:\n\nproduction of stationary wave in a medium;\nformation of interference wave patterns;\noccurrence of total internal reflection of a wave.\n\n(d) A ray of light is incident on one face of an equilateral glass prism.\n\nDraw a ray diagram to show the path of the ray through the prism.\nCalculate the refractive index of the glass if the angle of minimum deviation is 41\\(^o\\).', '(a) State two differences between a sound wave and a radio wave.\n(b) Explain why a vibrating tuning fork sounds louder when its stem is pressed against a table top than when held in air.\n(c)State two conditions necessary for the:\n\nproduction of stationary wave in a medium;\nformation of interference wave patterns;\noccurrence of total internal reflection of a wave.\n\n(d) A ray of light is incident on one face of an equilateral glass prism.\n\nDraw a ray diagram to show the path of the ray through the prism.\nCalculate the refractive index of the glass if the angle of minimum deviation is 41\\(^o\\).\n\n\n\n\n(a) State two differences between a sound wave and a radio wave.~(b) Explain why a vibrating tuning fork sounds louder when its stem is pressed against a table top than when held in air.\n(c)State two conditions necessary for the:\n\nproduction of stationary wave in a medium;\nformation of interference wave patterns;\noccurrence of total internal reflection of a wave.\n\n(d) A ray of light is incident on one face of an equilateral glass prism.\n\nDraw a ray diagram to show the path of the ray through the prism.\nCalculate the refractive index of the glass if the angle of minimum deviation is 41\\(^o\\).\n\n\n\n\n(a) State two differences between a sound wave and a radio wave.\n(b) Explain why a vibrating tuning fork sounds louder when its stem is pressed against a table top than when held in air.\n(c)State two conditions necessary for the:\n\nproduction of stationary wave in a medium;\nformation of interference wave patterns;\noccurrence of total internal reflection of a wave.\n\n(d) A ray of light is incident on one face of an equilateral glass prism.\n\nDraw a ray diagram to show the path of the ray through the prism.\nCalculate the refractive index of the glass if the angle of minimum deviation is 41\\(^o\\).~(c)State two conditions necessary for the:\n\nproduction of stationary wave in a medium;\nformation of interference wave patterns;\noccurrence of total internal reflection of a wave.\n\n(d) A ray of light is incident on one face of an equilateral glass prism.\n\nDraw a ray diagram to show the path of the ray through the prism.\nCalculate the refractive index of the glass if the angle of minimum deviation is 41\\(^o\\).\n\n\n\n\n(a) State two differences between a sound wave and a radio wave.\n(b) Explain why a vibrating tuning fork sounds louder when its stem is pressed against a table top than when held in air.\n(c)State two conditions necessary for the:\n\nproduction of stationary wave in a medium;\nformation of interference wave patterns;\noccurrence of total internal reflection of a wave.\n\n(d) A ray of light is incident on one face of an equilateral glass prism.\n\nDraw a ray diagram to show the path of the ray through the prism.\nCalculate the refractive index of the glass if the angle of minimum deviation is 41\\(^o\\).~(d) A ray of light is incident on one face of an equilateral glass prism.\n\nDraw a ray diagram to show the path of the ray through the prism.\nCalculate the refractive index of the glass if the angle of minimum deviation is 41\\(^o\\).\n\n\n\n\n(a) State two differences between a sound wave and a radio wave.\n(b) Explain why a vibrating tuning fork sounds louder when its stem is pressed against a table top than when held in air.\n(c)State two conditions necessary for the:\n\nproduction of stationary wave in a medium;\nformation of interference wave patterns;\noccurrence of total internal reflection of a wave.\n\n(d) A ray of light is incident on one face of an equilateral glass prism.\n\nDraw a ray diagram to show the path of the ray through the prism.\nCalculate the refractive index of the glass if the angle of minimum deviation is 41\\(^o\\).', NULL, '\n\nExplanation\n(a) Differences between a sound wave and a radiowave\n\n\n\nSound wave\nRadio wave\n\n\nis a mechanical wave\nis an electromagnetic wave\n\n\n is longitudinal\nis transverse\n\n\nhas speed less than speed of light.\nhas speed equal to that of light\n\n\nrequires a material medium for propagation/cannot travel through vacuum.\ndoes not require a material medium for propagation/can travel through vacuum.\n\n\n\n\n  \n(b) The vibrating tuning fork forces the table top to vibrate. The table top has a much larger area and is therefore in contact with a larger volume of air undergoing vibration. Hence, sound heard is amplified.\n \n(c) Conditions necessary for:\n (i) production of stationary wave\n- Two waves must be traveling in opposite directions.\n- Incident and reflected waves must have the same frequency/wavelength.\n- Incident and reflected waves must have equal amplitude.\n- There must be superimposition.\n- There must be a barrier\n(ii) formation of interference wave pattern\n- Two sources must be involved.\n- The two sources must be coherent/have a constant phase difference as well as the same frequency and amplitude.\n- The waves that are interfering must have the same amplitude\n- The distance between the sources must be of the order of the wavelengths of the waves.\n (iii) total internal reflection\n - The wave must be traveling from a denser medium to a less dense medium\n - The angle of incidence of the wave in the denser medium must be greater than the critical angle for the medium.\n\n \n(d)(i) Diagram\n\n Correct diagram\n (ii)  \\(\\frac{ Sin (A + dm)}{2}\\)\n n =\n  Sin (A/2)\n \n  Sin (60 + 41)/2\n          \n  Sin (60/2)\n  = 1.54\n\n', '', 'Explanation\n(a) Differences between a sound wave and a radiowave\n\n\n\nSound wave\nRadio wave\n\n\nis a mechanical wave\nis an electromagnetic wave\n\n\n is longitudinal\nis transverse\n\n\nhas speed less than speed of light.\nhas speed equal to that of light\n\n\nrequires a material medium for propagation/cannot travel through vacuum.\ndoes not require a material medium for propagation/can travel through vacuum.\n\n\n\n\n  \n(b) The vibrating tuning fork forces the table top to vibrate. The table top has a much larger area and is therefore in contact with a larger volume of air undergoing vibration. Hence, sound heard is amplified.\n \n(c) Conditions necessary for:\n (i) production of stationary wave\n- Two waves must be traveling in opposite directions.\n- Incident and reflected waves must have the same frequency/wavelength.\n- Incident and reflected waves must have equal amplitude.\n- There must be superimposition.\n- There must be a barrier\n(ii) formation of interference wave pattern\n- Two sources must be involved.\n- The two sources must be coherent/have a constant phase difference as well as the same frequency and amplitude.\n- The waves that are interfering must have the same amplitude\n- The distance between the sources must be of the order of the wavelengths of the waves.\n (iii) total internal reflection\n - The wave must be traveling from a denser medium to a less dense medium\n - The angle of incidence of the wave in the denser medium must be greater than the critical angle for the medium.\n\n \n(d)(i) Diagram\n\n Correct diagram\n (ii)  \\(\\frac{ Sin (A + dm)}{2}\\)\n n =\n  Sin (A/2)\n \n  Sin (60 + 41)/2\n          \n  Sin (60/2)\n  = 1.54\n\n', '(a) Differences between a sound wave and a radiowave\n\n\n\nSound wave\nRadio wave\n\n\nis a mechanical wave\nis an electromagnetic wave\n\n\n is longitudinal\nis transverse\n\n\nhas speed less than speed of light.\nhas speed equal to that of light\n\n\nrequires a material medium for propagation/cannot travel through vacuum.\ndoes not require a material medium for propagation/can travel through vacuum.\n\n\n\n\n  ~(b) The vibrating tuning fork forces the table top to vibrate. The table top has a much larger area and is therefore in contact with a larger volume of air undergoing vibration. Hence, sound heard is amplified.\n \n(c) Conditions necessary for:\n (i) production of stationary wave\n- Two waves must be traveling in opposite directions.\n- Incident and reflected waves must have the same frequency/wavelength.\n- Incident and reflected waves must have equal amplitude.\n- There must be superimposition.\n- There must be a barrier\n(ii) formation of interference wave pattern\n- Two sources must be involved.\n- The two sources must be coherent/have a constant phase difference as well as the same frequency and amplitude.\n- The waves that are interfering must have the same amplitude\n- The distance between the sources must be of the order of the wavelengths of the waves.\n (iii) total internal reflection\n - The wave must be traveling from a denser medium to a less dense medium\n - The angle of incidence of the wave in the denser medium must be greater than the critical angle for the medium.\n\n \n(d)(i) Diagram\n\n Correct diagram\n (ii)  \\(\\frac{ Sin (A + dm)}{2}\\)\n n =\n  Sin (A/2)\n \n  Sin (60 + 41)/2\n          \n  Sin (60/2)\n  = 1.54\n~(c) Conditions necessary for:\n (i) production of stationary wave\n- Two waves must be traveling in opposite directions.\n- Incident and reflected waves must have the same frequency/wavelength.\n- Incident and reflected waves must have equal amplitude.\n- There must be superimposition.\n- There must be a barrier\n(ii) formation of interference wave pattern\n- Two sources must be involved.\n- The two sources must be coherent/have a constant phase difference as well as the same frequency and amplitude.\n- The waves that are interfering must have the same amplitude\n- The distance between the sources must be of the order of the wavelengths of the waves.\n (iii) total internal reflection\n - The wave must be traveling from a denser medium to a less dense medium\n - The angle of incidence of the wave in the denser medium must be greater than the critical angle for the medium.\n\n \n(d)(i) Diagram\n\n Correct diagram\n (ii)  \\(\\frac{ Sin (A + dm)}{2}\\)\n n =\n  Sin (A/2)\n \n  Sin (60 + 41)/2\n          \n  Sin (60/2)\n  = 1.54\n~(d)(i) Diagram\n\n Correct diagram\n (ii)  \\(\\frac{ Sin (A + dm)}{2}\\)\n n =\n  Sin (A/2)\n \n  Sin (60 + 41)/2\n          \n  Sin (60/2)\n  = 1.54\n', 'WAEC-PK-1', 'WAEC-PHY', 1);
INSERT INTO `exam_simulation2` (`id`, `examYear`, `questions`, `parent_que`, `child_que`, `gen_child_que_desc`, `imageDir`, `explanation`, `parent_que_expl`, `child_que_exp`, `gen_child_que_expl`, `package_id`, `subject_id`, `status`) VALUES
(132, '2017', '\n\nQuestion\n\n(a) State two essential differences between a moving coil galvanometer and a d.c. generator.\n(b) Explain the term eddy currents and state two devices in which the currents are applied.\n(c) State the principle on which the potentiometer is based when it is functioning.\n(d) A source of e.m.f. 110 V and frequency 60Hz is connected to a resistor, an inductor and a capacitor in series. When the current in the capacitor is 2A, the potential differences across the resistor is 80 V and that across the inductor is 40 V. Draw the vector diagram of the potential differences across the inductor, the capacitor and the resistor.\nCalculate the:\n(i) potential difference across the capacitor;\n(ii) capacitance of the capacitor;\n(iii) inductance of the inductor. [π = 3.14]\n\n', '', '(a) State two essential differences between a moving coil galvanometer and a d.c. generator.\n(b) Explain the term eddy currents and state two devices in which the currents are applied.\n(c) State the principle on which the potentiometer is based when it is functioning.\n(d) A source of e.m.f. 110 V and frequency 60Hz is connected to a resistor, an inductor and a capacitor in series. When the current in the capacitor is 2A, the potential differences across the resistor is 80 V and that across the inductor is 40 V. Draw the vector diagram of the potential differences across the inductor, the capacitor and the resistor.\nCalculate the:\n(i) potential difference across the capacitor;\n(ii) capacitance of the capacitor;\n(iii) inductance of the inductor. [π = 3.14]', '(a) State two essential differences between a moving coil galvanometer and a d.c. generator.~(b) Explain the term eddy currents and state two devices in which the currents are applied.\n(c) State the principle on which the potentiometer is based when it is functioning.\n(d) A source of e.m.f. 110 V and frequency 60Hz is connected to a resistor, an inductor and a capacitor in series. When the current in the capacitor is 2A, the potential differences across the resistor is 80 V and that across the inductor is 40 V. Draw the vector diagram of the potential differences across the inductor, the capacitor and the resistor.\nCalculate the:\n(i) potential difference across the capacitor;\n(ii) capacitance of the capacitor;\n(iii) inductance of the inductor. [π = 3.14]~(c) State the principle on which the potentiometer is based when it is functioning.\n(d) A source of e.m.f. 110 V and frequency 60Hz is connected to a resistor, an inductor and a capacitor in series. When the current in the capacitor is 2A, the potential differences across the resistor is 80 V and that across the inductor is 40 V. Draw the vector diagram of the potential differences across the inductor, the capacitor and the resistor.\nCalculate the:\n(i) potential difference across the capacitor;\n(ii) capacitance of the capacitor;\n(iii) inductance of the inductor. [π = 3.14]~(d) A source of e.m.f. 110 V and frequency 60Hz is connected to a resistor, an inductor and a capacitor in series. When the current in the capacitor is 2A, the potential differences across the resistor is 80 V and that across the inductor is 40 V. Draw the vector diagram of the potential differences across the inductor, the capacitor and the resistor.\nCalculate the:\n(i) potential difference across the capacitor;\n(ii) capacitance of the capacitor;\n(iii) inductance of the inductor. [π = 3.14]', NULL, '\n\nExplanation\na) Differences between a d. c. generator and a moving coil galvanometer.\n\n\n\nD. C. generator\nmoving coil galvanometer\n\n\nconverts mechanical energy to electrical energy\nconverts electrical energy to mechanical energy.\n\n\nuses split rings or commutator\nuses hair springs.\n\n\nrotation of coil is continuous\nrotation of coil is incomplete\n\n\nuses carbon brushes (as terminals)\nuses jeweled bearings (as terminals)\n\n\n\n(b) Eddy currents are currents induced in a conductor when subjected to varying magnetic field.   \nAny valid additional information e.g.\n\neddy current flows in a circular path or closed loops;\neddy current generates heat;\neddy current cannot flow through gaps or slots;\nThe currents move in such a direction as to oppose the change producing them.\n\nDevices in which eddy currents are applied\n\npointers of sensitive electric meters\nsensitive mass balances\nbrakes in large electric motors\nspeedometers in automobiles\ndetection of cracks in railway tracks\ndetection of metals\n\n(c) Principle on which a potentiometer is based\nWhen a steady current is allowed to pass through a uniform wire, equal lengths of the wire will have equal potential differences.\n OR\nThe p.d across a length of a wire is (directly) proportional to the length provided the wire has a uniform cross section.\n(d)\n\n Vector diagram showing\n V\\(_L\\) with arrow\n V\\(_C\\) with arrow\n V\\(_R\\) with arrow\n\n(i) V\\(^2\\) = V\\(_R^2\\) + (V\\(_L\\) - V\\(_C\\))\\(^2\\) \n\n 110\\(^2\\)  = 80\\(^2\\) + (40 – V\\(_C\\))\\(^2\\)\n(40 – V\\(_C\\)) = \\(\\sqrt{(110 + 80) (110 – 80)}\\) \n 40 – V\\(_C\\) = + 75.5\n V\\(_C\\) = 40 + 75.5\n V\\(_C\\) = 115.5V\n  \n(ii) X\\(_c\\) \\(\\frac{V_c}{I} = \\frac{1}{2 \\pi f c}\\)\nC = \\(\\frac{1}{2 \\pi fV_c}\\)\nC \\(\\frac{2}{2 \\times 3.14 \\times 60 \\times 115.5} = 45.9\\muF\\) \n\n(iii) \\(V_L = IX_L = I \\times 2 \\pi fL}\\)\nL = \\(\\frac{V_l}{I \\times 1 \\times 2 \\pifVc}\\)\nC = \\(\\frac{2}{2 \\times 3.142 \\times 60 \\times 2}\\)\n= 0.053H\n\n', '', 'Explanation\na) Differences between a d. c. generator and a moving coil galvanometer.\n\n\n\nD. C. generator\nmoving coil galvanometer\n\n\nconverts mechanical energy to electrical energy\nconverts electrical energy to mechanical energy.\n\n\nuses split rings or commutator\nuses hair springs.\n\n\nrotation of coil is continuous\nrotation of coil is incomplete\n\n\nuses carbon brushes (as terminals)\nuses jeweled bearings (as terminals)\n\n\n\n(b) Eddy currents are currents induced in a conductor when subjected to varying magnetic field.   \nAny valid additional information e.g.\n\neddy current flows in a circular path or closed loops;\neddy current generates heat;\neddy current cannot flow through gaps or slots;\nThe currents move in such a direction as to oppose the change producing them.\n\nDevices in which eddy currents are applied\n\npointers of sensitive electric meters\nsensitive mass balances\nbrakes in large electric motors\nspeedometers in automobiles\ndetection of cracks in railway tracks\ndetection of metals\n\n(c) Principle on which a potentiometer is based\nWhen a steady current is allowed to pass through a uniform wire, equal lengths of the wire will have equal potential differences.\n OR\nThe p.d across a length of a wire is (directly) proportional to the length provided the wire has a uniform cross section.\n(d)\n\n Vector diagram showing\n V\\(_L\\) with arrow\n V\\(_C\\) with arrow\n V\\(_R\\) with arrow\n\n(i) V\\(^2\\) = V\\(_R^2\\) + (V\\(_L\\) - V\\(_C\\))\\(^2\\) \n\n 110\\(^2\\)  = 80\\(^2\\) + (40 – V\\(_C\\))\\(^2\\)\n(40 – V\\(_C\\)) = \\(\\sqrt{(110 + 80) (110 – 80)}\\) \n 40 – V\\(_C\\) = + 75.5\n V\\(_C\\) = 40 + 75.5\n V\\(_C\\) = 115.5V\n  \n(ii) X\\(_c\\) \\(\\frac{V_c}{I} = \\frac{1}{2 \\pi f c}\\)\nC = \\(\\frac{1}{2 \\pi fV_c}\\)\nC \\(\\frac{2}{2 \\times 3.14 \\times 60 \\times 115.5} = 45.9\\muF\\) \n\n(iii) \\(V_L = IX_L = I \\times 2 \\pi fL}\\)\nL = \\(\\frac{V_l}{I \\times 1 \\times 2 \\pifVc}\\)\nC = \\(\\frac{2}{2 \\times 3.142 \\times 60 \\times 2}\\)\n= 0.053H\n\n', 'a) Differences between a d. c. generator and a moving coil galvanometer.\n\n\n\nD. C. generator\nmoving coil galvanometer\n\n\nconverts mechanical energy to electrical energy\nconverts electrical energy to mechanical energy.\n\n\nuses split rings or commutator\nuses hair springs.\n\n\nrotation of coil is continuous\nrotation of coil is incomplete\n\n\nuses carbon brushes (as terminals)\nuses jeweled bearings (as terminals)\n\n\n~(b) Eddy currents are currents induced in a conductor when subjected to varying magnetic field.   \nAny valid additional information e.g.\n\neddy current flows in a circular path or closed loops;\neddy current generates heat;\neddy current cannot flow through gaps or slots;\nThe currents move in such a direction as to oppose the change producing them.\n\nDevices in which eddy currents are applied\n\npointers of sensitive electric meters\nsensitive mass balances\nbrakes in large electric motors\nspeedometers in automobiles\ndetection of cracks in railway tracks\ndetection of metals\n\n(c) Principle on which a potentiometer is based\nWhen a steady current is allowed to pass through a uniform wire, equal lengths of the wire will have equal potential differences.\n OR\nThe p.d across a length of a wire is (directly) proportional to the length provided the wire has a uniform cross section.\n(d)\n\n Vector diagram showing\n V\\(_L\\) with arrow\n V\\(_C\\) with arrow\n V\\(_R\\) with arrow\n\n(i) V\\(^2\\) = V\\(_R^2\\) + (V\\(_L\\) - V\\(_C\\))\\(^2\\) \n\n 110\\(^2\\)  = 80\\(^2\\) + (40 – V\\(_C\\))\\(^2\\)\n(40 – V\\(_C\\)) = \\(\\sqrt{(110 + 80) (110 – 80)}\\) \n 40 – V\\(_C\\) = + 75.5\n V\\(_C\\) = 40 + 75.5\n V\\(_C\\) = 115.5V\n  \n(ii) X\\(_c\\) \\(\\frac{V_c}{I} = \\frac{1}{2 \\pi f c}\\)\nC = \\(\\frac{1}{2 \\pi fV_c}\\)\nC \\(\\frac{2}{2 \\times 3.14 \\times 60 \\times 115.5} = 45.9\\muF\\) \n\n(iii) \\(V_L = IX_L = I \\times 2 \\pi fL}\\)\nL = \\(\\frac{V_l}{I \\times 1 \\times 2 \\pifVc}\\)\nC = \\(\\frac{2}{2 \\times 3.142 \\times 60 \\times 2}\\)\n= 0.053H\n~(c) Principle on which a potentiometer is based\nWhen a steady current is allowed to pass through a uniform wire, equal lengths of the wire will have equal potential differences.\n OR\nThe p.d across a length of a wire is (directly) proportional to the length provided the wire has a uniform cross section.\n(d)\n\n Vector diagram showing\n V\\(_L\\) with arrow\n V\\(_C\\) with arrow\n V\\(_R\\) with arrow\n\n(i) V\\(^2\\) = V\\(_R^2\\) + (V\\(_L\\) - V\\(_C\\))\\(^2\\) \n\n 110\\(^2\\)  = 80\\(^2\\) + (40 – V\\(_C\\))\\(^2\\)\n(40 – V\\(_C\\)) = \\(\\sqrt{(110 + 80) (110 – 80)}\\) \n 40 – V\\(_C\\) = + 75.5\n V\\(_C\\) = 40 + 75.5\n V\\(_C\\) = 115.5V\n  \n(ii) X\\(_c\\) \\(\\frac{V_c}{I} = \\frac{1}{2 \\pi f c}\\)\nC = \\(\\frac{1}{2 \\pi fV_c}\\)\nC \\(\\frac{2}{2 \\times 3.14 \\times 60 \\times 115.5} = 45.9\\muF\\) \n\n(iii) \\(V_L = IX_L = I \\times 2 \\pi fL}\\)\nL = \\(\\frac{V_l}{I \\times 1 \\times 2 \\pifVc}\\)\nC = \\(\\frac{2}{2 \\times 3.142 \\times 60 \\times 2}\\)\n= 0.053H\n~(d)\n\n Vector diagram showing\n V\\(_L\\) with arrow\n V\\(_C\\) with arrow\n V\\(_R\\) with arrow\n\n(i) V\\(^2\\) = V\\(_R^2\\) + (V\\(_L\\) - V\\(_C\\))\\(^2\\) \n\n 110\\(^2\\)  = 80\\(^2\\) + (40 – V\\(_C\\))\\(^2\\)\n(40 – V\\(_C\\)) = \\(\\sqrt{(110 + 80) (110 – 80)}\\) \n 40 – V\\(_C\\) = + 75.5\n V\\(_C\\) = 40 + 75.5\n V\\(_C\\) = 115.5V\n  \n(ii) X\\(_c\\) \\(\\frac{V_c}{I} = \\frac{1}{2 \\pi f c}\\)\nC = \\(\\frac{1}{2 \\pi fV_c}\\)\nC \\(\\frac{2}{2 \\times 3.14 \\times 60 \\times 115.5} = 45.9\\muF\\) \n\n(iii) \\(V_L = IX_L = I \\times 2 \\pi fL}\\)\nL = \\(\\frac{V_l}{I \\times 1 \\times 2 \\pifVc}\\)\nC = \\(\\frac{2}{2 \\times 3.142 \\times 60 \\times 2}\\)\n= 0.053H\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(133, '2017', '\n\nQuestion\n\n(a) Briefly explain the following terms:\n(i) emission line spectra;\n(ii) line absorption spectra.\n(b) Draw a labeled diagram showing the structure of a simple type of photocell and explain its mode of operation.\n(c) State two\n (i) reasons to show that x-rays are waves;\n (ii) uses of x-rays other than in medicine.\n(d) An electron jumps from an energy level of -1.6eV to one of -1.4 eV in an atom. Calculate the energy and wavelength of the emitted radiation. [ h = 6.6 x 10\\(^{-34}\\) Js; c = 3.00 x 10\\(^8\\) ms\\(^{-1}\\); eV = 1.6 x 10\\(^{-19}\\) J ]\n\n', '', '(a) Briefly explain the following terms:\n(i) emission line spectra;\n(ii) line absorption spectra.\n(b) Draw a labeled diagram showing the structure of a simple type of photocell and explain its mode of operation.\n(c) State two\n (i) reasons to show that x-rays are waves;\n (ii) uses of x-rays other than in medicine.\n(d) An electron jumps from an energy level of -1.6eV to one of -1.4 eV in an atom. Calculate the energy and wavelength of the emitted radiation. [ h = 6.6 x 10\\(^{-34}\\) Js; c = 3.00 x 10\\(^8\\) ms\\(^{-1}\\); eV = 1.6 x 10\\(^{-19}\\) J ]', '(a) Briefly explain the following terms:\n(i) emission line spectra;\n(ii) line absorption spectra.~(b) Draw a labeled diagram showing the structure of a simple type of photocell and explain its mode of operation.\n(c) State two\n (i) reasons to show that x-rays are waves;\n (ii) uses of x-rays other than in medicine.\n(d) An electron jumps from an energy level of -1.6eV to one of -1.4 eV in an atom. Calculate the energy and wavelength of the emitted radiation. [ h = 6.6 x 10\\(^{-34}\\) Js; c = 3.00 x 10\\(^8\\) ms\\(^{-1}\\); eV = 1.6 x 10\\(^{-19}\\) J ]~(c) State two\n (i) reasons to show that x-rays are waves;\n (ii) uses of x-rays other than in medicine.\n(d) An electron jumps from an energy level of -1.6eV to one of -1.4 eV in an atom. Calculate the energy and wavelength of the emitted radiation. [ h = 6.6 x 10\\(^{-34}\\) Js; c = 3.00 x 10\\(^8\\) ms\\(^{-1}\\); eV = 1.6 x 10\\(^{-19}\\) J ]~(d) An electron jumps from an energy level of -1.6eV to one of -1.4 eV in an atom. Calculate the energy and wavelength of the emitted radiation. [ h = 6.6 x 10\\(^{-34}\\) Js; c = 3.00 x 10\\(^8\\) ms\\(^{-1}\\); eV = 1.6 x 10\\(^{-19}\\) J ]', NULL, '\n\nExplanation\n(a)(i) Emission line spectra consist of distinct and separate bright lines of definite wavelengths on a dark background.  \nAny valid additional information e.g    \n-They are obtained when light from a luminous source undergoes dispersion and is observed directly.\n- When an electron moves from one energy level to another energy level, line spectra are observed\n - The colour of the spectra is a characteristic of the source.\n(ii) Line absorption spectra are obtained when light passes through a cool gas and certain wavelengths of the light are absorbed. This gives series of dark lines, each corresponding to one of the wavelengths absorbed.   \n(b) Diagram\n\nWhen radiation of appropriate frequency falls on the cathode (with photosensitive surface) electrons are emitted. Because the anode is positive with respect to the cathode, the electrons are attracted to the anode. The electrons flow completes the circuit and current flows.\n\n(c)(i) Wave properties of X-rays\n  - diffraction\n  - interference\n  - reflection\n  - refraction\n  - polarization\n(ii) Uses of X-rays other than in medicine\n\nto study crystal structures\nto detect presence of cracks in welded parts\nin artistic works for determining the authenticity of such works\nat security check points to scan\nX-ray spectroscopy for identifying isotopes of elements.\n\n(d)\n\n ΔE = E\\(_i\\) - E\\(_f\\)  \n  = - 1.6 – (- 10.4)\n  = 8.8 e V or 1.41 x 10\\(^{-18}\\) J\n  E = hf = hc/λ\n  or\n\n  λ = hc/E\n = 6. 6 x 10\\(^{-34}\\) x 3 x10\\(^{8}\\)\n \n 8.8 x 1.6 x 10\\(^{-19}\\)\n\nλ = 1.4 x 10\\(^{-7}\\) m\n\n', '', 'Explanation\n(a)(i) Emission line spectra consist of distinct and separate bright lines of definite wavelengths on a dark background.  \nAny valid additional information e.g    \n-They are obtained when light from a luminous source undergoes dispersion and is observed directly.\n- When an electron moves from one energy level to another energy level, line spectra are observed\n - The colour of the spectra is a characteristic of the source.\n(ii) Line absorption spectra are obtained when light passes through a cool gas and certain wavelengths of the light are absorbed. This gives series of dark lines, each corresponding to one of the wavelengths absorbed.   \n(b) Diagram\n\nWhen radiation of appropriate frequency falls on the cathode (with photosensitive surface) electrons are emitted. Because the anode is positive with respect to the cathode, the electrons are attracted to the anode. The electrons flow completes the circuit and current flows.\n\n(c)(i) Wave properties of X-rays\n  - diffraction\n  - interference\n  - reflection\n  - refraction\n  - polarization\n(ii) Uses of X-rays other than in medicine\n\nto study crystal structures\nto detect presence of cracks in welded parts\nin artistic works for determining the authenticity of such works\nat security check points to scan\nX-ray spectroscopy for identifying isotopes of elements.\n\n(d)\n\n ΔE = E\\(_i\\) - E\\(_f\\)  \n  = - 1.6 – (- 10.4)\n  = 8.8 e V or 1.41 x 10\\(^{-18}\\) J\n  E = hf = hc/λ\n  or\n\n  λ = hc/E\n = 6. 6 x 10\\(^{-34}\\) x 3 x10\\(^{8}\\)\n \n 8.8 x 1.6 x 10\\(^{-19}\\)\n\nλ = 1.4 x 10\\(^{-7}\\) m\n\n', '(a)(i) Emission line spectra consist of distinct and separate bright lines of definite wavelengths on a dark background.  \nAny valid additional information e.g    \n-They are obtained when light from a luminous source undergoes dispersion and is observed directly.\n- When an electron moves from one energy level to another energy level, line spectra are observed\n - The colour of the spectra is a characteristic of the source.\n(ii) Line absorption spectra are obtained when light passes through a cool gas and certain wavelengths of the light are absorbed. This gives series of dark lines, each corresponding to one of the wavelengths absorbed.   ~(b) Diagram\n\nWhen radiation of appropriate frequency falls on the cathode (with photosensitive surface) electrons are emitted. Because the anode is positive with respect to the cathode, the electrons are attracted to the anode. The electrons flow completes the circuit and current flows.\n\n(c)(i) Wave properties of X-rays\n  - diffraction\n  - interference\n  - reflection\n  - refraction\n  - polarization\n(ii) Uses of X-rays other than in medicine\n\nto study crystal structures\nto detect presence of cracks in welded parts\nin artistic works for determining the authenticity of such works\nat security check points to scan\nX-ray spectroscopy for identifying isotopes of elements.\n\n(d)\n\n ΔE = E\\(_i\\) - E\\(_f\\)  \n  = - 1.6 – (- 10.4)\n  = 8.8 e V or 1.41 x 10\\(^{-18}\\) J\n  E = hf = hc/λ\n  or\n\n  λ = hc/E\n = 6. 6 x 10\\(^{-34}\\) x 3 x10\\(^{8}\\)\n \n 8.8 x 1.6 x 10\\(^{-19}\\)\n\nλ = 1.4 x 10\\(^{-7}\\) m\n~(c)(i) Wave properties of X-rays\n  - diffraction\n  - interference\n  - reflection\n  - refraction\n  - polarization\n(ii) Uses of X-rays other than in medicine\n\nto study crystal structures\nto detect presence of cracks in welded parts\nin artistic works for determining the authenticity of such works\nat security check points to scan\nX-ray spectroscopy for identifying isotopes of elements.\n\n(d)\n\n ΔE = E\\(_i\\) - E\\(_f\\)  \n  = - 1.6 – (- 10.4)\n  = 8.8 e V or 1.41 x 10\\(^{-18}\\) J\n  E = hf = hc/λ\n  or\n\n  λ = hc/E\n = 6. 6 x 10\\(^{-34}\\) x 3 x10\\(^{8}\\)\n \n 8.8 x 1.6 x 10\\(^{-19}\\)\n\nλ = 1.4 x 10\\(^{-7}\\) m\n~(d)\n\n ΔE = E\\(_i\\) - E\\(_f\\)  \n  = - 1.6 – (- 10.4)\n  = 8.8 e V or 1.41 x 10\\(^{-18}\\) J\n  E = hf = hc/λ\n  or\n\n  λ = hc/E\n = 6. 6 x 10\\(^{-34}\\) x 3 x10\\(^{8}\\)\n \n 8.8 x 1.6 x 10\\(^{-19}\\)\n\nλ = 1.4 x 10\\(^{-7}\\) m\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(134, '2017', '\n\nQuestion\n\nGive three observations in support of de Broglie\'s assumption that moving particles behave like waves\n\n', 'Give three observations in support of de Broglie\'s assumption that moving particles behave like waves', '', '', NULL, '\n\nExplanation\n- Moving electrons exhibits wavelength/frequency\n- electrons show refraction\n- electrons show reflection\n- electrons show interference\n- electrons show diffraction patterns\n\n', '#NAME?', '', '', 'WAEC-PK-1', 'WAEC-PHY', 1),
(135, '2017', '\n\nQuestion\n\nState three methods of polirizing an unpolarized light\n\n', 'State three methods of polirizing an unpolarized light', '', '', NULL, '\n\nExplanation\nMethod of polarizing an unpolarized light;\n- by reflexion\n- by double refraction\n- selective absorption by crystals\n- polarizing by pile of plates\n- passing unpolarised light through a polariod\n\n', 'Method of polarizing an unpolarized light;\n- by reflexion\n- by double refraction\n- selective absorption by crystals\n- polarizing by pile of plates\n- passing unpolarised light through a polariod\n\n', '', '', 'WAEC-PK-1', 'WAEC-PHY', 1),
(136, '2017', '\n\nQuestion\n\nA wire gradually stretched by loading it until it snaps.\n(a) Sketch a load-extension graph for the wire\n(b) Indicate on the graph the\n(i) elastic limit (E);\n(ii) yield point (Y);\n(iii) breaking point (B).\n\n', '', 'A wire gradually stretched by loading it until it snaps.\n(a) Sketch a load-extension graph for the wire\n(b) Indicate on the graph the\n(i) elastic limit (E);\n(ii) yield point (Y);\n(iii) breaking point (B).', '(a) Sketch a load-extension graph for the wire~(b) Indicate on the graph the\n(i) elastic limit (E);\n(ii) yield point (Y);\n(iii) breaking point (B).', NULL, '\n\nExplanation\n', '', '', '', 'WAEC-PK-1', 'WAEC-PHY', 1),
(137, '2017', '\n\nQuestion\n\nIn an electrolysis experiment, the ammeter records a steady current of 1 A. The mass of copper deposited in 30 minutes is 0.66 g. Calculate the error in the ammeter reading. [ Electrochemical equivalent of copper = 0.00033 g C\\(^{-1}\\)]\n\n', 'In an electrolysis experiment, the ammeter records a steady current of 1 A. The mass of copper deposited in 30 minutes is 0.66 g. Calculate the error in the ammeter reading. [ Electrochemical equivalent of copper = 0.00033 g C\\(^{-1}\\)]', '', '', NULL, '\n\nExplanation\nI = \\(\\frac{M}{zt} = \\frac{0.66}{33 \\times 10^{-5} \\times 30 \\times 60}\\)\n= 1.11A\nError = 1.11 - 1 = 0.11A\n\n', 'I = \\(\\frac{M}{zt} = \\frac{0.66}{33 \\times 10^{-5} \\times 30 \\times 60}\\)\n= 1.11A\nError = 1.11 - 1 = 0.11A\n\n', '', '', 'WAEC-PK-1', 'WAEC-PHY', 1),
(138, '2017', '\n\nQuestion\n\nThe horizontal component of the initial speed of a particle projected at 30° to the horizontal is 50 m\\(^{-1}\\). If the acceleration of free fall due to gravity is 10 ms\\(^{-2}\\), determine its;\n(a) initial speed;\n(b) speed at the maximum height reached.\n\n', '', 'The horizontal component of the initial speed of a particle projected at 30° to the horizontal is 50 m\\(^{-1}\\). If the acceleration of free fall due to gravity is 10 ms\\(^{-2}\\), determine its;\n(a) initial speed;\n(b) speed at the maximum height reached.', '(a) initial speed;~(b) speed at the maximum height reached.', NULL, '\n\nExplanation\n(a) Horizontal component\n\\(U_x = U cos \\theta\\)\nU = \\(\\frac{U_x}{cos0} = \\frac{50}{cos 30^o}\\) = 57.7 m/s\n(b) At maximum height\n\\(v_y = O\\) and \\(V_x = U_x\\) = 50m/s\nV = \\(\\sqrt{V^2_x + V^2_v} = \\sqrt{50^2 + 0^2}\\)\n= 50m/s\n\n', '', 'Explanation\n(a) Horizontal component\n\\(U_x = U cos \\theta\\)\nU = \\(\\frac{U_x}{cos0} = \\frac{50}{cos 30^o}\\) = 57.7 m/s\n(b) At maximum height\n\\(v_y = O\\) and \\(V_x = U_x\\) = 50m/s\nV = \\(\\sqrt{V^2_x + V^2_v} = \\sqrt{50^2 + 0^2}\\)\n= 50m/s\n\n', '(a) Horizontal component\n\\(U_x = U cos \\theta\\)\nU = \\(\\frac{U_x}{cos0} = \\frac{50}{cos 30^o}\\) = 57.7 m/s~(b) At maximum height\n\\(v_y = O\\) and \\(V_x = U_x\\) = 50m/s\nV = \\(\\sqrt{V^2_x + V^2_v} = \\sqrt{50^2 + 0^2}\\)\n= 50m/s\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(139, '2017', '\n\nQuestion\n\nExplain plane polarized light\n\n', 'Explain plane polarized light', '', '', NULL, '\n\nExplanation\nPlane polarized light is a light wave vibrations that occur only in one plane perpendicular to the direction of propagation of the light. Light is a transverse wave, that can be plane polarized by reflection from surfaces.\n\n', 'Plane polarized light is a light wave vibrations that occur only in one plane perpendicular to the direction of propagation of the light. Light is a transverse wave, that can be plane polarized by reflection from surfaces.\n\n', '', '', 'WAEC-PK-1', 'WAEC-PHY', 1),
(140, '2017', '\n\nQuestion\n\nA ball bearing falls through a viscous liquid\n(a) Using a labelled diagram of a tall vessel, show all the forces acting on it\n(b) When will it attain terminal velocity?\n\n', '', 'A ball bearing falls through a viscous liquid\n(a) Using a labelled diagram of a tall vessel, show all the forces acting on it\n(b) When will it attain terminal velocity?', '(a) Using a labelled diagram of a tall vessel, show all the forces acting on it~(b) When will it attain terminal velocity?', NULL, '\n\nExplanation\n(a)\n\n(b) Terminal velocity is attained when W = U + V\n\n', '', 'Explanation\n(a)\n\n(b) Terminal velocity is attained when W = U + V\n\n', '(a)\n~(b) Terminal velocity is attained when W = U + V\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(141, '2017', '\n\nQuestion\n\nUse the diagram to answer question;\n\nCamphor is a chemical that sub-limes and interacts with water reducing the surface tension where it is put. Explain why the toy boat illustrated above would move forward with the camphor placed at the back of it but would be stationary before the placement of the camphor.\n\n', 'Use the diagram to answer question;\n\nCamphor is a chemical that sub-limes and interacts with water reducing the surface tension where it is put. Explain why the toy boat illustrated above would move forward with the camphor placed at the back of it but would be stationary before the placement of the camphor.', '', '', NULL, '\n\nExplanation\nAt the initial state, F\\(_1\\) = F\\(_1\\) . Therefore their resultant equal zero. At the placement of camphor F\\(_1\\) becomes less than F\\(_2\\) . Their resultant (F\\(_2\\) - F\\(_1\\)) makes the toy boat to move forward.\n\n', 'At the initial state, F\\(_1\\) = F\\(_1\\) . Therefore their resultant equal zero. At the placement of camphor F\\(_1\\) becomes less than F\\(_2\\) . Their resultant (F\\(_2\\) - F\\(_1\\)) makes the toy boat to move forward.\n\n', '', '', 'WAEC-PK-1', 'WAEC-PHY', 1),
(142, '2017', '\n\nQuestion\n\nAn electron of charge 1.60 x 10\\(^{-19}\\)C is accelerated under a potential difference of 1.0 x 10\\(^{5}\\) V. Calculate the energy of the electron in joules.\n\n', 'An electron of charge 1.60 x 10\\(^{-19}\\)C is accelerated under a potential difference of 1.0 x 10\\(^{5}\\) V. Calculate the energy of the electron in joules.', '', '', NULL, '\n\nExplanation\nEnergy of electron = eV = 1.60 x 10\\(^{-19}\\) x 1.0 x 10\\(^{5}\\)\n= 1.60 x 1 0\\(^{-14}\\)J\n\n', 'Energy of electron = eV = 1.60 x 10\\(^{-19}\\) x 1.0 x 10\\(^{5}\\)\n= 1.60 x 1 0\\(^{-14}\\)J\n\n', '', '', 'WAEC-PK-1', 'WAEC-PHY', 1),
(143, '2017', '\n\nQuestion\n\na) On what principle does lighting in a fluorescent tube operate?\n(b) State two factors which determine the colour of light produced in a fluorescent tube.\n\n', '', 'a) On what principle does lighting in a fluorescent tube operate?\n(b) State two factors which determine the colour of light produced in a fluorescent tube.', '~(b) State two factors which determine the colour of light produced in a fluorescent tube.', NULL, '\n\nExplanation\n(a) The princle on which lighting in a fluorescent operation is electrical discharge through a gas at low pressure and at high voltage.\n(b) The colour of light produced in a fluorescent tube depends on factors;\n(i) coating inside the tube; and (ii) the nature ype of gas in the tube.\n\n', '', 'Explanation\n(a) The princle on which lighting in a fluorescent operation is electrical discharge through a gas at low pressure and at high voltage.\n(b) The colour of light produced in a fluorescent tube depends on factors;\n(i) coating inside the tube; and (ii) the nature ype of gas in the tube.\n\n', '(a) The princle on which lighting in a fluorescent operation is electrical discharge through a gas at low pressure and at high voltage.~(b) The colour of light produced in a fluorescent tube depends on factors;\n(i) coating inside the tube; and (ii) the nature ype of gas in the tube.\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(144, '2017', '\n\nQuestion\n\n(a) Define electrolysis.\n(b) Classify each of the following substances as an electrolyte or a non-electrolyte\n(i) sugar solution;\n(ii) kerosene;\n(iii) alkaline solution\n(iv) lemon fruit juice.\n\n', '', '(a) Define electrolysis.\n(b) Classify each of the following substances as an electrolyte or a non-electrolyte\n(i) sugar solution;\n(ii) kerosene;\n(iii) alkaline solution\n(iv) lemon fruit juice.', '(a) Define electrolysis.~(b) Classify each of the following substances as an electrolyte or a non-electrolyte\n(i) sugar solution;\n(ii) kerosene;\n(iii) alkaline solution\n(iv) lemon fruit juice.', NULL, '\n\nExplanation\n(a) Electrolysis is the process through which solutions or molten substance is broken down into their component parts by the passage of electric current.\n(b)\n\n\n\n\nElectrolytes\n\n\nNon-electrolytes\n\n\n\n\n- alkaline\n- lemon fruit juice\n\n\n- sugar solution\n- kerosene\n\n\n\n\n\n\n', '', 'Explanation\n(a) Electrolysis is the process through which solutions or molten substance is broken down into their component parts by the passage of electric current.\n(b)\n\n\n\n\nElectrolytes\n\n\nNon-electrolytes\n\n\n\n\n- alkaline\n- lemon fruit juice\n\n\n- sugar solution\n- kerosene\n\n\n\n\n\n\n', '(a) Electrolysis is the process through which solutions or molten substance is broken down into their component parts by the passage of electric current.~(b)\n\n\n\n\nElectrolytes\n\n\nNon-electrolytes\n\n\n\n\n- alkaline\n- lemon fruit juice\n\n\n- sugar solution\n- kerosene\n\n\n\n\n\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(145, '2017', '\n\nQuestion\n\n(a) State Heisenberg\'s Uncertainty Principle.\n(b) State one phenomenon that can only be explained in terms of the wave nature of light.\n\n', '', '(a) State Heisenberg\'s Uncertainty Principle.\n(b) State one phenomenon that can only be explained in terms of the wave nature of light.', '(a) State Heisenberg\'s Uncertainty Principle.~(b) State one phenomenon that can only be explained in terms of the wave nature of light.', NULL, '\n\nExplanation\nHeiscnberg\'s uncertainty principle states that it ipossible to determine with accuracy both the ion and the momentum of a particle simultaneously i.e; \\(\\Delta p.\\Delta x \\geq \\frac{h}{2} \\pi\\) .\n(b) The phenomenon that can only explained in terms of the wave nature of light are ference and diffraction.\n\n', '', 'Explanation\nHeiscnberg\'s uncertainty principle states that it ipossible to determine with accuracy both the ion and the momentum of a particle simultaneously i.e; \\(\\Delta p.\\Delta x \\geq \\frac{h}{2} \\pi\\) .\n(b) The phenomenon that can only explained in terms of the wave nature of light are ference and diffraction.\n\n', '~(b) The phenomenon that can only explained in terms of the wave nature of light are ference and diffraction.\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(146, '2017', '\n\nQuestion\n\n(a) A particle moves on a straight path with an initial speed u and final speed v in time t. Show that the total distance X covered by the particle is given by\n\\(x = ut + \\frac{1}{2}at^2\\)\nwhere a is the magnitude of acceleration\n(b) State;\n(i) Newton\'s second law of motion\n(ii) the principle of conservation of energy\n(iii) the law of floatation.\n(c) Consider a balloon of mass 0.030 kg being inflated with a gas of density 0.54 kg m\\(^{-3}\\). What will be the volume of the balloon when it just begins to rise in air of density 1.29 kg m\\(^{-3}\\)? [ g = 10 ms\\(^{-2}\\)]\n\n', '', '(a) A particle moves on a straight path with an initial speed u and final speed v in time t. Show that the total distance X covered by the particle is given by\n\\(x = ut + \\frac{1}{2}at^2\\)\nwhere a is the magnitude of acceleration\n(b) State;\n(i) Newton\'s second law of motion\n(ii) the principle of conservation of energy\n(iii) the law of floatation.\n(c) Consider a balloon of mass 0.030 kg being inflated with a gas of density 0.54 kg m\\(^{-3}\\). What will be the volume of the balloon when it just begins to rise in air of density 1.29 kg m\\(^{-3}\\)? [ g = 10 ms\\(^{-2}\\)]', '(a) A particle moves on a straight path with an initial speed u and final speed v in time t. Show that the total distance X covered by the particle is given by\n\\(x = ut + \\frac{1}{2}at^2\\)\nwhere a is the magnitude of acceleration~(b) State;\n(i) Newton\'s second law of motion\n(ii) the principle of conservation of energy\n(iii) the law of floatation.\n(c) Consider a balloon of mass 0.030 kg being inflated with a gas of density 0.54 kg m\\(^{-3}\\). What will be the volume of the balloon when it just begins to rise in air of density 1.29 kg m\\(^{-3}\\)? [ g = 10 ms\\(^{-2}\\)]~(c) Consider a balloon of mass 0.030 kg being inflated with a gas of density 0.54 kg m\\(^{-3}\\). What will be the volume of the balloon when it just begins to rise in air of density 1.29 kg m\\(^{-3}\\)? [ g = 10 ms\\(^{-2}\\)]', NULL, '\n\nExplanation\na) Average velocity = \\(\\frac{u + u}{2}\\)\nFrom equation V = u + at\nAeverage velocity = \\(\\frac{u + u + at}{2}\\) = \\(u + \\frac{1}{2} at\\)\nDistance covered = Ave. vel. x time\nx = (u + \\(\\frac{1}{2}\\)at)t\nx = ut +s\n(b)i) Newton\'s second law of motion states that the of change of momentum is proportional to the lied force and takes place in tho direction of the force.\n(ii) The principle of conservation of energy es that energy is neither created nor destroyed can be changed from one form to another. The law of floatation states that an object wil t in a fluid when the upthrust exerted upon it by fluid in which it floats equals the weight of the\n(c) Mass of air displaced = mass of ballon on and contents V x density of A air\n= V x density of gas + content\nV x 1.29 = v x 0.54 + 0.030\nV = \\(\\frac{0.03}{0.75}\\) = 0.04m\\(^3}\\)\n\n', '', 'Explanation\na) Average velocity = \\(\\frac{u + u}{2}\\)\nFrom equation V = u + at\nAeverage velocity = \\(\\frac{u + u + at}{2}\\) = \\(u + \\frac{1}{2} at\\)\nDistance covered = Ave. vel. x time\nx = (u + \\(\\frac{1}{2}\\)at)t\nx = ut +s\n(b)i) Newton\'s second law of motion states that the of change of momentum is proportional to the lied force and takes place in tho direction of the force.\n(ii) The principle of conservation of energy es that energy is neither created nor destroyed can be changed from one form to another. The law of floatation states that an object wil t in a fluid when the upthrust exerted upon it by fluid in which it floats equals the weight of the\n(c) Mass of air displaced = mass of ballon on and contents V x density of A air\n= V x density of gas + content\nV x 1.29 = v x 0.54 + 0.030\nV = \\(\\frac{0.03}{0.75}\\) = 0.04m\\(^3}\\)\n\n', 'a) Average velocity = \\(\\frac{u + u}{2}\\)\nFrom equation V = u + at\nAeverage velocity = \\(\\frac{u + u + at}{2}\\) = \\(u + \\frac{1}{2} at\\)\nDistance covered = Ave. vel. x time\nx = (u + \\(\\frac{1}{2}\\)at)t\nx = ut +s~(b)i) Newton\'s second law of motion states that the of change of momentum is proportional to the lied force and takes place in tho direction of the force.\n(ii) The principle of conservation of energy es that energy is neither created nor destroyed can be changed from one form to another. The law of floatation states that an object wil t in a fluid when the upthrust exerted upon it by fluid in which it floats equals the weight of the\n(c) Mass of air displaced = mass of ballon on and contents V x density of A air\n= V x density of gas + content\nV x 1.29 = v x 0.54 + 0.030\nV = \\(\\frac{0.03}{0.75}\\) = 0.04m\\(^3}\\)\n~(c) Mass of air displaced = mass of ballon on and contents V x density of A air\n= V x density of gas + content\nV x 1.29 = v x 0.54 + 0.030\nV = \\(\\frac{0.03}{0.75}\\) = 0.04m\\(^3}\\)\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(147, '2017', '\n\nQuestion\n\n(a) State two differences between boiling and evaporation.\n(b) A closed inexpansibie vessel contains air saturated with water vapour at 77°C. The total pressure in the vessel is 1007 mmHg. Calculate the new pressure in the vessel if the temperature is reduced 1.2 27°C. [ The s.v.p. of water at 77°C and 27°C respectively are 314 mmHg and 27 mmHg. Treat the air in the vessel as an ideal gas]\n(c) The lengtn of a zinc rod at 23°C is 200 m. Calculate the increase in length of the rod when its temperature rises to 33°C. [ expansivity of zinc = 2.6 x 10\\(^{-5}\\) K\\(^{-1}\\)]\n(d) Explain with there is no temperature change when a solid being heated changes into liquid at its melting point.\n\n', '', '(a) State two differences between boiling and evaporation.\n(b) A closed inexpansibie vessel contains air saturated with water vapour at 77°C. The total pressure in the vessel is 1007 mmHg. Calculate the new pressure in the vessel if the temperature is reduced 1.2 27°C. [ The s.v.p. of water at 77°C and 27°C respectively are 314 mmHg and 27 mmHg. Treat the air in the vessel as an ideal gas]\n(c) The lengtn of a zinc rod at 23°C is 200 m. Calculate the increase in length of the rod when its temperature rises to 33°C. [ expansivity of zinc = 2.6 x 10\\(^{-5}\\) K\\(^{-1}\\)]\n(d) Explain with there is no temperature change when a solid being heated changes into liquid at its melting point.', '(a) State two differences between boiling and evaporation.~(b) A closed inexpansibie vessel contains air saturated with water vapour at 77°C. The total pressure in the vessel is 1007 mmHg. Calculate the new pressure in the vessel if the temperature is reduced 1.2 27°C. [ The s.v.p. of water at 77°C and 27°C respectively are 314 mmHg and 27 mmHg. Treat the air in the vessel as an ideal gas]\n(c) The lengtn of a zinc rod at 23°C is 200 m. Calculate the increase in length of the rod when its temperature rises to 33°C. [ expansivity of zinc = 2.6 x 10\\(^{-5}\\) K\\(^{-1}\\)]\n(d) Explain with there is no temperature change when a solid being heated changes into liquid at its melting point.~(c) The lengtn of a zinc rod at 23°C is 200 m. Calculate the increase in length of the rod when its temperature rises to 33°C. [ expansivity of zinc = 2.6 x 10\\(^{-5}\\) K\\(^{-1}\\)]\n(d) Explain with there is no temperature change when a solid being heated changes into liquid at its melting point.~(d) Explain with there is no temperature change when a solid being heated changes into liquid at its melting point.', NULL, '\n\nExplanation\na)\n\n\n\n\nBoiling\n\n\nEvaporation\n\n\n\n\nTakes place at a particular temperature\nTakes place through-out the whole volume of the liquid.\nTakes place at a steady temperature\n\n\n\nTakes place at all temperatures\nTakes place only on the surface of the liquid.\nCan take place at reduced temperature.\n\n\n\n\n\n\n(b) \\(P_1 = 1007 - 314 = 693 mmH_g\\)\n\\(\\frac{P_1}{T_1} - \\frac{P_2}{T_2} = \\frac{693}{350} = \\frac{P_2}{300}\\)\n\\(p_2 = \\frac{963}{350} \\times \\frac{300}{1}\\) = 594 mmH\\(_g\\)\nPressure in vessel at 27°C = pressure of air + S.V.P of water at 27°C\n= 594 + 27 = 621mm Hg\n\n(c) \\(\\alpha = \\frac{L_2 - L_1}{L_1(\\theta_1 - \\theta_2)}\\)\n= 2.6 x 10-5 x 200 (33-23) = 0.052m\n\n(d) At the solid state the molecules are tightly bounded, therefore at the melting point the heat supplied is absorbed and used to break the bonds holding the molecules together.\n\n', '', 'Explanation\na)\n\n\n\n\nBoiling\n\n\nEvaporation\n\n\n\n\nTakes place at a particular temperature\nTakes place through-out the whole volume of the liquid.\nTakes place at a steady temperature\n\n\n\nTakes place at all temperatures\nTakes place only on the surface of the liquid.\nCan take place at reduced temperature.\n\n\n\n\n\n\n(b) \\(P_1 = 1007 - 314 = 693 mmH_g\\)\n\\(\\frac{P_1}{T_1} - \\frac{P_2}{T_2} = \\frac{693}{350} = \\frac{P_2}{300}\\)\n\\(p_2 = \\frac{963}{350} \\times \\frac{300}{1}\\) = 594 mmH\\(_g\\)\nPressure in vessel at 27°C = pressure of air + S.V.P of water at 27°C\n= 594 + 27 = 621mm Hg\n\n(c) \\(\\alpha = \\frac{L_2 - L_1}{L_1(\\theta_1 - \\theta_2)}\\)\n= 2.6 x 10-5 x 200 (33-23) = 0.052m\n\n(d) At the solid state the molecules are tightly bounded, therefore at the melting point the heat supplied is absorbed and used to break the bonds holding the molecules together.\n\n', 'a)\n\n\n\n\nBoiling\n\n\nEvaporation\n\n\n\n\nTakes place at a particular temperature\nTakes place through-out the whole volume of the liquid.\nTakes place at a steady temperature\n\n\n\nTakes place at all temperatures\nTakes place only on the surface of the liquid.\nCan take place at reduced temperature.\n\n\n\n\n\n~(b) \\(P_1 = 1007 - 314 = 693 mmH_g\\)\n\\(\\frac{P_1}{T_1} - \\frac{P_2}{T_2} = \\frac{693}{350} = \\frac{P_2}{300}\\)\n\\(p_2 = \\frac{963}{350} \\times \\frac{300}{1}\\) = 594 mmH\\(_g\\)\nPressure in vessel at 27°C = pressure of air + S.V.P of water at 27°C\n= 594 + 27 = 621mm Hg\n\n(c) \\(\\alpha = \\frac{L_2 - L_1}{L_1(\\theta_1 - \\theta_2)}\\)\n= 2.6 x 10-5 x 200 (33-23) = 0.052m\n\n(d) At the solid state the molecules are tightly bounded, therefore at the melting point the heat supplied is absorbed and used to break the bonds holding the molecules together.\n~(c) \\(\\alpha = \\frac{L_2 - L_1}{L_1(\\theta_1 - \\theta_2)}\\)\n= 2.6 x 10-5 x 200 (33-23) = 0.052m\n\n(d) At the solid state the molecules are tightly bounded, therefore at the melting point the heat supplied is absorbed and used to break the bonds holding the molecules together.\n~(d) At the solid state the molecules are tightly bounded, therefore at the melting point the heat supplied is absorbed and used to break the bonds holding the molecules together.\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(148, '2017', '\n\nQuestion\n\n(a) The equation y = \\(\\alpha\\) sin (wt - kx) represents a plane wave travelling in a medium along the x-direction, being the displacement at the point x at time t.\n(i) Given that x is in metres and t is in seconds, state the units of k and w\n(ii) What physical quantity does \\(\\frac{W}{k}\\) represent? Justify your answer\n(iii) State whether the wave is travelling in the positive or negative x-direction.\n\n(b)(i) What are beats?\n(ii) A sonometer wire has a frequency of 259 Hz. It is sounded alongside a tuning fork of frequency 256 Hz. Calculate the beat frequency\n(iii) The sonometer wire in (b)(ii) above is under a tension of 1200 N. If a metre of the wire has a mass of 0.03 kg, calculate the length l of the wire when it is vibrating in the fundamental mode.\n\n(c) List two similarities between the human eye and the photographic camera.\n\n', '', '(a) The equation y = \\(\\alpha\\) sin (wt - kx) represents a plane wave travelling in a medium along the x-direction, being the displacement at the point x at time t.\n(i) Given that x is in metres and t is in seconds, state the units of k and w\n(ii) What physical quantity does \\(\\frac{W}{k}\\) represent? Justify your answer\n(iii) State whether the wave is travelling in the positive or negative x-direction.\n\n(b)(i) What are beats?\n(ii) A sonometer wire has a frequency of 259 Hz. It is sounded alongside a tuning fork of frequency 256 Hz. Calculate the beat frequency\n(iii) The sonometer wire in (b)(ii) above is under a tension of 1200 N. If a metre of the wire has a mass of 0.03 kg, calculate the length l of the wire when it is vibrating in the fundamental mode.\n\n(c) List two similarities between the human eye and the photographic camera.', '(a) The equation y = \\(\\alpha\\) sin (wt - kx) represents a plane wave travelling in a medium along the x-direction, being the displacement at the point x at time t.\n(i) Given that x is in metres and t is in seconds, state the units of k and w\n(ii) What physical quantity does \\(\\frac{W}{k}\\) represent? Justify your answer\n(iii) State whether the wave is travelling in the positive or negative x-direction.\n~(b)(i) What are beats?\n(ii) A sonometer wire has a frequency of 259 Hz. It is sounded alongside a tuning fork of frequency 256 Hz. Calculate the beat frequency\n(iii) The sonometer wire in (b)(ii) above is under a tension of 1200 N. If a metre of the wire has a mass of 0.03 kg, calculate the length l of the wire when it is vibrating in the fundamental mode.\n\n(c) List two similarities between the human eye and the photographic camera.~(c) List two similarities between the human eye and the photographic camera.', NULL, '\n\nExplanation\n(a)(i) K is measured in m\\(^{-1}\\) and w rad s\\(^{-1}\\)\n(ii) \\(\\frac{w}{k}\\) represents speed of wave\nw = 2\\(\\pi\\)f; and\nk = \\(\\frac{2 \\pi}{\\pi}\\)\n\\(\\frac{w}{k} \\frac{2 \\pi f}{2 \\pi/\\lambda} = f\\lambda\\)\nBut V = f\\(\\lambda\\)\n(iii) The direction of travel of the wave :s in the positive x - direction.\n\n(b)(i) Beats are periodic rise and fall in amplitude of the sound produced when two notes of nearly equal frequencies are sounded together.\n(ii) Beat frequency = \\(f_1 - f_2\\)\n= 259 - 256 = 3Hz\n(iii) f = \\(\\frac{1}{2L} \\sqrt{T}{m}\\)\n259 = \\(\\frac{1}{2L} \\sqrt{\\frac{1200}{0.03}}\\)\nL = 0.386m\n\n(c) 1. Both have converging or convex lenses.\n2. Their interiors are black.\n3. Both form real, inverted, and diminished image.\n4. Images of objects are formed on the retina of the eye and the film of the camera.\n\n', '', 'Explanation\n(a)(i) K is measured in m\\(^{-1}\\) and w rad s\\(^{-1}\\)\n(ii) \\(\\frac{w}{k}\\) represents speed of wave\nw = 2\\(\\pi\\)f; and\nk = \\(\\frac{2 \\pi}{\\pi}\\)\n\\(\\frac{w}{k} \\frac{2 \\pi f}{2 \\pi/\\lambda} = f\\lambda\\)\nBut V = f\\(\\lambda\\)\n(iii) The direction of travel of the wave :s in the positive x - direction.\n\n(b)(i) Beats are periodic rise and fall in amplitude of the sound produced when two notes of nearly equal frequencies are sounded together.\n(ii) Beat frequency = \\(f_1 - f_2\\)\n= 259 - 256 = 3Hz\n(iii) f = \\(\\frac{1}{2L} \\sqrt{T}{m}\\)\n259 = \\(\\frac{1}{2L} \\sqrt{\\frac{1200}{0.03}}\\)\nL = 0.386m\n\n(c) 1. Both have converging or convex lenses.\n2. Their interiors are black.\n3. Both form real, inverted, and diminished image.\n4. Images of objects are formed on the retina of the eye and the film of the camera.\n\n', '(a)(i) K is measured in m\\(^{-1}\\) and w rad s\\(^{-1}\\)\n(ii) \\(\\frac{w}{k}\\) represents speed of wave\nw = 2\\(\\pi\\)f; and\nk = \\(\\frac{2 \\pi}{\\pi}\\)\n\\(\\frac{w}{k} \\frac{2 \\pi f}{2 \\pi/\\lambda} = f\\lambda\\)\nBut V = f\\(\\lambda\\)\n(iii) The direction of travel of the wave :s in the positive x - direction.\n~(b)(i) Beats are periodic rise and fall in amplitude of the sound produced when two notes of nearly equal frequencies are sounded together.\n(ii) Beat frequency = \\(f_1 - f_2\\)\n= 259 - 256 = 3Hz\n(iii) f = \\(\\frac{1}{2L} \\sqrt{T}{m}\\)\n259 = \\(\\frac{1}{2L} \\sqrt{\\frac{1200}{0.03}}\\)\nL = 0.386m\n\n(c) 1. Both have converging or convex lenses.\n2. Their interiors are black.\n3. Both form real, inverted, and diminished image.\n4. Images of objects are formed on the retina of the eye and the film of the camera.\n~(c) 1. Both have converging or convex lenses.\n2. Their interiors are black.\n3. Both form real, inverted, and diminished image.\n4. Images of objects are formed on the retina of the eye and the film of the camera.\n', 'WAEC-PK-1', 'WAEC-PHY', 1);
INSERT INTO `exam_simulation2` (`id`, `examYear`, `questions`, `parent_que`, `child_que`, `gen_child_que_desc`, `imageDir`, `explanation`, `parent_que_expl`, `child_que_exp`, `gen_child_que_expl`, `package_id`, `subject_id`, `status`) VALUES
(149, '2017', '\n\nQuestion\n\n(a) What is a magnetic field?\n(b) With the aid of a labelled diagram describe an experiment to show that a magnetic field exists around a straight wire carrying current.\n(c) A 40 \\(\\mu\\)F capacitor in series with a 40 Q resistor is connected to a 100 V, 50 Hz a.c. supply.\n(i) Draw a circuit diagram of the arrangement\n(ii) Calculate the: I. impedance in the circuit; II. current in the circuit III. potential difference across the capacitor.\n\n', '', '(a) What is a magnetic field?\n(b) With the aid of a labelled diagram describe an experiment to show that a magnetic field exists around a straight wire carrying current.\n(c) A 40 \\(\\mu\\)F capacitor in series with a 40 Q resistor is connected to a 100 V, 50 Hz a.c. supply.\n(i) Draw a circuit diagram of the arrangement\n(ii) Calculate the: I. impedance in the circuit; II. current in the circuit III. potential difference across the capacitor.', '(a) What is a magnetic field?~(b) With the aid of a labelled diagram describe an experiment to show that a magnetic field exists around a straight wire carrying current.\n(c) A 40 \\(\\mu\\)F capacitor in series with a 40 Q resistor is connected to a 100 V, 50 Hz a.c. supply.\n(i) Draw a circuit diagram of the arrangement\n(ii) Calculate the: I. impedance in the circuit; II. current in the circuit III. potential difference across the capacitor.~(c) A 40 \\(\\mu\\)F capacitor in series with a 40 Q resistor is connected to a 100 V, 50 Hz a.c. supply.\n(i) Draw a circuit diagram of the arrangement\n(ii) Calculate the: I. impedance in the circuit; II. current in the circuit III. potential difference across the capacitor.', NULL, '\n\nExplanation\n(a) A magnetic field is a region or space where a magnetic force is experienced.\n(b)\n\n\nConnect a straight wire to the circuit as shown above, sprinkle iron fillings on the cardboard paper round the straight wire; and tap gently. Then close the key and gently tap again, the iron fillings form a concentric circles round the wire. This shows that there is magnetic field round a current carrying conductor.\n(c)(i)\n\n\\(x_c = \\frac{1}{2 \\pi fc} = \\frac{1}{2 \\pi \\times 50(40 \\times 10^{-6})}\\)\n= 79.567\\(\\Omega\\)\nz = \\(\\sqrt{R^2 + X^2_c} = \\sqrt{40^2 + 79,580^2}\\)\n(ii) I = \\(\\frac{v}{2} = \\frac{100}{89.1} = 1.12A\\)\n(iii) V = IX\\(_c\\) = 1.12 x 79.567\n= 89.1V\n\n', '', 'Explanation\n(a) A magnetic field is a region or space where a magnetic force is experienced.\n(b)\n\n\nConnect a straight wire to the circuit as shown above, sprinkle iron fillings on the cardboard paper round the straight wire; and tap gently. Then close the key and gently tap again, the iron fillings form a concentric circles round the wire. This shows that there is magnetic field round a current carrying conductor.\n(c)(i)\n\n\\(x_c = \\frac{1}{2 \\pi fc} = \\frac{1}{2 \\pi \\times 50(40 \\times 10^{-6})}\\)\n= 79.567\\(\\Omega\\)\nz = \\(\\sqrt{R^2 + X^2_c} = \\sqrt{40^2 + 79,580^2}\\)\n(ii) I = \\(\\frac{v}{2} = \\frac{100}{89.1} = 1.12A\\)\n(iii) V = IX\\(_c\\) = 1.12 x 79.567\n= 89.1V\n\n', '(a) A magnetic field is a region or space where a magnetic force is experienced.~(b)\n\n\nConnect a straight wire to the circuit as shown above, sprinkle iron fillings on the cardboard paper round the straight wire; and tap gently. Then close the key and gently tap again, the iron fillings form a concentric circles round the wire. This shows that there is magnetic field round a current carrying conductor.\n(c)(i)\n\n\\(x_c = \\frac{1}{2 \\pi fc} = \\frac{1}{2 \\pi \\times 50(40 \\times 10^{-6})}\\)\n= 79.567\\(\\Omega\\)\nz = \\(\\sqrt{R^2 + X^2_c} = \\sqrt{40^2 + 79,580^2}\\)\n(ii) I = \\(\\frac{v}{2} = \\frac{100}{89.1} = 1.12A\\)\n(iii) V = IX\\(_c\\) = 1.12 x 79.567\n= 89.1V\n~(c)(i)\n\n\\(x_c = \\frac{1}{2 \\pi fc} = \\frac{1}{2 \\pi \\times 50(40 \\times 10^{-6})}\\)\n= 79.567\\(\\Omega\\)\nz = \\(\\sqrt{R^2 + X^2_c} = \\sqrt{40^2 + 79,580^2}\\)\n(ii) I = \\(\\frac{v}{2} = \\frac{100}{89.1} = 1.12A\\)\n(iii) V = IX\\(_c\\) = 1.12 x 79.567\n= 89.1V\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(150, '2017', '\n\nQuestion\n\n(a)(i) State Einstein\'s equation of photoelectric effect\n(ii) What conservation principle does the equation represent?\n(b) List three applications of photocells\n(c) A photo emissive surface has a threshold frequency of 4.02 x 10\\(^{14}\\)Hz. If the surface is illuminated by light of frequency 5.0 x 10\\(^{15}\\)Hz, calculate the:\n(i) threshold wavelength;\n(ii) work function;\n(iii) kinetic energy of the emitted photoelectrons. [ c = 3.0 x 10\\(^{8}\\) ms\\(^{-1}\\), h = 6.63 x 10\\(^{-34}\\) Js]\n\n', '', '(a)(i) State Einstein\'s equation of photoelectric effect\n(ii) What conservation principle does the equation represent?\n(b) List three applications of photocells\n(c) A photo emissive surface has a threshold frequency of 4.02 x 10\\(^{14}\\)Hz. If the surface is illuminated by light of frequency 5.0 x 10\\(^{15}\\)Hz, calculate the:\n(i) threshold wavelength;\n(ii) work function;\n(iii) kinetic energy of the emitted photoelectrons. [ c = 3.0 x 10\\(^{8}\\) ms\\(^{-1}\\), h = 6.63 x 10\\(^{-34}\\) Js]', '(a)(i) State Einstein\'s equation of photoelectric effect\n(ii) What conservation principle does the equation represent?~(b) List three applications of photocells\n(c) A photo emissive surface has a threshold frequency of 4.02 x 10\\(^{14}\\)Hz. If the surface is illuminated by light of frequency 5.0 x 10\\(^{15}\\)Hz, calculate the:\n(i) threshold wavelength;\n(ii) work function;\n(iii) kinetic energy of the emitted photoelectrons. [ c = 3.0 x 10\\(^{8}\\) ms\\(^{-1}\\), h = 6.63 x 10\\(^{-34}\\) Js]~(c) A photo emissive surface has a threshold frequency of 4.02 x 10\\(^{14}\\)Hz. If the surface is illuminated by light of frequency 5.0 x 10\\(^{15}\\)Hz, calculate the:\n(i) threshold wavelength;\n(ii) work function;\n(iii) kinetic energy of the emitted photoelectrons. [ c = 3.0 x 10\\(^{8}\\) ms\\(^{-1}\\), h = 6.63 x 10\\(^{-34}\\) Js]', NULL, '\n\nExplanation\n(a)(i) E = W + K.E, where\nE = energy of incident radiation\nW = work function of metal\nK.E. = Kinetic energy of ejected electrons\n(ii) The equation represents the principle of conservation of energy.\n\n(b) Applications of photocells:\n1. for burglar alarm\n2. for television cameras\n3. automatic doors\n4. automatic street lighting\n5. automatic digital counters.\n\n(c)(i) V = \\(f_o\\lambda_o\\)\n\\(\\lambda_o = \\frac{v}{f_o} = \\frac{3 \\times10^8}{4.02 \\times 10^{14}}\\)\n= 7.46 x 10\\(^{-7}\\)m\n(ii) W = \\(hf_o = 6.63 \\times 10^{-34} \\times 4.02 \\times 10^{14}\\)\n= 2.67 x 10\\(^{-19}\\)J\n(iii) K.E. = \\(h (f - f_o) = 6.63 \\times10^{-34} (5.0 \\times 10^{15} - 4.02 \\times 10^{14})\\)\n= 3.05 x 10\\(^{-15}\\)J\n\n', '', 'Explanation\n(a)(i) E = W + K.E, where\nE = energy of incident radiation\nW = work function of metal\nK.E. = Kinetic energy of ejected electrons\n(ii) The equation represents the principle of conservation of energy.\n\n(b) Applications of photocells:\n1. for burglar alarm\n2. for television cameras\n3. automatic doors\n4. automatic street lighting\n5. automatic digital counters.\n\n(c)(i) V = \\(f_o\\lambda_o\\)\n\\(\\lambda_o = \\frac{v}{f_o} = \\frac{3 \\times10^8}{4.02 \\times 10^{14}}\\)\n= 7.46 x 10\\(^{-7}\\)m\n(ii) W = \\(hf_o = 6.63 \\times 10^{-34} \\times 4.02 \\times 10^{14}\\)\n= 2.67 x 10\\(^{-19}\\)J\n(iii) K.E. = \\(h (f - f_o) = 6.63 \\times10^{-34} (5.0 \\times 10^{15} - 4.02 \\times 10^{14})\\)\n= 3.05 x 10\\(^{-15}\\)J\n\n', '(a)(i) E = W + K.E, where\nE = energy of incident radiation\nW = work function of metal\nK.E. = Kinetic energy of ejected electrons\n(ii) The equation represents the principle of conservation of energy.\n~(b) Applications of photocells:\n1. for burglar alarm\n2. for television cameras\n3. automatic doors\n4. automatic street lighting\n5. automatic digital counters.\n\n(c)(i) V = \\(f_o\\lambda_o\\)\n\\(\\lambda_o = \\frac{v}{f_o} = \\frac{3 \\times10^8}{4.02 \\times 10^{14}}\\)\n= 7.46 x 10\\(^{-7}\\)m\n(ii) W = \\(hf_o = 6.63 \\times 10^{-34} \\times 4.02 \\times 10^{14}\\)\n= 2.67 x 10\\(^{-19}\\)J\n(iii) K.E. = \\(h (f - f_o) = 6.63 \\times10^{-34} (5.0 \\times 10^{15} - 4.02 \\times 10^{14})\\)\n= 3.05 x 10\\(^{-15}\\)J\n~(c)(i) V = \\(f_o\\lambda_o\\)\n\\(\\lambda_o = \\frac{v}{f_o} = \\frac{3 \\times10^8}{4.02 \\times 10^{14}}\\)\n= 7.46 x 10\\(^{-7}\\)m\n(ii) W = \\(hf_o = 6.63 \\times 10^{-34} \\times 4.02 \\times 10^{14}\\)\n= 2.67 x 10\\(^{-19}\\)J\n(iii) K.E. = \\(h (f - f_o) = 6.63 \\times10^{-34} (5.0 \\times 10^{15} - 4.02 \\times 10^{14})\\)\n= 3.05 x 10\\(^{-15}\\)J\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(151, '2017', '\n\nQuestion\n\nA particle is projected at an angle of 30° to the horizontal with a speed of 250 ms\\(^{-1}\\). Calculate the;\n(a) total time of flight of the particle;\n(b) speed of the particle at its maximum height. [ g = 10 ms\\(^{-2}\\)]\n\n', '', 'A particle is projected at an angle of 30° to the horizontal with a speed of 250 ms\\(^{-1}\\). Calculate the;\n(a) total time of flight of the particle;\n(b) speed of the particle at its maximum height. [ g = 10 ms\\(^{-2}\\)]', '(a) total time of flight of the particle;~(b) speed of the particle at its maximum height. [ g = 10 ms\\(^{-2}\\)]', NULL, '\n\nExplanation\n(a) T = \\(\\frac{2Usin\\theta}{g} = \\frac{2 \\times 250 \\times sin 30}{10}\\)\n= 25s\n(b) Speed at h\\(_{\\text{max}} Ucos \\theta\\)\n= 250 cos 30\\(^o\\) = 216m/s\n\n', '', 'Explanation\n(a) T = \\(\\frac{2Usin\\theta}{g} = \\frac{2 \\times 250 \\times sin 30}{10}\\)\n= 25s\n(b) Speed at h\\(_{\\text{max}} Ucos \\theta\\)\n= 250 cos 30\\(^o\\) = 216m/s\n\n', '(a) T = \\(\\frac{2Usin\\theta}{g} = \\frac{2 \\times 250 \\times sin 30}{10}\\)\n= 25s~(b) Speed at h\\(_{\\text{max}} Ucos \\theta\\)\n= 250 cos 30\\(^o\\) = 216m/s\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(152, '2017', '\n\nQuestion\n\nState;\n(a) the difference between plane polarized light and ordinary light;\n(b) two uses of polaroids.\n\n', '', 'State;\n(a) the difference between plane polarized light and ordinary light;\n(b) two uses of polaroids.', '(a) the difference between plane polarized light and ordinary light;~(b) two uses of polaroids.', NULL, '\n\nExplanation\n(a) In plane polarized light the vibration of the particles are confirmed to a single plane while in ordinary light, the particles vibrate in all planes perpendicular to the direction of propagation of the wave.\n(b) Uses of polaroids are:\n— stress analysis\n— measurement of concentration of sugar solution\n— used to eliminate sun glare\n— used in sun glasses to reduce the intensity of sunlight\n\n', '', 'Explanation\n(a) In plane polarized light the vibration of the particles are confirmed to a single plane while in ordinary light, the particles vibrate in all planes perpendicular to the direction of propagation of the wave.\n(b) Uses of polaroids are:\n— stress analysis\n— measurement of concentration of sugar solution\n— used to eliminate sun glare\n— used in sun glasses to reduce the intensity of sunlight\n\n', '(a) In plane polarized light the vibration of the particles are confirmed to a single plane while in ordinary light, the particles vibrate in all planes perpendicular to the direction of propagation of the wave.~(b) Uses of polaroids are:\n— stress analysis\n— measurement of concentration of sugar solution\n— used to eliminate sun glare\n— used in sun glasses to reduce the intensity of sunlight\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(153, '2017', '\n\nQuestion\n\nDefine\n(a) electrolysis;\n(b) electrolyte;\n(c) electrode.\n\n', '', 'Define\n(a) electrolysis;\n(b) electrolyte;\n(c) electrode.', '(a) electrolysis;~(b) electrolyte;\n(c) electrode.~(c) electrode.', NULL, '\n\nExplanation\n(a) Electrolysis is the process of decomposing certain solutions or liquids or molten substances into their component parts by the passage of electricity through them.\n(b) Electrolyte is the solution or liquid or molten substance which will conduct electricity and which can be decomposed into its component parts during electrolysis.\n(c) Electrodes are the conductors in contact with an electrolyte through which current enters or leaves the electrolyte.\n\n\n', '', 'Explanation\n(a) Electrolysis is the process of decomposing certain solutions or liquids or molten substances into their component parts by the passage of electricity through them.\n(b) Electrolyte is the solution or liquid or molten substance which will conduct electricity and which can be decomposed into its component parts during electrolysis.\n(c) Electrodes are the conductors in contact with an electrolyte through which current enters or leaves the electrolyte.\n\n\n', '(a) Electrolysis is the process of decomposing certain solutions or liquids or molten substances into their component parts by the passage of electricity through them.~(b) Electrolyte is the solution or liquid or molten substance which will conduct electricity and which can be decomposed into its component parts during electrolysis.\n(c) Electrodes are the conductors in contact with an electrolyte through which current enters or leaves the electrolyte.\n\n~(c) Electrodes are the conductors in contact with an electrolyte through which current enters or leaves the electrolyte.\n\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(154, '2017', '\n\nQuestion\n\nCopper of thickness d is plated on the cathode of a copper voltameter. If the total surface area of the cathode is 60 cm\\(^2\\) and a steady current of 5.0 A is maintained in the voltameter for 1 hour. calculate the value of d. [density of copper = 8.9 x 10\\(^3\\)kg m\\(^{-3}\\) electro chemical equivalent of copper = 3.3 x 10\\(^{-7}\\) kg C\\(^{-1}\\)\n\n', 'Copper of thickness d is plated on the cathode of a copper voltameter. If the total surface area of the cathode is 60 cm\\(^2\\) and a steady current of 5.0 A is maintained in the voltameter for 1 hour. calculate the value of d. [density of copper = 8.9 x 10\\(^3\\)kg m\\(^{-3}\\) electro chemical equivalent of copper = 3.3 x 10\\(^{-7}\\) kg C\\(^{-1}\\)', '', '', NULL, '\n\nExplanation\nMass of Cu deposited = Area x thickness x density = electrochemical equivalent x current x time.\nThickness = \\(\\frac{\\text{electrochemical equivalent} \\times \\text{current} \\times \\text{time}}{Area \\times density}\\)\n= \\(\\frac{3.3 \\times 10^{-7} \\times 5 \\times 3600}{60 \\times 10^{-4} \\times 8.9 \\times 10^{3}}\\)\n= 1.1 x 10\\(^{-4}\\) m\n\n', 'Mass of Cu deposited = Area x thickness x density = electrochemical equivalent x current x time.\nThickness = \\(\\frac{\\text{electrochemical equivalent} \\times \\text{current} \\times \\text{time}}{Area \\times density}\\)\n= \\(\\frac{3.3 \\times 10^{-7} \\times 5 \\times 3600}{60 \\times 10^{-4} \\times 8.9 \\times 10^{3}}\\)\n= 1.1 x 10\\(^{-4}\\) m\n\n', '', '', 'WAEC-PK-1', 'WAEC-PHY', 1),
(155, '2017', '\n\nQuestion\n\n(a) Explain thermiopic emission\n(b) State two applications of electrical conduction through gases.\n\n', '', '(a) Explain thermiopic emission\n(b) State two applications of electrical conduction through gases.', '(a) Explain thermiopic emission~(b) State two applications of electrical conduction through gases.', NULL, '\n\nExplanation\n(a) Thermionic emission is the release of electrons from the surface of heated mentals. It is a surface phenomenon and the rate of electron emission increases with temperature increase.\n(b) Applications of electrical conducting gases are:\n— indentification of gases\n— in advertising neon signs\n— in fluorescent tubes.\n\n', '', 'Explanation\n(a) Thermionic emission is the release of electrons from the surface of heated mentals. It is a surface phenomenon and the rate of electron emission increases with temperature increase.\n(b) Applications of electrical conducting gases are:\n— indentification of gases\n— in advertising neon signs\n— in fluorescent tubes.\n\n', '(a) Thermionic emission is the release of electrons from the surface of heated mentals. It is a surface phenomenon and the rate of electron emission increases with temperature increase.~(b) Applications of electrical conducting gases are:\n— indentification of gases\n— in advertising neon signs\n— in fluorescent tubes.\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(156, '2017', '\n\nQuestion\n\nA wire of length 5.0 m and diameter 2.0 mm extends by 0.25 mm when a force of 50 N was used to stretch it from its end. Calculate the;\n(a) stress on the wire;\n(b) strain in, the wire. [\\(\\pi = 3.142\\)]\n\n', '', 'A wire of length 5.0 m and diameter 2.0 mm extends by 0.25 mm when a force of 50 N was used to stretch it from its end. Calculate the;\n(a) stress on the wire;\n(b) strain in, the wire. [\\(\\pi = 3.142\\)]', '(a) stress on the wire;~(b) strain in, the wire. [\\(\\pi = 3.142\\)]', NULL, '\n\nExplanation\n(a) Stress = \\(\\frac{Force}{Area} = \\frac{50}{3.142(1 \\times 10^{-3})^2}\\)\n(b) Strain = \\(\\frac{extension}{length} = \\frac{2.5 \\times 10^{-5}}{5}\\)\n= 5 x 10\\(^{-5}\\)\n\n', '', 'Explanation\n(a) Stress = \\(\\frac{Force}{Area} = \\frac{50}{3.142(1 \\times 10^{-3})^2}\\)\n(b) Strain = \\(\\frac{extension}{length} = \\frac{2.5 \\times 10^{-5}}{5}\\)\n= 5 x 10\\(^{-5}\\)\n\n', '(a) Stress = \\(\\frac{Force}{Area} = \\frac{50}{3.142(1 \\times 10^{-3})^2}\\)~(b) Strain = \\(\\frac{extension}{length} = \\frac{2.5 \\times 10^{-5}}{5}\\)\n= 5 x 10\\(^{-5}\\)\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(157, '2017', '\n\nQuestion\n\n(a) List two factors that can affect the rate of diffusion\n(b) State two examples to illustrate the effects of surface tension.\n\n', '', '(a) List two factors that can affect the rate of diffusion\n(b) State two examples to illustrate the effects of surface tension.', '(a) List two factors that can affect the rate of diffusion~(b) State two examples to illustrate the effects of surface tension.', NULL, '\n\nExplanation\n(a) Factors that can affect the rate of diffusion are:\n— Mass — Temperature — Density.\n(b) Effects of surface tension are:\n— needle gently placed on the surface of water floats\n— water skaters are able to walk on the surface of water\n— spilled mercury forms spherical droplets\n— soap bubbles are spherical in shape.\n\n', '', 'Explanation\n(a) Factors that can affect the rate of diffusion are:\n— Mass — Temperature — Density.\n(b) Effects of surface tension are:\n— needle gently placed on the surface of water floats\n— water skaters are able to walk on the surface of water\n— spilled mercury forms spherical droplets\n— soap bubbles are spherical in shape.\n\n', '(a) Factors that can affect the rate of diffusion are:\n— Mass — Temperature — Density.~(b) Effects of surface tension are:\n— needle gently placed on the surface of water floats\n— water skaters are able to walk on the surface of water\n— spilled mercury forms spherical droplets\n— soap bubbles are spherical in shape.\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(158, '2017', '\n\nQuestion\n\nState two effects to show (a) the existence of matter waves;\n(b) that radiation behaves like particles.\n\n', '', 'State two effects to show (a) the existence of matter waves;\n(b) that radiation behaves like particles.', '(a) the existence of matter waves;~(b) that radiation behaves like particles.', NULL, '\n\nExplanation\na) Effects to show the existence of matter waves:\n- diffraction of electrons\n— diffraction on neutron from crystal surface\n— diffraction pattern by reflecting hydrogen and helium molecules from crystal surface.\n\n(b) Effects to show that radiation behaves like waves:\n— compton effect\n— photoelectric effect\n— black body radiation\n— emission/absorption spectra\n— quantization of radiation energy proposed by Planck.\n\n\n', '', 'Explanation\na) Effects to show the existence of matter waves:\n- diffraction of electrons\n— diffraction on neutron from crystal surface\n— diffraction pattern by reflecting hydrogen and helium molecules from crystal surface.\n\n(b) Effects to show that radiation behaves like waves:\n— compton effect\n— photoelectric effect\n— black body radiation\n— emission/absorption spectra\n— quantization of radiation energy proposed by Planck.\n\n\n', 'a) Effects to show the existence of matter waves:\n- diffraction of electrons\n— diffraction on neutron from crystal surface\n— diffraction pattern by reflecting hydrogen and helium molecules from crystal surface.\n~(b) Effects to show that radiation behaves like waves:\n— compton effect\n— photoelectric effect\n— black body radiation\n— emission/absorption spectra\n— quantization of radiation energy proposed by Planck.\n\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(159, '2017', '\n\nQuestion\n\nA stone projected horizontally from the top of a tower with a speed of 4 ms\\(^{-1}\\) lands on the level ground at a horizontal distance 25 m from the foot of the tower. Calculate the height of the tower. g = 10 ms\\(^{-2}\\)]\n\n', 'A stone projected horizontally from the top of a tower with a speed of 4 ms\\(^{-1}\\) lands on the level ground at a horizontal distance 25 m from the foot of the tower. Calculate the height of the tower. g = 10 ms\\(^{-2}\\)]', '', '', NULL, '\n\nExplanation\nRange R = U\\(\\sqrt{\\frac{2H}{g}}\\)\nH = \\(\\frac{gR^2}{2u^2} = \\frac{10 \\times 25^2}{2 \\times 4^2}\\)\n= 195.3m\n\n', 'Range R = U\\(\\sqrt{\\frac{2H}{g}}\\)\nH = \\(\\frac{gR^2}{2u^2} = \\frac{10 \\times 25^2}{2 \\times 4^2}\\)\n= 195.3m\n\n', '', '', 'WAEC-PK-1', 'WAEC-PHY', 1),
(160, '2017', '\n\nQuestion\n\nDefine (a) tensile Stress; (b) tensile strain; (c) yield point.\n\n', 'Define (a) tensile Stress; (b) tensile strain; (c) yield point.', '', '', NULL, '\n\nExplanation\n(a) Tensile stress is the force per unit cross-sectional area of a wire or rod.\n(b) Tensile strain is the extension per unit length.\n(c) Yield point is a point on a load versus extension graph that represents the least force beyond which plastic deformation occurs.\n\n', '', 'Explanation\n(a) Tensile stress is the force per unit cross-sectional area of a wire or rod.\n(b) Tensile strain is the extension per unit length.\n(c) Yield point is a point on a load versus extension graph that represents the least force beyond which plastic deformation occurs.\n\n', '(a) Tensile stress is the force per unit cross-sectional area of a wire or rod.~(b) Tensile strain is the extension per unit length.\n(c) Yield point is a point on a load versus extension graph that represents the least force beyond which plastic deformation occurs.\n~(c) Yield point is a point on a load versus extension graph that represents the least force beyond which plastic deformation occurs.\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(161, '2017', '\n\nQuestion\n\n(a) State the conditions of equilibrium for a number of coplanar parallel forces.\n(b) A metre rule is found to balance horizontally at the 48 cm mark. When a body of mass 60 g is suspended at the 6 cm mark, the balance point is found to be at the 30 cm mark. Calculate the;\n(i) mass of the metre rule;\n(ii) distance of the balance point from the zero end, if the body were moved to the 13 cm mark.\n(c) a man pulls up a box of mass 70 kg using an inclined plane of effective length 5 m unto a platform 2.5 m high at a uniform speed. If the frictional force between the box and the plane is 1000 N;\n(i) draw a diagram to illustrate all the forces acting on the box while in motion;\n(ii) calculate the I. minimum effort applied in pulling up the box; II. velocity ratio of the plane, if it is inclined at 30° to the horizontal; Ill. force ratio of the plane.\n\n', '', '(a) State the conditions of equilibrium for a number of coplanar parallel forces.\n(b) A metre rule is found to balance horizontally at the 48 cm mark. When a body of mass 60 g is suspended at the 6 cm mark, the balance point is found to be at the 30 cm mark. Calculate the;\n(i) mass of the metre rule;\n(ii) distance of the balance point from the zero end, if the body were moved to the 13 cm mark.\n(c) a man pulls up a box of mass 70 kg using an inclined plane of effective length 5 m unto a platform 2.5 m high at a uniform speed. If the frictional force between the box and the plane is 1000 N;\n(i) draw a diagram to illustrate all the forces acting on the box while in motion;\n(ii) calculate the I. minimum effort applied in pulling up the box; II. velocity ratio of the plane, if it is inclined at 30° to the horizontal; Ill. force ratio of the plane.', '(a) State the conditions of equilibrium for a number of coplanar parallel forces.~(b) A metre rule is found to balance horizontally at the 48 cm mark. When a body of mass 60 g is suspended at the 6 cm mark, the balance point is found to be at the 30 cm mark. Calculate the;\n(i) mass of the metre rule;\n(ii) distance of the balance point from the zero end, if the body were moved to the 13 cm mark.\n(c) a man pulls up a box of mass 70 kg using an inclined plane of effective length 5 m unto a platform 2.5 m high at a uniform speed. If the frictional force between the box and the plane is 1000 N;\n(i) draw a diagram to illustrate all the forces acting on the box while in motion;\n(ii) calculate the I. minimum effort applied in pulling up the box; II. velocity ratio of the plane, if it is inclined at 30° to the horizontal; Ill. force ratio of the plane.~(c) a man pulls up a box of mass 70 kg using an inclined plane of effective length 5 m unto a platform 2.5 m high at a uniform speed. If the frictional force between the box and the plane is 1000 N;\n(i) draw a diagram to illustrate all the forces acting on the box while in motion;\n(ii) calculate the I. minimum effort applied in pulling up the box; II. velocity ratio of the plane, if it is inclined at 30° to the horizontal; Ill. force ratio of the plane.', NULL, '\n\nExplanation\n(a) The sum of the clockwise moments about a point = the sum of the anticlockwise moments about the same point\n(b)(i)\n\nTake a moment about the fulcrum F\n60 x 24 = m x 18\nm = \\(\\frac{60 \\times 24}{18}\\)\n= 80g\n\n(ii)\n\nTake a momen about F, 60 x (\\(x\\) - 13) = 80 x (48 - \\(x\\))\n3(\\(x\\) - 13) = 4(48 - \\(x\\))\n3\\(x\\) - 39 = 192 - 4\\(x\\)\n7\\(x\\) = 231\n\\(x = \\frac{231}{7}\\) = 33cm\n\n(c)(i)\n\n(ii)\n(i) Minimum effort E = F + mgsin\\(\\theta\\) = 1000 + 700 sin \\(\\theta\\) = 1350N\n(ii) V.R = \\(\\frac{1}{sin \\theta} = \\frac{1}{sin 30}\\) = 2\n(ii) M.A = \\(\\frac{L}{E} = \\frac{700}{1350}\\) = 0.52\n\n', '', 'Explanation\n(a) The sum of the clockwise moments about a point = the sum of the anticlockwise moments about the same point\n(b)(i)\n\nTake a moment about the fulcrum F\n60 x 24 = m x 18\nm = \\(\\frac{60 \\times 24}{18}\\)\n= 80g\n\n(ii)\n\nTake a momen about F, 60 x (\\(x\\) - 13) = 80 x (48 - \\(x\\))\n3(\\(x\\) - 13) = 4(48 - \\(x\\))\n3\\(x\\) - 39 = 192 - 4\\(x\\)\n7\\(x\\) = 231\n\\(x = \\frac{231}{7}\\) = 33cm\n\n(c)(i)\n\n(ii)\n(i) Minimum effort E = F + mgsin\\(\\theta\\) = 1000 + 700 sin \\(\\theta\\) = 1350N\n(ii) V.R = \\(\\frac{1}{sin \\theta} = \\frac{1}{sin 30}\\) = 2\n(ii) M.A = \\(\\frac{L}{E} = \\frac{700}{1350}\\) = 0.52\n\n', '(a) The sum of the clockwise moments about a point = the sum of the anticlockwise moments about the same point~(b)(i)\n\nTake a moment about the fulcrum F\n60 x 24 = m x 18\nm = \\(\\frac{60 \\times 24}{18}\\)\n= 80g\n\n(ii)\n\nTake a momen about F, 60 x (\\(x\\) - 13) = 80 x (48 - \\(x\\))\n3(\\(x\\) - 13) = 4(48 - \\(x\\))\n3\\(x\\) - 39 = 192 - 4\\(x\\)\n7\\(x\\) = 231\n\\(x = \\frac{231}{7}\\) = 33cm\n\n(c)(i)\n\n(ii)\n(i) Minimum effort E = F + mgsin\\(\\theta\\) = 1000 + 700 sin \\(\\theta\\) = 1350N\n(ii) V.R = \\(\\frac{1}{sin \\theta} = \\frac{1}{sin 30}\\) = 2\n(ii) M.A = \\(\\frac{L}{E} = \\frac{700}{1350}\\) = 0.52\n~(c)(i)\n\n(ii)\n(i) Minimum effort E = F + mgsin\\(\\theta\\) = 1000 + 700 sin \\(\\theta\\) = 1350N\n(ii) V.R = \\(\\frac{1}{sin \\theta} = \\frac{1}{sin 30}\\) = 2\n(ii) M.A = \\(\\frac{L}{E} = \\frac{700}{1350}\\) = 0.52\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(162, '2017', '\n\nQuestion\n\n(a) Define the boiling point of a liquid.\n(b) With the aid of a sketch diagram, describe an experiment to determine the boiling point of a small quantity of a Iiquid\n(c) A piece of copper of mass 300 g at a temperature of 950°C is quickly transferred into a vessel of negligible thermal capacity containing 250 g of water at 25°C. If the final steady temperature of the mixture is 100°C, calculate the mass of water that will boil away. Especific heat capacity of copper = 4.0 x 10\\(^2\\) J kg\\(^{-1}\\)K\\(^{-1}\\) specific heat capacity of water = 4.2 x 10\\(^{3}\\) J kg\\(^{-1}\\)K\\(^{-1}\\) specific latent heat of vaporization of steam = 2.26 x 10\\(^6\\)J kg\\(^{-1}\\)\n\n', '', '(a) Define the boiling point of a liquid.\n(b) With the aid of a sketch diagram, describe an experiment to determine the boiling point of a small quantity of a Iiquid\n(c) A piece of copper of mass 300 g at a temperature of 950°C is quickly transferred into a vessel of negligible thermal capacity containing 250 g of water at 25°C. If the final steady temperature of the mixture is 100°C, calculate the mass of water that will boil away. Especific heat capacity of copper = 4.0 x 10\\(^2\\) J kg\\(^{-1}\\)K\\(^{-1}\\) specific heat capacity of water = 4.2 x 10\\(^{3}\\) J kg\\(^{-1}\\)K\\(^{-1}\\) specific latent heat of vaporization of steam = 2.26 x 10\\(^6\\)J kg\\(^{-1}\\)', '(a) Define the boiling point of a liquid.~(b) With the aid of a sketch diagram, describe an experiment to determine the boiling point of a small quantity of a Iiquid\n(c) A piece of copper of mass 300 g at a temperature of 950°C is quickly transferred into a vessel of negligible thermal capacity containing 250 g of water at 25°C. If the final steady temperature of the mixture is 100°C, calculate the mass of water that will boil away. Especific heat capacity of copper = 4.0 x 10\\(^2\\) J kg\\(^{-1}\\)K\\(^{-1}\\) specific heat capacity of water = 4.2 x 10\\(^{3}\\) J kg\\(^{-1}\\)K\\(^{-1}\\) specific latent heat of vaporization of steam = 2.26 x 10\\(^6\\)J kg\\(^{-1}\\)~(c) A piece of copper of mass 300 g at a temperature of 950°C is quickly transferred into a vessel of negligible thermal capacity containing 250 g of water at 25°C. If the final steady temperature of the mixture is 100°C, calculate the mass of water that will boil away. Especific heat capacity of copper = 4.0 x 10\\(^2\\) J kg\\(^{-1}\\)K\\(^{-1}\\) specific heat capacity of water = 4.2 x 10\\(^{3}\\) J kg\\(^{-1}\\)K\\(^{-1}\\) specific latent heat of vaporization of steam = 2.26 x 10\\(^6\\)J kg\\(^{-1}\\)', NULL, '\n\nExplanation\n(a) Boiling point of a liquid is the temperature at which the saturated vapour pressure of the liquid is equal to the external atmospheric pressure.\n(b)\n\nAdd and trap a few drops of the liquid in the closed limb of the J-tube above the mercury. Insert into a beaker of water the J-tube, stirrer and thermometer. Heat the beaker and content and stir continuously. Then read and record the temperature at which the mercury levels in the two limbs are the same. Heat the beaker a little further and remove heat source. Also read and record the temperature at which the mercury levels are once again the same. The average of the two temperature values is the boiling point of the liquid.\n(c) Heat lost by copper = Heat gained by water\n0.3 x 400 (950 - 100) = 0.25 x 4200 x (100 - 25) + 2.26 x 10\\(^6\\) x m 102000\n= 78750 + 2.26 x 10\\(^6\\)m\nm = \\(\\frac{102000 — 78750}{2.26 \\times 10^6}\\)\n= 0.0102876 kg or 10.3g\n\n', '', 'Explanation\n(a) Boiling point of a liquid is the temperature at which the saturated vapour pressure of the liquid is equal to the external atmospheric pressure.\n(b)\n\nAdd and trap a few drops of the liquid in the closed limb of the J-tube above the mercury. Insert into a beaker of water the J-tube, stirrer and thermometer. Heat the beaker and content and stir continuously. Then read and record the temperature at which the mercury levels in the two limbs are the same. Heat the beaker a little further and remove heat source. Also read and record the temperature at which the mercury levels are once again the same. The average of the two temperature values is the boiling point of the liquid.\n(c) Heat lost by copper = Heat gained by water\n0.3 x 400 (950 - 100) = 0.25 x 4200 x (100 - 25) + 2.26 x 10\\(^6\\) x m 102000\n= 78750 + 2.26 x 10\\(^6\\)m\nm = \\(\\frac{102000 — 78750}{2.26 \\times 10^6}\\)\n= 0.0102876 kg or 10.3g\n\n', '(a) Boiling point of a liquid is the temperature at which the saturated vapour pressure of the liquid is equal to the external atmospheric pressure.~(b)\n\nAdd and trap a few drops of the liquid in the closed limb of the J-tube above the mercury. Insert into a beaker of water the J-tube, stirrer and thermometer. Heat the beaker and content and stir continuously. Then read and record the temperature at which the mercury levels in the two limbs are the same. Heat the beaker a little further and remove heat source. Also read and record the temperature at which the mercury levels are once again the same. The average of the two temperature values is the boiling point of the liquid.\n(c) Heat lost by copper = Heat gained by water\n0.3 x 400 (950 - 100) = 0.25 x 4200 x (100 - 25) + 2.26 x 10\\(^6\\) x m 102000\n= 78750 + 2.26 x 10\\(^6\\)m\nm = \\(\\frac{102000 — 78750}{2.26 \\times 10^6}\\)\n= 0.0102876 kg or 10.3g\n~(c) Heat lost by copper = Heat gained by water\n0.3 x 400 (950 - 100) = 0.25 x 4200 x (100 - 25) + 2.26 x 10\\(^6\\) x m 102000\n= 78750 + 2.26 x 10\\(^6\\)m\nm = \\(\\frac{102000 — 78750}{2.26 \\times 10^6}\\)\n= 0.0102876 kg or 10.3g\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(163, '2017', '\n\nQuestion\n\n(a)(i) What is a wave motion?\n(ii) State two differences between a radio wave and a sound wave.\n(b)(i) Given that you are provided with a tuning fork, a burette and other necessary apparatus, describe with the aid of a diagram, an experiment to determine the frequency of a note emitted by a source of sound. [assume the velocity of sound in air is known]\n(ii)State two precautions necessary to obtain accurate result in the experiment described in (b)(i) above\n(c) A pipe closed at one end is 100 cm long. If the air in the pipe is set into vibration and a fundamental note is produced, calculate the frequency of the note. [ velocity of sound in air = 340 ms\\(^{-1}\\)]\n\n', '', '(a)(i) What is a wave motion?\n(ii) State two differences between a radio wave and a sound wave.\n(b)(i) Given that you are provided with a tuning fork, a burette and other necessary apparatus, describe with the aid of a diagram, an experiment to determine the frequency of a note emitted by a source of sound. [assume the velocity of sound in air is known]\n(ii)State two precautions necessary to obtain accurate result in the experiment described in (b)(i) above\n(c) A pipe closed at one end is 100 cm long. If the air in the pipe is set into vibration and a fundamental note is produced, calculate the frequency of the note. [ velocity of sound in air = 340 ms\\(^{-1}\\)]', '(a)(i) What is a wave motion?\n(ii) State two differences between a radio wave and a sound wave.~(b)(i) Given that you are provided with a tuning fork, a burette and other necessary apparatus, describe with the aid of a diagram, an experiment to determine the frequency of a note emitted by a source of sound. [assume the velocity of sound in air is known]\n(ii)State two precautions necessary to obtain accurate result in the experiment described in (b)(i) above\n(c) A pipe closed at one end is 100 cm long. If the air in the pipe is set into vibration and a fundamental note is produced, calculate the frequency of the note. [ velocity of sound in air = 340 ms\\(^{-1}\\)]~(c) A pipe closed at one end is 100 cm long. If the air in the pipe is set into vibration and a fundamental note is produced, calculate the frequency of the note. [ velocity of sound in air = 340 ms\\(^{-1}\\)]', NULL, '\n\nExplanation\n(a)(i) A wave motion is a process of transferring a disturbance in form of K.E from one point to another in a medium without any transfer of the particles of the medium.\n(ii)\n\n\n\n\nRadio wave\n\n\nSound wave\n\n\n\n\nIt is electro-magnetic wave.\nIt is a transverse wave\nRequires no medium for propagation.\nHas velocity equal to that of light.\n\n\nIt is a mechanical wave\nIt is longitudinal wave\n\nRequires medium for propagation.\n\nIts velocity is less than that of light.\n\n\n\n\n\n(b) (i)\n\nFill the burette with water. Set the prongs of the fork into vibration and bring it near the open end of the burette, then gradually open the tap to run out some water, there-by increasing the length of the air column until a loud sound is heard. Measure and record the length L\\(_2\\) of the air column. Further run out the water while le vibrating tuning fork is still held over open end of the burette until a second loud sound is heard, then read and record the length L\\(_2\\) of the air column. Last repeat the experiment to obtain L L\\(_1\\) and L L\\(_2\\) again and find the mean values of L L\\(_1\\) and L L\\(_2\\). Obtain the frequency of the tuning fork using\nf = \\(\\frac{v}{2(L_2 - L_1)}\\)\nii) Precautions: - repeat the experiment to obtain nean values\n- tap tuning fork on a pad not on a hard surface\n- hold tuning fork over the burette to avoid contact\n- obtain two lengths of air column to remove rnd correction.\n\n(c) \\(f_o = \\frac{v}{4L} = \\frac{340}{4 \\times 100 \\times 10^{-2}}\\) = 851Hz\n\n', '', 'Explanation\n(a)(i) A wave motion is a process of transferring a disturbance in form of K.E from one point to another in a medium without any transfer of the particles of the medium.\n(ii)\n\n\n\n\nRadio wave\n\n\nSound wave\n\n\n\n\nIt is electro-magnetic wave.\nIt is a transverse wave\nRequires no medium for propagation.\nHas velocity equal to that of light.\n\n\nIt is a mechanical wave\nIt is longitudinal wave\n\nRequires medium for propagation.\n\nIts velocity is less than that of light.\n\n\n\n\n\n(b) (i)\n\nFill the burette with water. Set the prongs of the fork into vibration and bring it near the open end of the burette, then gradually open the tap to run out some water, there-by increasing the length of the air column until a loud sound is heard. Measure and record the length L\\(_2\\) of the air column. Further run out the water while le vibrating tuning fork is still held over open end of the burette until a second loud sound is heard, then read and record the length L\\(_2\\) of the air column. Last repeat the experiment to obtain L L\\(_1\\) and L L\\(_2\\) again and find the mean values of L L\\(_1\\) and L L\\(_2\\). Obtain the frequency of the tuning fork using\nf = \\(\\frac{v}{2(L_2 - L_1)}\\)\nii) Precautions: - repeat the experiment to obtain nean values\n- tap tuning fork on a pad not on a hard surface\n- hold tuning fork over the burette to avoid contact\n- obtain two lengths of air column to remove rnd correction.\n\n(c) \\(f_o = \\frac{v}{4L} = \\frac{340}{4 \\times 100 \\times 10^{-2}}\\) = 851Hz\n\n', '(a)(i) A wave motion is a process of transferring a disturbance in form of K.E from one point to another in a medium without any transfer of the particles of the medium.\n(ii)\n\n\n\n\nRadio wave\n\n\nSound wave\n\n\n\n\nIt is electro-magnetic wave.\nIt is a transverse wave\nRequires no medium for propagation.\nHas velocity equal to that of light.\n\n\nIt is a mechanical wave\nIt is longitudinal wave\n\nRequires medium for propagation.\n\nIts velocity is less than that of light.\n\n\n\n\n~(b) (i)\n\nFill the burette with water. Set the prongs of the fork into vibration and bring it near the open end of the burette, then gradually open the tap to run out some water, there-by increasing the length of the air column until a loud sound is heard. Measure and record the length L\\(_2\\) of the air column. Further run out the water while le vibrating tuning fork is still held over open end of the burette until a second loud sound is heard, then read and record the length L\\(_2\\) of the air column. Last repeat the experiment to obtain L L\\(_1\\) and L L\\(_2\\) again and find the mean values of L L\\(_1\\) and L L\\(_2\\). Obtain the frequency of the tuning fork using\nf = \\(\\frac{v}{2(L_2 - L_1)}\\)\nii) Precautions: - repeat the experiment to obtain nean values\n- tap tuning fork on a pad not on a hard surface\n- hold tuning fork over the burette to avoid contact\n- obtain two lengths of air column to remove rnd correction.\n\n(c) \\(f_o = \\frac{v}{4L} = \\frac{340}{4 \\times 100 \\times 10^{-2}}\\) = 851Hz\n~(c) \\(f_o = \\frac{v}{4L} = \\frac{340}{4 \\times 100 \\times 10^{-2}}\\) = 851Hz\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(164, '2017', '\n\nQuestion\n\n(a) With the aid of a simple diagram, explain how a step down transformer works.\n(b)(i) State three ways by which energy is lost in a transformer\n(ii) Mention how each of the losses in (b)(i) above can be minimized\n(c) A 95% efficient transformer is used to operate a lamp rated 60W, 220 V from a 4400 V a.c supply. Calculate the;\n(i) ratio of the number of turns in the primary coil to the number of turns in the secondary coil of the transformer\n(ii) current taken from the main circuit.\n\n', '', '(a) With the aid of a simple diagram, explain how a step down transformer works.\n(b)(i) State three ways by which energy is lost in a transformer\n(ii) Mention how each of the losses in (b)(i) above can be minimized\n(c) A 95% efficient transformer is used to operate a lamp rated 60W, 220 V from a 4400 V a.c supply. Calculate the;\n(i) ratio of the number of turns in the primary coil to the number of turns in the secondary coil of the transformer\n(ii) current taken from the main circuit.', '(a) With the aid of a simple diagram, explain how a step down transformer works.~(b)(i) State three ways by which energy is lost in a transformer\n(ii) Mention how each of the losses in (b)(i) above can be minimized\n(c) A 95% efficient transformer is used to operate a lamp rated 60W, 220 V from a 4400 V a.c supply. Calculate the;\n(i) ratio of the number of turns in the primary coil to the number of turns in the secondary coil of the transformer\n(ii) current taken from the main circuit.~(c) A 95% efficient transformer is used to operate a lamp rated 60W, 220 V from a 4400 V a.c supply. Calculate the;\n(i) ratio of the number of turns in the primary coil to the number of turns in the secondary coil of the transformer\n(ii) current taken from the main circuit.', NULL, '\n\nExplanation\n(a)\n\nIf the primary coil is connected to an A.C source, the A.C passing through the coil sets up a changing ternating magnetic flux in the soft iron core. The secondary coil is linked with the changing flux which Juces an e.m.f in it. Since the number of turns in a secondary is less than that in the primary, the induced e.m.f is less than e.m.f in the primary. For s reason the A.C output is less than the A.0 in primary.\n(b)(i) Ways in which heat is lost in a transformer:\n- Eddy current\n- Flux leakages\n- Hysteresis losses\n- Copper or joules losses\n\n(ii)\n\n\n\n\nLosses\n\n\nMinimizing method\n\n\n\n\nEddy current losses.\nFlux leakage,\nHysteresis losses.\nCopper or joules losses.\n\n\nLaminating the core.\nDesign an efficient core.\nCore made of soft magnetic material of soft iron.\nUsing low resistance coil.\n\n\n\n\n\n(c) (i) \\(\\frac{N_s}{N_p} = \\frac{E_s}{E_p} = \\frac{220}{4400}\\)\n\\(N_p : N_s = 20 : 1\\)\n(ii) Efficiency = \\frac{output}{input} = \\frac{V_s \\times I_s}{V_p \\times I_p}\\)\n= \\(\\frac{95}{100} = \\frac{60}{4400 \\times I_p}\\)\n\\(I_p = \\frac{100 \\times 60}{95 \\times 440} = 0.014A\\)\n\n\n', '', 'Explanation\n(a)\n\nIf the primary coil is connected to an A.C source, the A.C passing through the coil sets up a changing ternating magnetic flux in the soft iron core. The secondary coil is linked with the changing flux which Juces an e.m.f in it. Since the number of turns in a secondary is less than that in the primary, the induced e.m.f is less than e.m.f in the primary. For s reason the A.C output is less than the A.0 in primary.\n(b)(i) Ways in which heat is lost in a transformer:\n- Eddy current\n- Flux leakages\n- Hysteresis losses\n- Copper or joules losses\n\n(ii)\n\n\n\n\nLosses\n\n\nMinimizing method\n\n\n\n\nEddy current losses.\nFlux leakage,\nHysteresis losses.\nCopper or joules losses.\n\n\nLaminating the core.\nDesign an efficient core.\nCore made of soft magnetic material of soft iron.\nUsing low resistance coil.\n\n\n\n\n\n(c) (i) \\(\\frac{N_s}{N_p} = \\frac{E_s}{E_p} = \\frac{220}{4400}\\)\n\\(N_p : N_s = 20 : 1\\)\n(ii) Efficiency = \\frac{output}{input} = \\frac{V_s \\times I_s}{V_p \\times I_p}\\)\n= \\(\\frac{95}{100} = \\frac{60}{4400 \\times I_p}\\)\n\\(I_p = \\frac{100 \\times 60}{95 \\times 440} = 0.014A\\)\n\n\n', '(a)\n\nIf the primary coil is connected to an A.C source, the A.C passing through the coil sets up a changing ternating magnetic flux in the soft iron core. The secondary coil is linked with the changing flux which Juces an e.m.f in it. Since the number of turns in a secondary is less than that in the primary, the induced e.m.f is less than e.m.f in the primary. For s reason the A.C output is less than the A.0 in primary.~(b)(i) Ways in which heat is lost in a transformer:\n- Eddy current\n- Flux leakages\n- Hysteresis losses\n- Copper or joules losses\n\n(ii)\n\n\n\n\nLosses\n\n\nMinimizing method\n\n\n\n\nEddy current losses.\nFlux leakage,\nHysteresis losses.\nCopper or joules losses.\n\n\nLaminating the core.\nDesign an efficient core.\nCore made of soft magnetic material of soft iron.\nUsing low resistance coil.\n\n\n\n\n\n(c) (i) \\(\\frac{N_s}{N_p} = \\frac{E_s}{E_p} = \\frac{220}{4400}\\)\n\\(N_p : N_s = 20 : 1\\)\n(ii) Efficiency = \\frac{output}{input} = \\frac{V_s \\times I_s}{V_p \\times I_p}\\)\n= \\(\\frac{95}{100} = \\frac{60}{4400 \\times I_p}\\)\n\\(I_p = \\frac{100 \\times 60}{95 \\times 440} = 0.014A\\)\n\n~(c) (i) \\(\\frac{N_s}{N_p} = \\frac{E_s}{E_p} = \\frac{220}{4400}\\)\n\\(N_p : N_s = 20 : 1\\)\n(ii) Efficiency = \\frac{output}{input} = \\frac{V_s \\times I_s}{V_p \\times I_p}\\)\n= \\(\\frac{95}{100} = \\frac{60}{4400 \\times I_p}\\)\n\\(I_p = \\frac{100 \\times 60}{95 \\times 440} = 0.014A\\)\n\n', 'WAEC-PK-1', 'WAEC-PHY', 1);
INSERT INTO `exam_simulation2` (`id`, `examYear`, `questions`, `parent_que`, `child_que`, `gen_child_que_desc`, `imageDir`, `explanation`, `parent_que_expl`, `child_que_exp`, `gen_child_que_expl`, `package_id`, `subject_id`, `status`) VALUES
(165, '2017', '\n\nQuestion\n\n(a)(i) With the aid of a labelled diagram describe the mode of operation of a modern X-ray tube.\n(ii)State the energy transformations that take place during the operation of the X-ray tube.\n\n(b) Define, as applied to X-rays, the following terms:\n(i) hardness;\n(ii) intensity.\n\n(c) State (i) four uses of X-rays;\n(ii) one hazard of over-exposure to X-rays in a radiological laboratory.\n\n', '', '(a)(i) With the aid of a labelled diagram describe the mode of operation of a modern X-ray tube.\n(ii)State the energy transformations that take place during the operation of the X-ray tube.\n\n(b) Define, as applied to X-rays, the following terms:\n(i) hardness;\n(ii) intensity.\n\n(c) State (i) four uses of X-rays;\n(ii) one hazard of over-exposure to X-rays in a radiological laboratory.', '(a)(i) With the aid of a labelled diagram describe the mode of operation of a modern X-ray tube.\n(ii)State the energy transformations that take place during the operation of the X-ray tube.\n~(b) Define, as applied to X-rays, the following terms:\n(i) hardness;\n(ii) intensity.\n\n(c) State (i) four uses of X-rays;\n(ii) one hazard of over-exposure to X-rays in a radiological laboratory.~(c) State (i) four uses of X-rays;\n(ii) one hazard of over-exposure to X-rays in a radiological laboratory.', NULL, '\n\nExplanation\n(a)(i)\n\n\n\nThermally generated electrons from the filament are accelerated through high P.d and focused by the focusing cathode to a target. The energy of the impending electrons is converted mostly into heat which conducted through the copper anode. A small percentage of the energy is converted into X-rays which are radiated from the target through the window.\n(ii) Energy transformation:\n\n\n(b)(i) Hardness is the strength or penetrating power of the x-ray.\n(ii) Intensity is the energy radiated per unit time per unit area.\n\n(c)(i) Uses of x-rays:\n- Cure of cancer\n- Radio therapy\n- Detection of fracture in bones\n- Radiography\n- In metals to detect defect and fractures\n- In agriculture to kill germs\n- Analytical tool in crystallography\n- To detect metallic weapons\n\n(ii) Hazzards of over-exposure to x-rays; causes\n- leukemia\n- tissue damage\n- cataracts\n- skin burn or cancer\n- genetic mutations with undesirable heredity\n- baldness\n- graying.\n\n\n\n', '', 'Explanation\n(a)(i)\n\n\n\nThermally generated electrons from the filament are accelerated through high P.d and focused by the focusing cathode to a target. The energy of the impending electrons is converted mostly into heat which conducted through the copper anode. A small percentage of the energy is converted into X-rays which are radiated from the target through the window.\n(ii) Energy transformation:\n\n\n(b)(i) Hardness is the strength or penetrating power of the x-ray.\n(ii) Intensity is the energy radiated per unit time per unit area.\n\n(c)(i) Uses of x-rays:\n- Cure of cancer\n- Radio therapy\n- Detection of fracture in bones\n- Radiography\n- In metals to detect defect and fractures\n- In agriculture to kill germs\n- Analytical tool in crystallography\n- To detect metallic weapons\n\n(ii) Hazzards of over-exposure to x-rays; causes\n- leukemia\n- tissue damage\n- cataracts\n- skin burn or cancer\n- genetic mutations with undesirable heredity\n- baldness\n- graying.\n\n\n\n', '(a)(i)\n\n\n\nThermally generated electrons from the filament are accelerated through high P.d and focused by the focusing cathode to a target. The energy of the impending electrons is converted mostly into heat which conducted through the copper anode. A small percentage of the energy is converted into X-rays which are radiated from the target through the window.\n(ii) Energy transformation:\n\n~(b)(i) Hardness is the strength or penetrating power of the x-ray.\n(ii) Intensity is the energy radiated per unit time per unit area.\n\n(c)(i) Uses of x-rays:\n- Cure of cancer\n- Radio therapy\n- Detection of fracture in bones\n- Radiography\n- In metals to detect defect and fractures\n- In agriculture to kill germs\n- Analytical tool in crystallography\n- To detect metallic weapons\n\n(ii) Hazzards of over-exposure to x-rays; causes\n- leukemia\n- tissue damage\n- cataracts\n- skin burn or cancer\n- genetic mutations with undesirable heredity\n- baldness\n- graying.\n\n\n~(c)(i) Uses of x-rays:\n- Cure of cancer\n- Radio therapy\n- Detection of fracture in bones\n- Radiography\n- In metals to detect defect and fractures\n- In agriculture to kill germs\n- Analytical tool in crystallography\n- To detect metallic weapons\n\n(ii) Hazzards of over-exposure to x-rays; causes\n- leukemia\n- tissue damage\n- cataracts\n- skin burn or cancer\n- genetic mutations with undesirable heredity\n- baldness\n- graying.\n\n\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(166, '2017', '\n\nQuestion\n\n(a) Explain the statement the acceleration of free fall cohesion.\n(b) State two factors that can affect the value of the narrow glass tube has a concave meniscus while acceleration of free fall at a place.\n\n', '', '(a) Explain the statement the acceleration of free fall cohesion.\n(b) State two factors that can affect the value of the narrow glass tube has a concave meniscus while acceleration of free fall at a place.', '(a) Explain the statement the acceleration of free fall cohesion.~(b) State two factors that can affect the value of the narrow glass tube has a concave meniscus while acceleration of free fall at a place.', NULL, '\n\nExplanation\n(a) This means that at the equator, the force of gravity on a body causes its velocity to increase at the rate of 9.78m/s per second.\n(b)(i) The rotation of the earth on its axis.\n(ii) The earth is not a perfect sphere.\n(iii) The altitude of a place\n\n', '', 'Explanation\n(a) This means that at the equator, the force of gravity on a body causes its velocity to increase at the rate of 9.78m/s per second.\n(b)(i) The rotation of the earth on its axis.\n(ii) The earth is not a perfect sphere.\n(iii) The altitude of a place\n\n', '(a) This means that at the equator, the force of gravity on a body causes its velocity to increase at the rate of 9.78m/s per second.~(b)(i) The rotation of the earth on its axis.\n(ii) The earth is not a perfect sphere.\n(iii) The altitude of a place\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(167, '2017', '\n\nQuestion\n\nA particle is projected horizontally at 10 ms\\(^{-1}\\) from a height of 45m. Calculate the horizontal distance covered by the particle before hitting the ground. [g = 10ms\\(^{-1}\\)]\n\n', 'A particle is projected horizontally at 10 ms\\(^{-1}\\) from a height of 45m. Calculate the horizontal distance covered by the particle before hitting the ground. [g = 10ms\\(^{-1}\\)]', '', '', NULL, '\n\nExplanation\n\\(h = \\frac{1}{2} gt^2\\) t = \\(\\sqrt{\\frac{2h}{g}}\\)\n= \\(\\sqrt{\\frac{2 \\times 45}{10}}\\) = 3s\nHorinzntal distance = vt = 10 x 3 = 30m\n\n', '\\(h = \\frac{1}{2} gt^2\\) t = \\(\\sqrt{\\frac{2h}{g}}\\)\n= \\(\\sqrt{\\frac{2 \\times 45}{10}}\\) = 3s\nHorinzntal distance = vt = 10 x 3 = 30m\n\n', '', '', 'WAEC-PK-1', 'WAEC-PHY', 1),
(168, '2017', '\n\nQuestion\n\nA particle dropped from a vertical height and falls freely for a time interval t. Sketch and explain a graph to show how h varies with (a) t (b) t\\(^{2}\\).\n\n', 'A particle dropped from a vertical height and falls freely for a time interval t. Sketch and explain a graph to show how h varies with (a) t (b) t\\(^{2}\\).', '', '', NULL, '\n\nExplanation\na)\n\n\nb)\n\nThe graph is a straight line and of negative slope.\n\n\n', 'a)\n\n\nb)\n\nThe graph is a straight line and of negative slope.\n\n\n', '', '', 'WAEC-PK-1', 'WAEC-PHY', 1),
(169, '2017', '\n\nQuestion\n\nThe diagram below represents the graph of the force applied in stretching a spiral spring against the corresponding extension produced within its elastic limit.\n\nUsing the notations on the graph, determine the:\n(a) force constant of the spring;\n(b) work done in stretching the spring from 10 x 10\\(^{-2}\\)m to 20 x 10\\(^{-2}\\)m.\n\n', '', 'The diagram below represents the graph of the force applied in stretching a spiral spring against the corresponding extension produced within its elastic limit.\n\nUsing the notations on the graph, determine the:\n(a) force constant of the spring;\n(b) work done in stretching the spring from 10 x 10\\(^{-2}\\)m to 20 x 10\\(^{-2}\\)m.', '(a) force constant of the spring;~(b) work done in stretching the spring from 10 x 10\\(^{-2}\\)m to 20 x 10\\(^{-2}\\)m.', NULL, '\n\nExplanation\na) Force constant of the spring = slope of the graph = \\(\\frac{\\Delta force}{\\Delta Extension}\\)\n\n= \\(\\frac{60 — 0}{20 - 0 \\times 10^{-2}} = \\frac{60}{20 \\times 10^{-2}}\\)\n = 300Nm\\(^{-2}\\)\n\n(b) Work done = Area under the trapezium ABCD\n= 1/2 (30 + 60) x (20 — 10) x 10\\(^{-2}\\) = 4.5J\n\n', '', 'Explanation\na) Force constant of the spring = slope of the graph = \\(\\frac{\\Delta force}{\\Delta Extension}\\)\n\n= \\(\\frac{60 — 0}{20 - 0 \\times 10^{-2}} = \\frac{60}{20 \\times 10^{-2}}\\)\n = 300Nm\\(^{-2}\\)\n\n(b) Work done = Area under the trapezium ABCD\n= 1/2 (30 + 60) x (20 — 10) x 10\\(^{-2}\\) = 4.5J\n\n', 'a) Force constant of the spring = slope of the graph = \\(\\frac{\\Delta force}{\\Delta Extension}\\)\n\n= \\(\\frac{60 — 0}{20 - 0 \\times 10^{-2}} = \\frac{60}{20 \\times 10^{-2}}\\)\n = 300Nm\\(^{-2}\\)\n~(b) Work done = Area under the trapezium ABCD\n= 1/2 (30 + 60) x (20 — 10) x 10\\(^{-2}\\) = 4.5J\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(170, '2017', '\n\nQuestion\n\n(a) Distinguish between the forces of adhesion and cohension\n(b) Give one example each of the forces of adhesion and cohesion.\n\n\n', '', '(a) Distinguish between the forces of adhesion and cohension\n(b) Give one example each of the forces of adhesion and cohesion.\n', '(a) Distinguish between the forces of adhesion and cohension~(b) Give one example each of the forces of adhesion and cohesion.\n', NULL, '\n\nExplanation\n(a) Forces of adhesion are forces between molecules of different substances while forcea of cohesion are forces between molecules of the same substance.\n(b) Examples: Adhesion — force between water molecules and glass molecules.\nCohesion — force between mercury molecules.\n\n', '', 'Explanation\n(a) Forces of adhesion are forces between molecules of different substances while forcea of cohesion are forces between molecules of the same substance.\n(b) Examples: Adhesion — force between water molecules and glass molecules.\nCohesion — force between mercury molecules.\n\n', '(a) Forces of adhesion are forces between molecules of different substances while forcea of cohesion are forces between molecules of the same substance.~(b) Examples: Adhesion — force between water molecules and glass molecules.\nCohesion — force between mercury molecules.\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(171, '2017', '\n\nQuestion\n\nExplain why water in a narrow glass tube has a concave meniscus while mecury in the same tube, has a convex meniscus\n\n', 'Explain why water in a narrow glass tube has a concave meniscus while mecury in the same tube, has a convex meniscus', '', '', NULL, '\n\nExplanation\nFor water in a glass tube, the forces of adhesion between the water and glass molecules are greater than the force of cohesion between the water molecules hence water clings to glass forming a concave meniscus. On the other hand, the forces of cohesion between mercury molecules are greater than those between mercury molecules and glass molecules, so mercury molecules cling to each other forming convex meniscus.\n\n', 'For water in a glass tube, the forces of adhesion between the water and glass molecules are greater than the force of cohesion between the water molecules hence water clings to glass forming a concave meniscus. On the other hand, the forces of cohesion between mercury molecules are greater than those between mercury molecules and glass molecules, so mercury molecules cling to each other forming convex meniscus.\n\n', '', '', 'WAEC-PK-1', 'WAEC-PHY', 1),
(172, '2017', '\n\nQuestion\n\n(a) State two applications of electrolysis.\n(b) Explain what is meant by the electrochemical equivalent of copper is 3.3 x 10\\(^{-7}\\) kgC\\(^{-1}\\)\n\n', '', '(a) State two applications of electrolysis.\n(b) Explain what is meant by the electrochemical equivalent of copper is 3.3 x 10\\(^{-7}\\) kgC\\(^{-1}\\)', '(a) State two applications of electrolysis.~(b) Explain what is meant by the electrochemical equivalent of copper is 3.3 x 10\\(^{-7}\\) kgC\\(^{-1}\\)', NULL, '\n\nExplanation\n(a) Applications of electrolysis:\n— Electrotyping\n— Electroplating\n— Purification of metals\n— Extraction of metals\n— Callibration of ammeter\n\n(b) It means that when a charge of one coulomb IC is passed through a copper voltameter 3.3 x 10\\(^{-7}\\) kg of copper is on the cathode.\n\n', '', 'Explanation\n(a) Applications of electrolysis:\n— Electrotyping\n— Electroplating\n— Purification of metals\n— Extraction of metals\n— Callibration of ammeter\n\n(b) It means that when a charge of one coulomb IC is passed through a copper voltameter 3.3 x 10\\(^{-7}\\) kg of copper is on the cathode.\n\n', '(a) Applications of electrolysis:\n— Electrotyping\n— Electroplating\n— Purification of metals\n— Extraction of metals\n— Callibration of ammeter\n~(b) It means that when a charge of one coulomb IC is passed through a copper voltameter 3.3 x 10\\(^{-7}\\) kg of copper is on the cathode.\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(173, '2017', '\n\nQuestion\n\nThe uncertinty in determining the duration during which an electron remains in a particular energy level before returning to the ground state is 2.0 x 10\\(^{-9}\\)s. Calculate the uncertainty in determining its energy at that level [Take \\(\\frac{h}{2\\pi} = h = 1.054 \\times 10^{-34}\\) Js]\n\n\n', 'The uncertinty in determining the duration during which an electron remains in a particular energy level before returning to the ground state is 2.0 x 10\\(^{-9}\\)s. Calculate the uncertainty in determining its energy at that level [Take \\(\\frac{h}{2\\pi} = h = 1.054 \\times 10^{-34}\\) Js]\n', '', '', NULL, '\n\nExplanation\n\\(\\Delta E \\geq \\frac{h}{2\\pi} \\times \\frac{1}{2t} = 1.054 x 10^{-37} \\times \\frac{1}{2.0 \\times 10^{-9}}\\)\n\\(\\Delta E \\geq 5.27 \\times 10^{-25}\\)J\n\n', '\\(\\Delta E \\geq \\frac{h}{2\\pi} \\times \\frac{1}{2t} = 1.054 x 10^{-37} \\times \\frac{1}{2.0 \\times 10^{-9}}\\)\n\\(\\Delta E \\geq 5.27 \\times 10^{-25}\\)J\n\n', '', '', 'WAEC-PK-1', 'WAEC-PHY', 1),
(174, '2017', '\n\nQuestion\n\nA paralIel beam of unpolarized light is incident on a plane glass of refractive index 1.60 at an angle to the normal. If the reflected beam is completely polarized, calculate the angle of incidence of the beam.\n\n', 'A paralIel beam of unpolarized light is incident on a plane glass of refractive index 1.60 at an angle to the normal. If the reflected beam is completely polarized, calculate the angle of incidence of the beam.', '', '', NULL, '\n\nExplanation\ntan\\(\\theta = \\mu \\text{(Brewsters law)}\\)\n\\(\\theta tan^{-1} = \\mu = tan{-1} 1.60 = 58^o\\)\n\n', 'tan\\(\\theta = \\mu \\text{(Brewsters law)}\\)\n\\(\\theta tan^{-1} = \\mu = tan{-1} 1.60 = 58^o\\)\n\n', '', '', 'WAEC-PK-1', 'WAEC-PHY', 1),
(175, '2017', '\n\nQuestion\n\nList three observations in support of the de-Broglie\'s assumptions that moving particles behave like waves.\n\n', 'List three observations in support of the de-Broglie\'s assumptions that moving particles behave like waves.', '', '', NULL, '\n\nExplanation\n— Electrons show diffraction patterns\n— They have frequencies or wavelengths\n— They show fraction or reflection\n— The molecular velocities can determined in gases\n\n', '— Electrons show diffraction patterns\n— They have frequencies or wavelengths\n— They show fraction or reflection\n— The molecular velocities can determined in gases\n\n', '', '', 'WAEC-PK-1', 'WAEC-PHY', 1),
(176, '2017', '\n\nQuestion\n\n(a) State two;\n(i) laws of solid friction;\n(ii) advantages of friction;\n(iii) methods of reducing friction.\n(b) Draw and label a diagram of a pulley system with velocity ratio of 5.\n(c)(i) Show that the efficiency L, force ratio M.A. and the velocity ratio V.R. of a machine are related by the equation \\(E = \\frac{M.A.}{V.R}\\) x 100%.\n(ii) The efficiency of a machine is 80%. Calcuate the work done by a person using the machine to raise a load of 300 kg through a height of 4 m.[ g = 10 ms\\(^{-2}\\) ]\n\n', '', '(a) State two;\n(i) laws of solid friction;\n(ii) advantages of friction;\n(iii) methods of reducing friction.\n(b) Draw and label a diagram of a pulley system with velocity ratio of 5.\n(c)(i) Show that the efficiency L, force ratio M.A. and the velocity ratio V.R. of a machine are related by the equation \\(E = \\frac{M.A.}{V.R}\\) x 100%.\n(ii) The efficiency of a machine is 80%. Calcuate the work done by a person using the machine to raise a load of 300 kg through a height of 4 m.[ g = 10 ms\\(^{-2}\\) ]', '(a) State two;\n(i) laws of solid friction;\n(ii) advantages of friction;\n(iii) methods of reducing friction.~(b) Draw and label a diagram of a pulley system with velocity ratio of 5.\n(c)(i) Show that the efficiency L, force ratio M.A. and the velocity ratio V.R. of a machine are related by the equation \\(E = \\frac{M.A.}{V.R}\\) x 100%.\n(ii) The efficiency of a machine is 80%. Calcuate the work done by a person using the machine to raise a load of 300 kg through a height of 4 m.[ g = 10 ms\\(^{-2}\\) ]~(c)(i) Show that the efficiency L, force ratio M.A. and the velocity ratio V.R. of a machine are related by the equation \\(E = \\frac{M.A.}{V.R}\\) x 100%.\n(ii) The efficiency of a machine is 80%. Calcuate the work done by a person using the machine to raise a load of 300 kg through a height of 4 m.[ g = 10 ms\\(^{-2}\\) ]', NULL, '\n\nExplanation\n(a)(i) Laws of solid friction states that:\n— friction opposes motion\n— friction depends on the nature of e surfaces\n— friction is independent of the area of the surfaces in contact\n— frictional force F is proportional to the normal reaction R\n— friction is dependent of relative velocity between the surfaces.\n\nii) Advantages of friction:\n— it prevents man from ipping when walking\n— it provides gripping effect of beIts in machines\n— for grinding when using grinding ones\n— used for the production of electric charges used in the brake systems of vehicles etc.\n\niii) Methods of reducing friction:\n— by lubrication by streamlining\n— by smoothing\n— by using ball bearings or rollers.\n\nb)\n \n\n(c)(i) Efficiency = \\(\\frac{\\text{Work output}}{\\text{work output}}\\) x 100%\n= \\(\\frac{\\text{load x distance by load}}{\\text{effort x distance moved by effort}}\\)\n= M.A x \\(\\frac{1}{VR} \\times \\frac{100 %}{1}\\)\n\n(ii) \\(e = \\frac{\\text{Work output}}{\\text{Work input}}\\) x 100\nWork input = \\(\\frac{\\text{Work output}}{E}\\) x 100\n= \\(\\frac{300 \\times 10 \\times 4 \\times 100}{80}\\)\n= 15,000J or 15KJ\n\n', '', 'Explanation\n(a)(i) Laws of solid friction states that:\n— friction opposes motion\n— friction depends on the nature of e surfaces\n— friction is independent of the area of the surfaces in contact\n— frictional force F is proportional to the normal reaction R\n— friction is dependent of relative velocity between the surfaces.\n\nii) Advantages of friction:\n— it prevents man from ipping when walking\n— it provides gripping effect of beIts in machines\n— for grinding when using grinding ones\n— used for the production of electric charges used in the brake systems of vehicles etc.\n\niii) Methods of reducing friction:\n— by lubrication by streamlining\n— by smoothing\n— by using ball bearings or rollers.\n\nb)\n \n\n(c)(i) Efficiency = \\(\\frac{\\text{Work output}}{\\text{work output}}\\) x 100%\n= \\(\\frac{\\text{load x distance by load}}{\\text{effort x distance moved by effort}}\\)\n= M.A x \\(\\frac{1}{VR} \\times \\frac{100 %}{1}\\)\n\n(ii) \\(e = \\frac{\\text{Work output}}{\\text{Work input}}\\) x 100\nWork input = \\(\\frac{\\text{Work output}}{E}\\) x 100\n= \\(\\frac{300 \\times 10 \\times 4 \\times 100}{80}\\)\n= 15,000J or 15KJ\n\n', '~b)\n \n\n(c)(i) Efficiency = \\(\\frac{\\text{Work output}}{\\text{work output}}\\) x 100%\n= \\(\\frac{\\text{load x distance by load}}{\\text{effort x distance moved by effort}}\\)\n= M.A x \\(\\frac{1}{VR} \\times \\frac{100 %}{1}\\)\n\n(ii) \\(e = \\frac{\\text{Work output}}{\\text{Work input}}\\) x 100\nWork input = \\(\\frac{\\text{Work output}}{E}\\) x 100\n= \\(\\frac{300 \\times 10 \\times 4 \\times 100}{80}\\)\n= 15,000J or 15KJ\n~(c)(i) Efficiency = \\(\\frac{\\text{Work output}}{\\text{work output}}\\) x 100%\n= \\(\\frac{\\text{load x distance by load}}{\\text{effort x distance moved by effort}}\\)\n= M.A x \\(\\frac{1}{VR} \\times \\frac{100 %}{1}\\)\n\n(ii) \\(e = \\frac{\\text{Work output}}{\\text{Work input}}\\) x 100\nWork input = \\(\\frac{\\text{Work output}}{E}\\) x 100\n= \\(\\frac{300 \\times 10 \\times 4 \\times 100}{80}\\)\n= 15,000J or 15KJ\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(177, '2017', '\n\nQuestion\n\n(a) With the aid of ray diagrams, explain total internal reflection.\n(b) Describe, with the aid of a labelled diagram, the essential features of an astronomical telescope in normal adjustment.\n(c) A converging lens forms a real image of a real object. If the magnification is 2 and the distance between the image and the object is 90.0 cm, determine the\n(i) focal length of the lens;\n(ii) object distance for which the image would be the same size as the object.\n\n', '', '(a) With the aid of ray diagrams, explain total internal reflection.\n(b) Describe, with the aid of a labelled diagram, the essential features of an astronomical telescope in normal adjustment.\n(c) A converging lens forms a real image of a real object. If the magnification is 2 and the distance between the image and the object is 90.0 cm, determine the\n(i) focal length of the lens;\n(ii) object distance for which the image would be the same size as the object.', '(a) With the aid of ray diagrams, explain total internal reflection.~(b) Describe, with the aid of a labelled diagram, the essential features of an astronomical telescope in normal adjustment.\n(c) A converging lens forms a real image of a real object. If the magnification is 2 and the distance between the image and the object is 90.0 cm, determine the\n(i) focal length of the lens;\n(ii) object distance for which the image would be the same size as the object.~(c) A converging lens forms a real image of a real object. If the magnification is 2 and the distance between the image and the object is 90.0 cm, determine the\n(i) focal length of the lens;\n(ii) object distance for which the image would be the same size as the object.', NULL, '\n\nExplanation\n(a) Consider a ray of light travelling from a denser medium (glass) to a less dense medium (air) at a small angle of incidence.\n(i)\n\n\nAs angle of incidence increases i, the angle of refraction r also increases, at a point when r = 90°, i = C (the critical angle)\n\nIf the angle of incidence i is made greater than the critical angle C no refraction in the less dense medium, the ray is totally internally reflected.\n\n(b)\n\nAn astronomical telescope consists of two converging lenses i.e an objective lens of long focal length fo and an eye piece of short focal length fe. In normal adjustment the principal foci of the lenses coincide, the separation of the lenses d = fo + fe.\n\n(c)(i) But M = \\(\\frac{v}{u} = 2 = \\frac{v}{u}\\)\nv = 2u\nu + v = 90\nu + 2u = 90\nu = 30cm\nf = \\(\\frac{uv}{u + v} = \\frac{u \\times 2u}{u + 2u} = \\frac{2u}{3}\\)\n= \\(\\frac{2 \\times 30}{3}\\) = 20cm\n\n(iii) U = 2f = 2 x 20\n= 40cm\n\n\n\n\n\n', '', 'Explanation\n(a) Consider a ray of light travelling from a denser medium (glass) to a less dense medium (air) at a small angle of incidence.\n(i)\n\n\nAs angle of incidence increases i, the angle of refraction r also increases, at a point when r = 90°, i = C (the critical angle)\n\nIf the angle of incidence i is made greater than the critical angle C no refraction in the less dense medium, the ray is totally internally reflected.\n\n(b)\n\nAn astronomical telescope consists of two converging lenses i.e an objective lens of long focal length fo and an eye piece of short focal length fe. In normal adjustment the principal foci of the lenses coincide, the separation of the lenses d = fo + fe.\n\n(c)(i) But M = \\(\\frac{v}{u} = 2 = \\frac{v}{u}\\)\nv = 2u\nu + v = 90\nu + 2u = 90\nu = 30cm\nf = \\(\\frac{uv}{u + v} = \\frac{u \\times 2u}{u + 2u} = \\frac{2u}{3}\\)\n= \\(\\frac{2 \\times 30}{3}\\) = 20cm\n\n(iii) U = 2f = 2 x 20\n= 40cm\n\n\n\n\n\n', '(a) Consider a ray of light travelling from a denser medium (glass) to a less dense medium (air) at a small angle of incidence.\n(i)\n\n\nAs angle of incidence increases i, the angle of refraction r also increases, at a point when r = 90°, i = C (the critical angle)\n\nIf the angle of incidence i is made greater than the critical angle C no refraction in the less dense medium, the ray is totally internally reflected.\n~(b)\n\nAn astronomical telescope consists of two converging lenses i.e an objective lens of long focal length fo and an eye piece of short focal length fe. In normal adjustment the principal foci of the lenses coincide, the separation of the lenses d = fo + fe.\n\n(c)(i) But M = \\(\\frac{v}{u} = 2 = \\frac{v}{u}\\)\nv = 2u\nu + v = 90\nu + 2u = 90\nu = 30cm\nf = \\(\\frac{uv}{u + v} = \\frac{u \\times 2u}{u + 2u} = \\frac{2u}{3}\\)\n= \\(\\frac{2 \\times 30}{3}\\) = 20cm\n\n(iii) U = 2f = 2 x 20\n= 40cm\n\n\n\n\n~(c)(i) But M = \\(\\frac{v}{u} = 2 = \\frac{v}{u}\\)\nv = 2u\nu + v = 90\nu + 2u = 90\nu = 30cm\nf = \\(\\frac{uv}{u + v} = \\frac{u \\times 2u}{u + 2u} = \\frac{2u}{3}\\)\n= \\(\\frac{2 \\times 30}{3}\\) = 20cm\n\n(iii) U = 2f = 2 x 20\n= 40cm\n\n\n\n\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(178, '2017', '\n\nQuestion\n\n(a)(i) Name and explain the common defects of a primary cell.\n(ii) State two advantages of a secondary cell over a primary cell.\n(b) Draw a labelled diagram to show the essential parts of a dry leclanche cell.\n(c)(i) Explain why six accumulators each of e.m.f 2V connected in series can be used to start the engine of a car whereas eight dry cells each of e.m.f 1.5 V connected in series cannot be used.\n(ii) Name the materials used for the positive terminal, the negative terminal and the electrolyte in a\nI. leclanche cell;\nII. charged lead acid accumulator.\n\n', '', '(a)(i) Name and explain the common defects of a primary cell.\n(ii) State two advantages of a secondary cell over a primary cell.\n(b) Draw a labelled diagram to show the essential parts of a dry leclanche cell.\n(c)(i) Explain why six accumulators each of e.m.f 2V connected in series can be used to start the engine of a car whereas eight dry cells each of e.m.f 1.5 V connected in series cannot be used.\n(ii) Name the materials used for the positive terminal, the negative terminal and the electrolyte in a\nI. leclanche cell;\nII. charged lead acid accumulator.', '(a)(i) Name and explain the common defects of a primary cell.\n(ii) State two advantages of a secondary cell over a primary cell.~(b) Draw a labelled diagram to show the essential parts of a dry leclanche cell.\n(c)(i) Explain why six accumulators each of e.m.f 2V connected in series can be used to start the engine of a car whereas eight dry cells each of e.m.f 1.5 V connected in series cannot be used.\n(ii) Name the materials used for the positive terminal, the negative terminal and the electrolyte in a\nI. leclanche cell;\nII. charged lead acid accumulator.~(c)(i) Explain why six accumulators each of e.m.f 2V connected in series can be used to start the engine of a car whereas eight dry cells each of e.m.f 1.5 V connected in series cannot be used.\n(ii) Name the materials used for the positive terminal, the negative terminal and the electrolyte in a\nI. leclanche cell;\nII. charged lead acid accumulator.', NULL, '\n\nExplanation\n(a)(i) Primary cell defects:\n— Polarization\n— Local action Polarization\n— This is due to hydrogen bubbles covering the positive electrode (copper plate). Local action\n— This is due to the use of impure zinc or the gradual wearing away of the negative electrode zinc plate.\n(ii) Advantages of secondary cell over a primary cell:\n— a secondary cell can be recharged while primary cell cannot.\n— secondary cell can supply current for a longer period while the primary cell has a short life-span.\n\n(b)\n\n\n(c)(i) The internal resistance of the dry cells is much larger than that of the accumulators, hence accumulators will produce larger currents needed to start the engine.\n(ii)\n\n\n\n\nPositive terminal\nNegative terminal\nElectrolyte\n\n\nLeclanche cell\nCarbon rod\nZinc rod\nAmmonium chloride (NH\\(_4\\)\n\n\n\ncharged lead acid accumulator\n\nLead (IV) oxide or Lead peroxide\nLead\n\nLead Tetra oxo-sulphate (VI) acid (H\\(_2\\)SO\\(_4\\))\n\n\n\n\n\n\n', '', 'Explanation\n(a)(i) Primary cell defects:\n— Polarization\n— Local action Polarization\n— This is due to hydrogen bubbles covering the positive electrode (copper plate). Local action\n— This is due to the use of impure zinc or the gradual wearing away of the negative electrode zinc plate.\n(ii) Advantages of secondary cell over a primary cell:\n— a secondary cell can be recharged while primary cell cannot.\n— secondary cell can supply current for a longer period while the primary cell has a short life-span.\n\n(b)\n\n\n(c)(i) The internal resistance of the dry cells is much larger than that of the accumulators, hence accumulators will produce larger currents needed to start the engine.\n(ii)\n\n\n\n\nPositive terminal\nNegative terminal\nElectrolyte\n\n\nLeclanche cell\nCarbon rod\nZinc rod\nAmmonium chloride (NH\\(_4\\)\n\n\n\ncharged lead acid accumulator\n\nLead (IV) oxide or Lead peroxide\nLead\n\nLead Tetra oxo-sulphate (VI) acid (H\\(_2\\)SO\\(_4\\))\n\n\n\n\n\n\n', '(a)(i) Primary cell defects:\n— Polarization\n— Local action Polarization\n— This is due to hydrogen bubbles covering the positive electrode (copper plate). Local action\n— This is due to the use of impure zinc or the gradual wearing away of the negative electrode zinc plate.\n(ii) Advantages of secondary cell over a primary cell:\n— a secondary cell can be recharged while primary cell cannot.\n— secondary cell can supply current for a longer period while the primary cell has a short life-span.\n~(b)\n\n\n(c)(i) The internal resistance of the dry cells is much larger than that of the accumulators, hence accumulators will produce larger currents needed to start the engine.\n(ii)\n\n\n\n\nPositive terminal\nNegative terminal\nElectrolyte\n\n\nLeclanche cell\nCarbon rod\nZinc rod\nAmmonium chloride (NH\\(_4\\)\n\n\n\ncharged lead acid accumulator\n\nLead (IV) oxide or Lead peroxide\nLead\n\nLead Tetra oxo-sulphate (VI) acid (H\\(_2\\)SO\\(_4\\))\n\n\n\n\n\n~(c)(i) The internal resistance of the dry cells is much larger than that of the accumulators, hence accumulators will produce larger currents needed to start the engine.\n(ii)\n\n\n\n\nPositive terminal\nNegative terminal\nElectrolyte\n\n\nLeclanche cell\nCarbon rod\nZinc rod\nAmmonium chloride (NH\\(_4\\)\n\n\n\ncharged lead acid accumulator\n\nLead (IV) oxide or Lead peroxide\nLead\n\nLead Tetra oxo-sulphate (VI) acid (H\\(_2\\)SO\\(_4\\))\n\n\n\n\n\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(179, '2017', '\n\nQuestion\n\n(a) Define\n(i) proton number;\n(ii) nucleon number;\n(iii) isotopes.\n(b) A nuclide \\(^A_ZX\\) emits \\(\\beta\\)-particle to form a daughter nuclide Y. Write a nuclear equation to illustrate the charge conservation.\n(c) The radioactive nuclei \\(^{210}_{84}P_o\\) emits an \\(\\alpha\\) - particle to produce \\(^{206}_{82}P_b\\). Calculate the energy, in MeV, released in each disintegration.\n\nTake the masses of \\(^{210}_{84}P_o\\) = 209.936730 u;\n\\(^{206}_{82}P_b\\) = 205.929421 u;\n\\(^{4}_{2}He\\) = 4.001504 u;\nand that 1u = 931 MeV\n\n', '', '(a) Define\n(i) proton number;\n(ii) nucleon number;\n(iii) isotopes.\n(b) A nuclide \\(^A_ZX\\) emits \\(\\beta\\)-particle to form a daughter nuclide Y. Write a nuclear equation to illustrate the charge conservation.\n(c) The radioactive nuclei \\(^{210}_{84}P_o\\) emits an \\(\\alpha\\) - particle to produce \\(^{206}_{82}P_b\\). Calculate the energy, in MeV, released in each disintegration.\n\nTake the masses of \\(^{210}_{84}P_o\\) = 209.936730 u;\n\\(^{206}_{82}P_b\\) = 205.929421 u;\n\\(^{4}_{2}He\\) = 4.001504 u;\nand that 1u = 931 MeV', '(a) Define\n(i) proton number;\n(ii) nucleon number;\n(iii) isotopes.~(b) A nuclide \\(^A_ZX\\) emits \\(\\beta\\)-particle to form a daughter nuclide Y. Write a nuclear equation to illustrate the charge conservation.\n(c) The radioactive nuclei \\(^{210}_{84}P_o\\) emits an \\(\\alpha\\) - particle to produce \\(^{206}_{82}P_b\\). Calculate the energy, in MeV, released in each disintegration.\n\nTake the masses of \\(^{210}_{84}P_o\\) = 209.936730 u;\n\\(^{206}_{82}P_b\\) = 205.929421 u;\n\\(^{4}_{2}He\\) = 4.001504 u;\nand that 1u = 931 MeV~(c) The radioactive nuclei \\(^{210}_{84}P_o\\) emits an \\(\\alpha\\) - particle to produce \\(^{206}_{82}P_b\\). Calculate the energy, in MeV, released in each disintegration.\n\nTake the masses of \\(^{210}_{84}P_o\\) = 209.936730 u;\n\\(^{206}_{82}P_b\\) = 205.929421 u;\n\\(^{4}_{2}He\\) = 4.001504 u;\nand that 1u = 931 MeV', NULL, '\n\nExplanation\n(a)(i) Proton number\n— the number of protons in the nucleus of an atom\n(ii) Nucleon number — the total number of protons and neutrons in the nucleus of are atom.\n(iii) Isotopes — are atoms or nuclides of the samc element having the same number of protons or atomic number but different neutrons or mass numbers\n(b) \\(^A_Z \\to ^A_{Z+}Y + ^0_{-1}\\)\n\\(^{210}_{84}Po \\to ^{206}_{82}Pb + ^4_2He + Energy\\)\nMass detect = 209.936730u - (205.929421 + 4.001504)u\n209.9367330u - 209.930925u\n= 5.805 x 10\\(^{-3}\\)U\nEnergy released = 5.805 x 10\\(^{-3}\\) x 931\n= 5.40MeV\n\n', '', 'Explanation\n(a)(i) Proton number\n— the number of protons in the nucleus of an atom\n(ii) Nucleon number — the total number of protons and neutrons in the nucleus of are atom.\n(iii) Isotopes — are atoms or nuclides of the samc element having the same number of protons or atomic number but different neutrons or mass numbers\n(b) \\(^A_Z \\to ^A_{Z+}Y + ^0_{-1}\\)\n\\(^{210}_{84}Po \\to ^{206}_{82}Pb + ^4_2He + Energy\\)\nMass detect = 209.936730u - (205.929421 + 4.001504)u\n209.9367330u - 209.930925u\n= 5.805 x 10\\(^{-3}\\)U\nEnergy released = 5.805 x 10\\(^{-3}\\) x 931\n= 5.40MeV\n\n', '(a)(i) Proton number\n— the number of protons in the nucleus of an atom\n(ii) Nucleon number — the total number of protons and neutrons in the nucleus of are atom.\n(iii) Isotopes — are atoms or nuclides of the samc element having the same number of protons or atomic number but different neutrons or mass numbers~(b) \\(^A_Z \\to ^A_{Z+}Y + ^0_{-1}\\)\n\\(^{210}_{84}Po \\to ^{206}_{82}Pb + ^4_2He + Energy\\)\nMass detect = 209.936730u - (205.929421 + 4.001504)u\n209.9367330u - 209.930925u\n= 5.805 x 10\\(^{-3}\\)U\nEnergy released = 5.805 x 10\\(^{-3}\\) x 931\n= 5.40MeV\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(180, '2017', '\n\nQuestion\n\n(a) State the conditions for the equilibrium of a rigid body acted upon by parallel forces.\n(b)(i) Describe an experiment to determine the mass of a metre rule using the principle of moments.\n(ii) State two precautions necessary to obtain accurate results in the experiment described in (b)(i) above.\n(c) A bullet of mass 120 g is fired horizontally into a fixed wooden block with a speed of 20 ms\\(^{-1}\\). If the bullet is brought to rest in the block in 0.1s by a constant resistance, calculate the (i) magnitude of the resistance; (ii) distance moved by the bullet in the wood.\n\n', '', '(a) State the conditions for the equilibrium of a rigid body acted upon by parallel forces.\n(b)(i) Describe an experiment to determine the mass of a metre rule using the principle of moments.\n(ii) State two precautions necessary to obtain accurate results in the experiment described in (b)(i) above.\n(c) A bullet of mass 120 g is fired horizontally into a fixed wooden block with a speed of 20 ms\\(^{-1}\\). If the bullet is brought to rest in the block in 0.1s by a constant resistance, calculate the (i) magnitude of the resistance; (ii) distance moved by the bullet in the wood.', '(a) State the conditions for the equilibrium of a rigid body acted upon by parallel forces.~(b)(i) Describe an experiment to determine the mass of a metre rule using the principle of moments.\n(ii) State two precautions necessary to obtain accurate results in the experiment described in (b)(i) above.\n(c) A bullet of mass 120 g is fired horizontally into a fixed wooden block with a speed of 20 ms\\(^{-1}\\). If the bullet is brought to rest in the block in 0.1s by a constant resistance, calculate the (i) magnitude of the resistance; (ii) distance moved by the bullet in the wood.~(c) A bullet of mass 120 g is fired horizontally into a fixed wooden block with a speed of 20 ms\\(^{-1}\\). If the bullet is brought to rest in the block in 0.1s by a constant resistance, calculate the (i) magnitude of the resistance; (ii) distance moved by the bullet in the wood.', NULL, '\n\nExplanation\n(a) — The sum of clockwise moments about any point is equal to the sum of anticlockwise moments about the same point\n— The sum of the forces in one recticn is equal to the sum of the forces in the opposite direction.\nb)\n\nMethod: Suspend the metre rule on a thread loop from clamp. Adjust its position until the metre rule balances horizontally and obtain its centre of gravity G. Suspend known mass M at a point on one side of the rule and just the position of the loop until the rule balances horizontally. Read and record the distances y and x in the diagram above. sing the principle of moments Mgx = Mgy\nm = \\(\\frac{My}{x}\\)\nTherefore the mass M of the metre rule is calculated.\nii) Precautions:\n— Firmly fixed the loop\n— Avoid draught\n— Avoid parallax when reading metre rule.\n.\nc)(i) Declaration a = \\(\\frac{v - u}{t} = \\frac{0 — 20}{0.1} = — 200m/s^2\\)\n Resistance F = Ma = 0.12 x 200 = 24N\nii) Distance S = \\(\\frac{v^2 - u^2}{2a} = \\frac{0^2 - 20^0}{2 \\times 200}\\)\n= \\(\\frac{400}{400} = 1.0m\\)\n\n', '(a) — The sum of clockwise moments about any point is equal to the sum of anticlockwise moments about the same point\n— The sum of the forces in one recticn is equal to the sum of the forces in the opposite direction.\nb)\n\nMethod: Suspend the metre rule on a thread loop from clamp. Adjust its position until the metre rule balances horizontally and obtain its centre of gravity G. Suspend known mass M at a point on one side of the rule and just the position of the loop until the rule balances horizontally. Read and record the distances y and x in the diagram above. sing the principle of moments Mgx = Mgy\nm = \\(\\frac{My}{x}\\)\nTherefore the mass M of the metre rule is calculated.\nii) Precautions:\n— Firmly fixed the loop\n— Avoid draught\n— Avoid parallax when reading metre rule.\n.\nc)(i) Declaration a = \\(\\frac{v - u}{t} = \\frac{0 — 20}{0.1} = — 200m/s^2\\)\n Resistance F = Ma = 0.12 x 200 = 24N\nii) Distance S = \\(\\frac{v^2 - u^2}{2a} = \\frac{0^2 - 20^0}{2 \\times 200}\\)\n= \\(\\frac{400}{400} = 1.0m\\)\n\n', '', '', 'WAEC-PK-1', 'WAEC-PHY', 1),
(181, '2018', '\n\nQuestion\n\nA stone thrown horizontally from the top of a vertical ice is completely melted. Determine the total mass of wall with a velocity of 15ms\\(^{-1}\\), hits the horizontal ground water in the container at a point 45m from the base of the wall. Calculate the specific latent heat of steam = 2.3 x 10\\(^3\\) Jg\\(^{-1}\\) height of the wall. [g = 10ms\\(^{-2}\\)]\n\n', 'A stone thrown horizontally from the top of a vertical ice is completely melted. Determine the total mass of wall with a velocity of 15ms\\(^{-1}\\), hits the horizontal ground water in the container at a point 45m from the base of the wall. Calculate the specific latent heat of steam = 2.3 x 10\\(^3\\) Jg\\(^{-1}\\) height of the wall. [g = 10ms\\(^{-2}\\)]', '', '', NULL, '\n\nExplanation\nR = \\(U\\sqrt{\\frac{2H}{g}}\\)\nH = \\(\\frac{gR^2}{2U^2} = \\frac{10 \\times 45^2}{2 \\times 15^2}\\)\n\n\n', 'R = \\(U\\sqrt{\\frac{2H}{g}}\\)\nH = \\(\\frac{gR^2}{2U^2} = \\frac{10 \\times 45^2}{2 \\times 15^2}\\)\n\n\n', '', '', 'WAEC-PK-1', 'WAEC-PHY', 1),
(182, '2018', '\n\nQuestion\n\nA ball is projected specific latent heat of ice = 3.4 x 10\\(^{2}\\) Jg\\(^{-1}\\) horizontally from a height of 20m above the ground specific heat capacity of water = 4.2 Jg\\(^{-1}\\) K\\(^{-1}\\) with an initial velocity of 0.4ms\\(^{-1}\\). Calculate the horizontal distance moved by the ball before hitting the ground. [g = 10ms\\(^{-1}\\)]\n\n', 'A ball is projected specific latent heat of ice = 3.4 x 10\\(^{2}\\) Jg\\(^{-1}\\) horizontally from a height of 20m above the ground specific heat capacity of water = 4.2 Jg\\(^{-1}\\) K\\(^{-1}\\) with an initial velocity of 0.4ms\\(^{-1}\\). Calculate the horizontal distance moved by the ball before hitting the ground. [g = 10ms\\(^{-1}\\)]', '', '', NULL, '\n\nExplanation\nR = \\(U\\sqrt{\\frac{2H}{g}}\\)\n= 0.4 \\(\\sqrt{\\frac{2 \\times 20}{10}}\\)\n= 0.8m\n\n', 'R = \\(U\\sqrt{\\frac{2H}{g}}\\)\n= 0.4 \\(\\sqrt{\\frac{2 \\times 20}{10}}\\)\n= 0.8m\n\n', '', '', 'WAEC-PK-1', 'WAEC-PHY', 1),
(183, '2018', '\n\nQuestion\n\nDescribe, with the aid of a diagram, how a wave can be plane polarized.\n\n', 'Describe, with the aid of a diagram, how a wave can be plane polarized.', '', '', NULL, '\n\nExplanation\n(i) unpolarized wave            (ii)\n \n\nPlaner Polarized Wave\n\n\n\n\n', '(i) unpolarized wave            (ii)\n \n\nPlaner Polarized Wave\n\n\n\n\n', '', '', 'WAEC-PK-1', 'WAEC-PHY', 1),
(184, '2018', '\n\nQuestion\n\n(a) Explain cations (b) Draw and label an electrolytic cell.\n\n', '(a) Explain cations (b) Draw and label an electrolytic cell.', '', '', NULL, '\n\nExplanation\n(a) Cations are positive ions that are attracted to the cathode during relectrolysis e.g. cu\\(^{2+}\\), Ag\\(^+\\) , H\\(^+\\), Na\\(^+\\) etc.\n(b)\n\n\n', '', 'Explanation\n(a) Cations are positive ions that are attracted to the cathode during relectrolysis e.g. cu\\(^{2+}\\), Ag\\(^+\\) , H\\(^+\\), Na\\(^+\\) etc.\n(b)\n\n\n', '(a) Cations are positive ions that are attracted to the cathode during relectrolysis e.g. cu\\(^{2+}\\), Ag\\(^+\\) , H\\(^+\\), Na\\(^+\\) etc.~(b)\n\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(185, '2018', '\n\nQuestion\n\nState (a) the principle upon which the lightening in fluorescent tubes operate;\n(b) two factors on which the colour of light from a fluorescent tube depends.\n\n', '', 'State (a) the principle upon which the lightening in fluorescent tubes operate;\n(b) two factors on which the colour of light from a fluorescent tube depends.', '(a) the principle upon which the lightening in fluorescent tubes operate;~(b) two factors on which the colour of light from a fluorescent tube depends.', NULL, '\n\nExplanation\n(a) Lightening in fluorescent tubes operate on the principle of electric discharge through low pressure gas.\n(b) Colour of light from fluorescent tube depend on (i) Nature of the coating inside the tube (ii) Nature of the gas in the tube.\n\n', '', 'Explanation\n(a) Lightening in fluorescent tubes operate on the principle of electric discharge through low pressure gas.\n(b) Colour of light from fluorescent tube depend on (i) Nature of the coating inside the tube (ii) Nature of the gas in the tube.\n\n', '(a) Lightening in fluorescent tubes operate on the principle of electric discharge through low pressure gas.~(b) Colour of light from fluorescent tube depend on (i) Nature of the coating inside the tube (ii) Nature of the gas in the tube.\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(186, '2018', '\n\nQuestion\n\n(a) Define Young modulus of elasticity;\n(b) A spiral spring extends from a length of 10.0 cm to 10.01 cm when a force of 20 N is applied on it. Calculate the force constant of the spring.\n\n', '', '(a) Define Young modulus of elasticity;\n(b) A spiral spring extends from a length of 10.0 cm to 10.01 cm when a force of 20 N is applied on it. Calculate the force constant of the spring.', '(a) Define Young modulus of elasticity;~(b) A spiral spring extends from a length of 10.0 cm to 10.01 cm when a force of 20 N is applied on it. Calculate the force constant of the spring.', NULL, '\n\nExplanation\n(a) Young\'s modulus of electricity is the ratio of tensile or compressive stress to tensile or compressive strain.\n(b) F = Ke\nK = \\(\\frac{f}{e} = \\frac{20}{(10.01 - 10.00)^{-2}}\\)\n= 2.0 x 10 N/M\n\n', '', 'Explanation\n(a) Young\'s modulus of electricity is the ratio of tensile or compressive stress to tensile or compressive strain.\n(b) F = Ke\nK = \\(\\frac{f}{e} = \\frac{20}{(10.01 - 10.00)^{-2}}\\)\n= 2.0 x 10 N/M\n\n', '(a) Young\'s modulus of electricity is the ratio of tensile or compressive stress to tensile or compressive strain.~(b) F = Ke\nK = \\(\\frac{f}{e} = \\frac{20}{(10.01 - 10.00)^{-2}}\\)\n= 2.0 x 10 N/M\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(187, '2018', '\n\nQuestion\n\n(a) Define elastic limit\n(b) State Hookes laws of elasticity\n\n', '', '(a) Define elastic limit\n(b) State Hookes laws of elasticity', '(a) Define elastic limit~(b) State Hookes laws of elasticity', NULL, '\n\nExplanation\n(a) Elastic limit is the point on a load/extension graph beyond which Hooke\'s law is no longer obeyed.\n(b) Hooke\'s law of elasticity states that the extension (e) in an elastic material (wire) is directly proportional to the force (load) applied, provided the elastic limit is not exceeded.\n\n', '', 'Explanation\n(a) Elastic limit is the point on a load/extension graph beyond which Hooke\'s law is no longer obeyed.\n(b) Hooke\'s law of elasticity states that the extension (e) in an elastic material (wire) is directly proportional to the force (load) applied, provided the elastic limit is not exceeded.\n\n', '(a) Elastic limit is the point on a load/extension graph beyond which Hooke\'s law is no longer obeyed.~(b) Hooke\'s law of elasticity states that the extension (e) in an elastic material (wire) is directly proportional to the force (load) applied, provided the elastic limit is not exceeded.\n~d) applied, provided the elastic limit is not exceeded.\n~(e) in an elastic material (wire) is directly proportional to the force (load) applied, provided the elastic limit is not exceeded.\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(188, '2018', '\n\nQuestion\n\n(a) Define surface tension\n(b) State two methods by which the surface tension of a liquid can be reduced\n\n', '', '(a) Define surface tension\n(b) State two methods by which the surface tension of a liquid can be reduced', '(a) Define surface tension~(b) State two methods by which the surface tension of a liquid can be reduced', NULL, '\n\nExplanation\n(a) Surface tension is the property of a liquid which makes its surface behave as though covered by an elastic skin.\n(b) The surface tension of a liquid can be reduced by (i) Heating the liquid (ii) Addition of impurities such as oil, alcohol, soap, camphor, kerosene etc.\n\n', '', 'Explanation\n(a) Surface tension is the property of a liquid which makes its surface behave as though covered by an elastic skin.\n(b) The surface tension of a liquid can be reduced by (i) Heating the liquid (ii) Addition of impurities such as oil, alcohol, soap, camphor, kerosene etc.\n\n', '(a) Surface tension is the property of a liquid which makes its surface behave as though covered by an elastic skin.~(b) The surface tension of a liquid can be reduced by (i) Heating the liquid (ii) Addition of impurities such as oil, alcohol, soap, camphor, kerosene etc.\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(189, '2018', '\n\nQuestion\n\nUsing the kinetic theory of matter, explain the definite structure of solids.\n\n\n', 'Using the kinetic theory of matter, explain the definite structure of solids.\n', '', '', NULL, '\n\nExplanation\nMatter is made up of molecules which are held together by intermolecular forces which make the molecules to vibrate about their mean positions giving the solids definite shape.\n\n', 'Matter is made up of molecules which are held together by intermolecular forces which make the molecules to vibrate about their mean positions giving the solids definite shape.\n\n', '', '', 'WAEC-PK-1', 'WAEC-PHY', 1);
INSERT INTO `exam_simulation2` (`id`, `examYear`, `questions`, `parent_que`, `child_que`, `gen_child_que_desc`, `imageDir`, `explanation`, `parent_que_expl`, `child_que_exp`, `gen_child_que_expl`, `package_id`, `subject_id`, `status`) VALUES
(190, '2018', '\n\nQuestion\n\n(a) Sketch a diagram of a simple pendulum performing simple harmonic motion and indicate positions of maximum potential energy and kinetic energy.\n(b) A body moving with simple harmonic motion in a straight line has velocity, v and acceleration, a, when the instantaneous displacement, x in cm, from its maximum position is given by x = 2.5 sin 0.4 \\(\\pi t\\), where t is in seconds. Determine the magnitude of the maximum (i) veloxity; (ii) acceleration\n(c) A mass m attached to a light spiral is caused to perform simple harmonic motion of frequency\nf = \\(\\frac{1}{2 \\pi} \\sqrt{\\frac{k}{m}}\\), where k is the force constant of the spring.\n(i) Explain the physical significance of \\(\\sqrt{\\frac{k}{m}}\\).\n(ii) If m = 0.30 kg, k = 30Nm\\(^{-1}\\) and the maximum position is 0.015m, calculate the maximum;\n(i) kinetic energy\n(ii) tension in the spring during the motion [g = 10 ms\\(^{-1}\\), \\(\\pi\\) = 3.142]\n\n', '', '(a) Sketch a diagram of a simple pendulum performing simple harmonic motion and indicate positions of maximum potential energy and kinetic energy.\n(b) A body moving with simple harmonic motion in a straight line has velocity, v and acceleration, a, when the instantaneous displacement, x in cm, from its maximum position is given by x = 2.5 sin 0.4 \\(\\pi t\\), where t is in seconds. Determine the magnitude of the maximum (i) veloxity; (ii) acceleration\n(c) A mass m attached to a light spiral is caused to perform simple harmonic motion of frequency\nf = \\(\\frac{1}{2 \\pi} \\sqrt{\\frac{k}{m}}\\), where k is the force constant of the spring.\n(i) Explain the physical significance of \\(\\sqrt{\\frac{k}{m}}\\).\n(ii) If m = 0.30 kg, k = 30Nm\\(^{-1}\\) and the maximum position is 0.015m, calculate the maximum;\n(i) kinetic energy\n(ii) tension in the spring during the motion [g = 10 ms\\(^{-1}\\), \\(\\pi\\) = 3.142]', '(a) Sketch a diagram of a simple pendulum performing simple harmonic motion and indicate positions of maximum potential energy and kinetic energy.~(b) A body moving with simple harmonic motion in a straight line has velocity, v and acceleration, a, when the instantaneous displacement, x in cm, from its maximum position is given by x = 2.5 sin 0.4 \\(\\pi t\\), where t is in seconds. Determine the magnitude of the maximum (i) veloxity; (ii) acceleration\n(c) A mass m attached to a light spiral is caused to perform simple harmonic motion of frequency\nf = \\(\\frac{1}{2 \\pi} \\sqrt{\\frac{k}{m}}\\), where k is the force constant of the spring.\n(i) Explain the physical significance of \\(\\sqrt{\\frac{k}{m}}\\).\n(ii) If m = 0.30 kg, k = 30Nm\\(^{-1}\\) and the maximum position is 0.015m, calculate the maximum;\n(i) kinetic energy\n(ii) tension in the spring during the motion [g = 10 ms\\(^{-1}\\), \\(\\pi\\) = 3.142]~(c) A mass m attached to a light spiral is caused to perform simple harmonic motion of frequency\nf = \\(\\frac{1}{2 \\pi} \\sqrt{\\frac{k}{m}}\\), where k is the force constant of the spring.\n(i) Explain the physical significance of \\(\\sqrt{\\frac{k}{m}}\\).\n(ii) If m = 0.30 kg, k = 30Nm\\(^{-1}\\) and the maximum position is 0.015m, calculate the maximum;\n(i) kinetic energy\n(ii) tension in the spring during the motion [g = 10 ms\\(^{-1}\\), \\(\\pi\\) = 3.142]', NULL, '\n\nExplanation\n\n(a)\n\n(b)(i) V = rw = 2.5 x 10\\(^{-2}\\) x 0.4 \\(\\pi\\)\n= 3.142 x 10\\(^{-2}\\)m/s\n(ii) a = rw\\(^2\\) = 2.5 x 1.0\\(^{-2}\\) x (0.4 \\(\\pi^2\\))\n\n(c) (i) \\(\\frac{k}{m}\\) = w. H is the rate of change of angular displacement. H is measured in radians per second\n(ii) (1) K.E\\(_{\\text{max}} \\frac{1}{2}MV^2_{\\text{max}} = \\frac{1}{2} mr^2w^2\\)\nBut w = \\(\\sqrt{\\frac{k}{m}} = w^2 = \\frac{k}{m} = \\frac{30}{0.3} = 100 rad/s^2\\)\nK.E\\(_{\\text{max}} = \\frac{1}{2}mr^2w^2 = \\frac{1}{2} \\times 0.30 \\times (0.015)^2\\) x 100\n= 3.38 x 10\\(^{3}\\)J\nT\\(_{\\text{max}}\\) = Weight + Ke = Mg + Ke\n= 0.30 x 10 + 30 x 0.015\n= 3.45N\n\n', '', 'Explanation\n\n(a)\n\n(b)(i) V = rw = 2.5 x 10\\(^{-2}\\) x 0.4 \\(\\pi\\)\n= 3.142 x 10\\(^{-2}\\)m/s\n(ii) a = rw\\(^2\\) = 2.5 x 1.0\\(^{-2}\\) x (0.4 \\(\\pi^2\\))\n\n(c) (i) \\(\\frac{k}{m}\\) = w. H is the rate of change of angular displacement. H is measured in radians per second\n(ii) (1) K.E\\(_{\\text{max}} \\frac{1}{2}MV^2_{\\text{max}} = \\frac{1}{2} mr^2w^2\\)\nBut w = \\(\\sqrt{\\frac{k}{m}} = w^2 = \\frac{k}{m} = \\frac{30}{0.3} = 100 rad/s^2\\)\nK.E\\(_{\\text{max}} = \\frac{1}{2}mr^2w^2 = \\frac{1}{2} \\times 0.30 \\times (0.015)^2\\) x 100\n= 3.38 x 10\\(^{3}\\)J\nT\\(_{\\text{max}}\\) = Weight + Ke = Mg + Ke\n= 0.30 x 10 + 30 x 0.015\n= 3.45N\n\n', '(a)\n~(b)(i) V = rw = 2.5 x 10\\(^{-2}\\) x 0.4 \\(\\pi\\)\n= 3.142 x 10\\(^{-2}\\)m/s\n(ii) a = rw\\(^2\\) = 2.5 x 1.0\\(^{-2}\\) x (0.4 \\(\\pi^2\\))\n\n(c) (i) \\(\\frac{k}{m}\\) = w. H is the rate of change of angular displacement. H is measured in radians per second\n(ii) (1) K.E\\(_{\\text{max}} \\frac{1}{2}MV^2_{\\text{max}} = \\frac{1}{2} mr^2w^2\\)\nBut w = \\(\\sqrt{\\frac{k}{m}} = w^2 = \\frac{k}{m} = \\frac{30}{0.3} = 100 rad/s^2\\)\nK.E\\(_{\\text{max}} = \\frac{1}{2}mr^2w^2 = \\frac{1}{2} \\times 0.30 \\times (0.015)^2\\) x 100\n= 3.38 x 10\\(^{3}\\)J\nT\\(_{\\text{max}}\\) = Weight + Ke = Mg + Ke\n= 0.30 x 10 + 30 x 0.015\n= 3.45N\n~(c) (i) \\(\\frac{k}{m}\\) = w. H is the rate of change of angular displacement. H is measured in radians per second\n(ii) (1) K.E\\(_{\\text{max}} \\frac{1}{2}MV^2_{\\text{max}} = \\frac{1}{2} mr^2w^2\\)\nBut w = \\(\\sqrt{\\frac{k}{m}} = w^2 = \\frac{k}{m} = \\frac{30}{0.3} = 100 rad/s^2\\)\nK.E\\(_{\\text{max}} = \\frac{1}{2}mr^2w^2 = \\frac{1}{2} \\times 0.30 \\times (0.015)^2\\) x 100\n= 3.38 x 10\\(^{3}\\)J\nT\\(_{\\text{max}}\\) = Weight + Ke = Mg + Ke\n= 0.30 x 10 + 30 x 0.015\n= 3.45N\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(191, '2018', '\n\nQuestion\n\n(a) Explain specific latent heat\n(b)(i) Describe how the specific latent heat of fusion of ice can be determined by the method of mixtures.\n(ii) State two precautions to be taken to ensure accurate results.\n(c) Steam, at 100°C, is passed into a container of negligible heat capacity containing 20 g of ice and 100 g of water at 0°C, until the ice is completely melted. Determine the total mass of water in the container. [Specific latent heat of steam = 2.3 x 10\\(^3\\) Jg\\(^{-1}\\), specific latent heat of ice = 3.4 x 10\\(^{2}\\) Jg\\(^{-1}\\), specifit heat capacity of water = 4.2 Jg\\(^{-1}\\) K\\(^{-1}\\)]\n\n\n', '', '(a) Explain specific latent heat\n(b)(i) Describe how the specific latent heat of fusion of ice can be determined by the method of mixtures.\n(ii) State two precautions to be taken to ensure accurate results.\n(c) Steam, at 100°C, is passed into a container of negligible heat capacity containing 20 g of ice and 100 g of water at 0°C, until the ice is completely melted. Determine the total mass of water in the container. [Specific latent heat of steam = 2.3 x 10\\(^3\\) Jg\\(^{-1}\\), specific latent heat of ice = 3.4 x 10\\(^{2}\\) Jg\\(^{-1}\\), specifit heat capacity of water = 4.2 Jg\\(^{-1}\\) K\\(^{-1}\\)]\n', '(a) Explain specific latent heat~(b)(i) Describe how the specific latent heat of fusion of ice can be determined by the method of mixtures.\n(ii) State two precautions to be taken to ensure accurate results.\n(c) Steam, at 100°C, is passed into a container of negligible heat capacity containing 20 g of ice and 100 g of water at 0°C, until the ice is completely melted. Determine the total mass of water in the container. [Specific latent heat of steam = 2.3 x 10\\(^3\\) Jg\\(^{-1}\\), specific latent heat of ice = 3.4 x 10\\(^{2}\\) Jg\\(^{-1}\\), specifit heat capacity of water = 4.2 Jg\\(^{-1}\\) K\\(^{-1}\\)]\n~(c) Steam, at 100°C, is passed into a container of negligible heat capacity containing 20 g of ice and 100 g of water at 0°C, until the ice is completely melted. Determine the total mass of water in the container. [Specific latent heat of steam = 2.3 x 10\\(^3\\) Jg\\(^{-1}\\), specific latent heat of ice = 3.4 x 10\\(^{2}\\) Jg\\(^{-1}\\), specifit heat capacity of water = 4.2 Jg\\(^{-1}\\) K\\(^{-1}\\)]\n', NULL, '\n\nExplanation\n(a) Specific latent heat of a substance is the quantity of heat energy required to change a unit mass of a substance from one state to another without change in temperature. The unit specific latent heat is Jkg\\(^{-1}\\).\n(b)(i) Mass of empty calorimeter = M;\nMass of cal. + water = M\\(_2\\)\nMass of cal. + water + ice = M\\(_3\\);\nTemp. of warm water = \\(\\theta_1\\)\nFinal temp. when all the is has melted = \\(\\theta_2\\);\nSp. heat capacity of cal = Cc\nSp. heat capacity of water = C\\(_w\\)\nlatent heat of ice = L\nMass of water = M\\(_2\\) - M\\(_1\\)\nMass of ice = M\\(_3\\) — M\\(_2\\)\nHeat lost by cal. and water = (m\\(_1\\)Cc + (M\\(_2\\) - M\\(_1\\))C x ( \\(\\theta_1\\) - \\(\\theta_2\\))\nHeat gained by ice in melting to water of 0°C = (M\\(_3 - M_2\\))L\nHeat gained by ice water from 0°C = (M\\(^3\\) - M\\(_2\\))C\\(_ w \\theta_2\\),\nTotal heat gained = Heat lost (M\\(_3\\) - M\\(_2\\)L + (M\\(_3\\) - M\\(_2\\))\nC\\(_w \\theta_2\\) = (M\\(_1\\)Cc + (M\\(_2\\) - M\\(_1\\)) C\\(_w\\) (\\(\\theta_2 - \\theta_2\\)\nL = \\(\\frac{M_1C_c + (M_2 - M_1) C_w(\\theta_1 - \\theta_2) - (M_3 M_2)C_w\\theta_2}{(M_3 - M_2)}\\)\n\n(ii) Precautions\n- Stir gently\n- Ensure that ice is dry before adding to water\n- Avoid parallax error in reading the balance and thermometer.\n\n(c) If M\\(_s\\) = mass of steam ; Heat lost by steam at 100°C = M\\(_s\\)l\\(_v\\) = M\\(_s\\) x 2.3 x 10\\(^3\\)(J)\nHeat lost by hot water from 100°C = 0°C ;\n= \\(M_sC_w \\Delta \\theta\\) = \\(M_s\\) x 4.2 x 100 (J) Heat gained by ice at 0\\(^oC\\)\n= M\\(_{\\text{ice}}\\) L\\(_f\\) = 20 x 3.4 x 10\\(^2\\) = 6800J;\nHeat lost by steam + hot water = Heat gained by ice;\nM\\(_sl_v\\) + M\\(_s\\)C\\(_w\\)\\(\\Delta\\) = M\\(_{\\text{ice}}\\)L\\(_f\\);\nM\\(_s\\) (2.3 x 10\\(^3\\) + 4.2 x 100) = 6800;\nM\\(_s\\) = \\(\\frac{6800}{2720}\\) = 2.5g\nMass of water in container = mass of melted ice + mass of water + mass of condensed steam.\n= 20 + 100 + 2.5 = 122.5g\n\n', '', 'Explanation\n(a) Specific latent heat of a substance is the quantity of heat energy required to change a unit mass of a substance from one state to another without change in temperature. The unit specific latent heat is Jkg\\(^{-1}\\).\n(b)(i) Mass of empty calorimeter = M;\nMass of cal. + water = M\\(_2\\)\nMass of cal. + water + ice = M\\(_3\\);\nTemp. of warm water = \\(\\theta_1\\)\nFinal temp. when all the is has melted = \\(\\theta_2\\);\nSp. heat capacity of cal = Cc\nSp. heat capacity of water = C\\(_w\\)\nlatent heat of ice = L\nMass of water = M\\(_2\\) - M\\(_1\\)\nMass of ice = M\\(_3\\) — M\\(_2\\)\nHeat lost by cal. and water = (m\\(_1\\)Cc + (M\\(_2\\) - M\\(_1\\))C x ( \\(\\theta_1\\) - \\(\\theta_2\\))\nHeat gained by ice in melting to water of 0°C = (M\\(_3 - M_2\\))L\nHeat gained by ice water from 0°C = (M\\(^3\\) - M\\(_2\\))C\\(_ w \\theta_2\\),\nTotal heat gained = Heat lost (M\\(_3\\) - M\\(_2\\)L + (M\\(_3\\) - M\\(_2\\))\nC\\(_w \\theta_2\\) = (M\\(_1\\)Cc + (M\\(_2\\) - M\\(_1\\)) C\\(_w\\) (\\(\\theta_2 - \\theta_2\\)\nL = \\(\\frac{M_1C_c + (M_2 - M_1) C_w(\\theta_1 - \\theta_2) - (M_3 M_2)C_w\\theta_2}{(M_3 - M_2)}\\)\n\n(ii) Precautions\n- Stir gently\n- Ensure that ice is dry before adding to water\n- Avoid parallax error in reading the balance and thermometer.\n\n(c) If M\\(_s\\) = mass of steam ; Heat lost by steam at 100°C = M\\(_s\\)l\\(_v\\) = M\\(_s\\) x 2.3 x 10\\(^3\\)(J)\nHeat lost by hot water from 100°C = 0°C ;\n= \\(M_sC_w \\Delta \\theta\\) = \\(M_s\\) x 4.2 x 100 (J) Heat gained by ice at 0\\(^oC\\)\n= M\\(_{\\text{ice}}\\) L\\(_f\\) = 20 x 3.4 x 10\\(^2\\) = 6800J;\nHeat lost by steam + hot water = Heat gained by ice;\nM\\(_sl_v\\) + M\\(_s\\)C\\(_w\\)\\(\\Delta\\) = M\\(_{\\text{ice}}\\)L\\(_f\\);\nM\\(_s\\) (2.3 x 10\\(^3\\) + 4.2 x 100) = 6800;\nM\\(_s\\) = \\(\\frac{6800}{2720}\\) = 2.5g\nMass of water in container = mass of melted ice + mass of water + mass of condensed steam.\n= 20 + 100 + 2.5 = 122.5g\n\n', '(a) Specific latent heat of a substance is the quantity of heat energy required to change a unit mass of a substance from one state to another without change in temperature. The unit specific latent heat is Jkg\\(^{-1}\\).~(b)(i) Mass of empty calorimeter = M;\nMass of cal. + water = M\\(_2\\)\nMass of cal. + water + ice = M\\(_3\\);\nTemp. of warm water = \\(\\theta_1\\)\nFinal temp. when all the is has melted = \\(\\theta_2\\);\nSp. heat capacity of cal = Cc\nSp. heat capacity of water = C\\(_w\\)\nlatent heat of ice = L\nMass of water = M\\(_2\\) - M\\(_1\\)\nMass of ice = M\\(_3\\) — M\\(_2\\)\nHeat lost by cal. and water = (m\\(_1\\)Cc + (M\\(_2\\) - M\\(_1\\))C x ( \\(\\theta_1\\) - \\(\\theta_2\\))\nHeat gained by ice in melting to water of 0°C = (M\\(_3 - M_2\\))L\nHeat gained by ice water from 0°C = (M\\(^3\\) - M\\(_2\\))C\\(_ w \\theta_2\\),\nTotal heat gained = Heat lost (M\\(_3\\) - M\\(_2\\)L + (M\\(_3\\) - M\\(_2\\))\nC\\(_w \\theta_2\\) = (M\\(_1\\)Cc + (M\\(_2\\) - M\\(_1\\)) C\\(_w\\) (\\(\\theta_2 - \\theta_2\\)\nL = \\(\\frac{M_1C_c + (M_2 - M_1) C_w(\\theta_1 - \\theta_2) - (M_3 M_2)C_w\\theta_2}{(M_3 - M_2)}\\)\n\n(ii) Precautions\n- Stir gently\n- Ensure that ice is dry before adding to water\n- Avoid parallax error in reading the balance and thermometer.\n\n(c) If M\\(_s\\) = mass of steam ; Heat lost by steam at 100°C = M\\(_s\\)l\\(_v\\) = M\\(_s\\) x 2.3 x 10\\(^3\\)(J)\nHeat lost by hot water from 100°C = 0°C ;\n= \\(M_sC_w \\Delta \\theta\\) = \\(M_s\\) x 4.2 x 100 (J) Heat gained by ice at 0\\(^oC\\)\n= M\\(_{\\text{ice}}\\) L\\(_f\\) = 20 x 3.4 x 10\\(^2\\) = 6800J;\nHeat lost by steam + hot water = Heat gained by ice;\nM\\(_sl_v\\) + M\\(_s\\)C\\(_w\\)\\(\\Delta\\) = M\\(_{\\text{ice}}\\)L\\(_f\\);\nM\\(_s\\) (2.3 x 10\\(^3\\) + 4.2 x 100) = 6800;\nM\\(_s\\) = \\(\\frac{6800}{2720}\\) = 2.5g\nMass of water in container = mass of melted ice + mass of water + mass of condensed steam.\n= 20 + 100 + 2.5 = 122.5g\n~(c) If M\\(_s\\) = mass of steam ; Heat lost by steam at 100°C = M\\(_s\\)l\\(_v\\) = M\\(_s\\) x 2.3 x 10\\(^3\\)(J)\nHeat lost by hot water from 100°C = 0°C ;\n= \\(M_sC_w \\Delta \\theta\\) = \\(M_s\\) x 4.2 x 100 (J) Heat gained by ice at 0\\(^oC\\)\n= M\\(_{\\text{ice}}\\) L\\(_f\\) = 20 x 3.4 x 10\\(^2\\) = 6800J;\nHeat lost by steam + hot water = Heat gained by ice;\nM\\(_sl_v\\) + M\\(_s\\)C\\(_w\\)\\(\\Delta\\) = M\\(_{\\text{ice}}\\)L\\(_f\\);\nM\\(_s\\) (2.3 x 10\\(^3\\) + 4.2 x 100) = 6800;\nM\\(_s\\) = \\(\\frac{6800}{2720}\\) = 2.5g\nMass of water in container = mass of melted ice + mass of water + mass of condensed steam.\n= 20 + 100 + 2.5 = 122.5g\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(192, '2018', '\n\nQuestion\n\n(a) Explain (i) refraction of a wave;\n(ii) critical angle\n\n(b) State two conditions necessary for (i) total internal reflection of a wave to occur\n(ii) interference wave patterns to be formed\n\n(c) The distance between two successive crests of a water wave travelling at 3.6ms\\(^{-1}\\) is 0.45m, calculate the frequency of the wave\n\n(d) A ray of light is incident at an angle of 30° at an air-glass interface.\n(i) Draw a ray diagram to show the deviation of the ray in the glass.\n(ii) Determine the angle of deviation. [Refractive index of glass = 1.50]\n\n', '', '(a) Explain (i) refraction of a wave;\n(ii) critical angle\n\n(b) State two conditions necessary for (i) total internal reflection of a wave to occur\n(ii) interference wave patterns to be formed\n\n(c) The distance between two successive crests of a water wave travelling at 3.6ms\\(^{-1}\\) is 0.45m, calculate the frequency of the wave\n\n(d) A ray of light is incident at an angle of 30° at an air-glass interface.\n(i) Draw a ray diagram to show the deviation of the ray in the glass.\n(ii) Determine the angle of deviation. [Refractive index of glass = 1.50]', '(a) Explain (i) refraction of a wave;\n(ii) critical angle\n~(b) State two conditions necessary for (i) total internal reflection of a wave to occur\n(ii) interference wave patterns to be formed\n\n(c) The distance between two successive crests of a water wave travelling at 3.6ms\\(^{-1}\\) is 0.45m, calculate the frequency of the wave\n\n(d) A ray of light is incident at an angle of 30° at an air-glass interface.\n(i) Draw a ray diagram to show the deviation of the ray in the glass.\n(ii) Determine the angle of deviation. [Refractive index of glass = 1.50]~(c) The distance between two successive crests of a water wave travelling at 3.6ms\\(^{-1}\\) is 0.45m, calculate the frequency of the wave\n\n(d) A ray of light is incident at an angle of 30° at an air-glass interface.\n(i) Draw a ray diagram to show the deviation of the ray in the glass.\n(ii) Determine the angle of deviation. [Refractive index of glass = 1.50]~(d) A ray of light is incident at an angle of 30° at an air-glass interface.\n(i) Draw a ray diagram to show the deviation of the ray in the glass.\n(ii) Determine the angle of deviation. [Refractive index of glass = 1.50]', NULL, '\n\nExplanation\n(a)(i) Refraction of a wave is the change in the direction of the wave as it moves from one medium to another of different density.\n(ii) Critical angle is the angle of incidence in the denser medium beyond which total internal reflection occurs.\n\n(b) (i) The conditions necessary for total internal reflection of wave to occur are:-\n- The wave must be travelling from a denser medium to a less dense medium\n- The angle of incidence of the wave in the denser medium must be greater than the critical angle of the medium.\n(ii) The conditions necessary for interference pattern to be formed are:-\n- Two or more sources must be involved.\n- The distance between the sources must be of the order of the wave length of the waves\n\n(c) V = \\(f \\lambda\\)\nF = \\(\\frac{v}{\\lambda} = \\frac{3.6}{0.45}\\)\n= 8.0Hz\n\n(d) (i)\n\n(ii) n = \\(\\frac{sin i}{sin r}\\)\nr = sin\\(^{-1} (\\frac{sin i}{n})\\)\n= sin\\(^{-1} (\\frac{sin 30}{1.5})\\)\n= 19.5\\(^o\\)\nDeviation = i - r = 30 - 19.5\n= 10.5\\(^o\\)\n\n\n\n\n', '', 'Explanation\n(a)(i) Refraction of a wave is the change in the direction of the wave as it moves from one medium to another of different density.\n(ii) Critical angle is the angle of incidence in the denser medium beyond which total internal reflection occurs.\n\n(b) (i) The conditions necessary for total internal reflection of wave to occur are:-\n- The wave must be travelling from a denser medium to a less dense medium\n- The angle of incidence of the wave in the denser medium must be greater than the critical angle of the medium.\n(ii) The conditions necessary for interference pattern to be formed are:-\n- Two or more sources must be involved.\n- The distance between the sources must be of the order of the wave length of the waves\n\n(c) V = \\(f \\lambda\\)\nF = \\(\\frac{v}{\\lambda} = \\frac{3.6}{0.45}\\)\n= 8.0Hz\n\n(d) (i)\n\n(ii) n = \\(\\frac{sin i}{sin r}\\)\nr = sin\\(^{-1} (\\frac{sin i}{n})\\)\n= sin\\(^{-1} (\\frac{sin 30}{1.5})\\)\n= 19.5\\(^o\\)\nDeviation = i - r = 30 - 19.5\n= 10.5\\(^o\\)\n\n\n\n\n', '(a)(i) Refraction of a wave is the change in the direction of the wave as it moves from one medium to another of different density.\n(ii) Critical angle is the angle of incidence in the denser medium beyond which total internal reflection occurs.\n~(b) (i) The conditions necessary for total internal reflection of wave to occur are:-\n- The wave must be travelling from a denser medium to a less dense medium\n- The angle of incidence of the wave in the denser medium must be greater than the critical angle of the medium.\n(ii) The conditions necessary for interference pattern to be formed are:-\n- Two or more sources must be involved.\n- The distance between the sources must be of the order of the wave length of the waves\n\n(c) V = \\(f \\lambda\\)\nF = \\(\\frac{v}{\\lambda} = \\frac{3.6}{0.45}\\)\n= 8.0Hz\n\n(d) (i)\n\n(ii) n = \\(\\frac{sin i}{sin r}\\)\nr = sin\\(^{-1} (\\frac{sin i}{n})\\)\n= sin\\(^{-1} (\\frac{sin 30}{1.5})\\)\n= 19.5\\(^o\\)\nDeviation = i - r = 30 - 19.5\n= 10.5\\(^o\\)\n\n\n\n~(c) V = \\(f \\lambda\\)\nF = \\(\\frac{v}{\\lambda} = \\frac{3.6}{0.45}\\)\n= 8.0Hz\n\n(d) (i)\n\n(ii) n = \\(\\frac{sin i}{sin r}\\)\nr = sin\\(^{-1} (\\frac{sin i}{n})\\)\n= sin\\(^{-1} (\\frac{sin 30}{1.5})\\)\n= 19.5\\(^o\\)\nDeviation = i - r = 30 - 19.5\n= 10.5\\(^o\\)\n\n\n\n~(d) (i)\n\n(ii) n = \\(\\frac{sin i}{sin r}\\)\nr = sin\\(^{-1} (\\frac{sin i}{n})\\)\n= sin\\(^{-1} (\\frac{sin 30}{1.5})\\)\n= 19.5\\(^o\\)\nDeviation = i - r = 30 - 19.5\n= 10.5\\(^o\\)\n\n\n\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(193, '2018', '\n\nQuestion\n\n(a) (i) Explain (I) electric potential; (II) electric potential energy\n(ii) State the SI unit of each of the term in (a)(i) above\n(b) An isolated electrically charged sphere of radius, r, and charge, Q, is supported on an insulator in air of permitivity, \\(\\varepsilon_o\\). Write down;\n(i) an expression for the electric field intensity on the surface of the sphere;\n(ii) an expression for the electric potential at the surface of the sphere;\n(iii) a relationship between the electric field intensity and the electric potential at the surface of the sphere\n\n(c) The plates of a parallel plate capacitor, 5.0 x 10\\(^{-3}\\) m apart are maintained at a potential difference of 5.0 x 10\\(^{4}\\) V. Calculate the magnitude f the\n(i) electric field intensity between the plates\n(ii) force on the electron\n(iii) acceleration of the elctron\n[electronic charge = 1.60 x 10\\(^{-19}\\)C, mass of electron = 9.1 \\times 10\\(^{-31}\\) kg]\n\n', '', '(a) (i) Explain (I) electric potential; (II) electric potential energy\n(ii) State the SI unit of each of the term in (a)(i) above\n(b) An isolated electrically charged sphere of radius, r, and charge, Q, is supported on an insulator in air of permitivity, \\(\\varepsilon_o\\). Write down;\n(i) an expression for the electric field intensity on the surface of the sphere;\n(ii) an expression for the electric potential at the surface of the sphere;\n(iii) a relationship between the electric field intensity and the electric potential at the surface of the sphere\n\n(c) The plates of a parallel plate capacitor, 5.0 x 10\\(^{-3}\\) m apart are maintained at a potential difference of 5.0 x 10\\(^{4}\\) V. Calculate the magnitude f the\n(i) electric field intensity between the plates\n(ii) force on the electron\n(iii) acceleration of the elctron\n[electronic charge = 1.60 x 10\\(^{-19}\\)C, mass of electron = 9.1 \\times 10\\(^{-31}\\) kg]', '(a) (i) Explain (I) electric potential; (II) electric potential energy\n(ii) State the SI unit of each of the term in (a)(i) above~(b) An isolated electrically charged sphere of radius, r, and charge, Q, is supported on an insulator in air of permitivity, \\(\\varepsilon_o\\). Write down;\n(i) an expression for the electric field intensity on the surface of the sphere;\n(ii) an expression for the electric potential at the surface of the sphere;\n(iii) a relationship between the electric field intensity and the electric potential at the surface of the sphere\n\n(c) The plates of a parallel plate capacitor, 5.0 x 10\\(^{-3}\\) m apart are maintained at a potential difference of 5.0 x 10\\(^{4}\\) V. Calculate the magnitude f the\n(i) electric field intensity between the plates\n(ii) force on the electron\n(iii) acceleration of the elctron\n[electronic charge = 1.60 x 10\\(^{-19}\\)C, mass of electron = 9.1 \\times 10\\(^{-31}\\) kg]~(c) The plates of a parallel plate capacitor, 5.0 x 10\\(^{-3}\\) m apart are maintained at a potential difference of 5.0 x 10\\(^{4}\\) V. Calculate the magnitude f the\n(i) electric field intensity between the plates\n(ii) force on the electron\n(iii) acceleration of the elctron\n[electronic charge = 1.60 x 10\\(^{-19}\\)C, mass of electron = 9.1 \\times 10\\(^{-31}\\) kg]', NULL, '\n\nExplanation\n(a)(i) Electric potential is the work done in moving a unit positive charge from infinity to a point in an electric field. H is measured in volts or J/C.\n(ii) Electric potential energy is the work done or energy required to move a charged particle from infinity to a point in an electric field. The unit is Joules.\n(ii) Electric potential unit is in volts v. Electric potential energy is measured in Joules (J)\n\n(b) (i) Electric field intensity E = \\(\\frac{Q}{4 \\pi \\varepsilon_or^2}\\)\n(ii) Electric potential V = \\(\\frac{Q}{4 \\pi \\varepsilon_or}\\)\n(iii) \\(\\frac{E}{v} = \\frac{I}{r}\\) OR E = \\(\\frac{Y}{4 \\pi \\varepsilon_o r}\\)\n\n(c) E = \\(\\frac{v}{d} = \\frac{5 \\times 10^4}{5 \\times 10^{-3}} = 10^7 v/m\\)\n(ii) F = qE = 1.6 x 10\\(^{-19}\\) x 10\\(^{7}\\) = 1.6 x 10\\(^{-12}\\) N\n(iii) F = ma\na = \\(\\frac{F}{m}\\) = \\(\\frac{1.6 \\times 10^{-12}}{9.1 \\times 10^{-31}}\\)\n= 1.76 x 10\\(^{18}\\) m/s\\(^2\\)\n\n', '', 'Explanation\n(a)(i) Electric potential is the work done in moving a unit positive charge from infinity to a point in an electric field. H is measured in volts or J/C.\n(ii) Electric potential energy is the work done or energy required to move a charged particle from infinity to a point in an electric field. The unit is Joules.\n(ii) Electric potential unit is in volts v. Electric potential energy is measured in Joules (J)\n\n(b) (i) Electric field intensity E = \\(\\frac{Q}{4 \\pi \\varepsilon_or^2}\\)\n(ii) Electric potential V = \\(\\frac{Q}{4 \\pi \\varepsilon_or}\\)\n(iii) \\(\\frac{E}{v} = \\frac{I}{r}\\) OR E = \\(\\frac{Y}{4 \\pi \\varepsilon_o r}\\)\n\n(c) E = \\(\\frac{v}{d} = \\frac{5 \\times 10^4}{5 \\times 10^{-3}} = 10^7 v/m\\)\n(ii) F = qE = 1.6 x 10\\(^{-19}\\) x 10\\(^{7}\\) = 1.6 x 10\\(^{-12}\\) N\n(iii) F = ma\na = \\(\\frac{F}{m}\\) = \\(\\frac{1.6 \\times 10^{-12}}{9.1 \\times 10^{-31}}\\)\n= 1.76 x 10\\(^{18}\\) m/s\\(^2\\)\n\n', '(a)(i) Electric potential is the work done in moving a unit positive charge from infinity to a point in an electric field. H is measured in volts or J/C.\n(ii) Electric potential energy is the work done or energy required to move a charged particle from infinity to a point in an electric field. The unit is Joules.\n(ii) Electric potential unit is in volts v. Electric potential energy is measured in Joules (J)\n~(b) (i) Electric field intensity E = \\(\\frac{Q}{4 \\pi \\varepsilon_or^2}\\)\n(ii) Electric potential V = \\(\\frac{Q}{4 \\pi \\varepsilon_or}\\)\n(iii) \\(\\frac{E}{v} = \\frac{I}{r}\\) OR E = \\(\\frac{Y}{4 \\pi \\varepsilon_o r}\\)\n\n(c) E = \\(\\frac{v}{d} = \\frac{5 \\times 10^4}{5 \\times 10^{-3}} = 10^7 v/m\\)\n(ii) F = qE = 1.6 x 10\\(^{-19}\\) x 10\\(^{7}\\) = 1.6 x 10\\(^{-12}\\) N\n(iii) F = ma\na = \\(\\frac{F}{m}\\) = \\(\\frac{1.6 \\times 10^{-12}}{9.1 \\times 10^{-31}}\\)\n= 1.76 x 10\\(^{18}\\) m/s\\(^2\\)\n~(c) E = \\(\\frac{v}{d} = \\frac{5 \\times 10^4}{5 \\times 10^{-3}} = 10^7 v/m\\)\n(ii) F = qE = 1.6 x 10\\(^{-19}\\) x 10\\(^{7}\\) = 1.6 x 10\\(^{-12}\\) N\n(iii) F = ma\na = \\(\\frac{F}{m}\\) = \\(\\frac{1.6 \\times 10^{-12}}{9.1 \\times 10^{-31}}\\)\n= 1.76 x 10\\(^{18}\\) m/s\\(^2\\)\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(194, '2018', '\n\nQuestion\n\nExplain wave-particle paradox\n\n', 'Explain wave-particle paradox', '', '', NULL, '\n\nExplanation\nThe wave particle paradox supposes that light exhibits wave properties such as reflection, diffraction, interference etc. and can also exist as a particle called photons. However light cannot behave as a wave and as particle at the same time.\n\n', 'The wave particle paradox supposes that light exhibits wave properties such as reflection, diffraction, interference etc. and can also exist as a particle called photons. However light cannot behave as a wave and as particle at the same time.\n\n', '', '', 'WAEC-PK-1', 'WAEC-PHY', 1),
(195, '2018', '\n\nQuestion\n\n(a) State two;\n(i) differences between nuclear fusion and nuclear fission;\n(ii) peaceful uses of atomic energy\n\n(b)(i) Explain chain reaction\n(ii) State (I) one condition necessary for chain reaction to occur.\n(II) two components in a nuclear reactor used to control chain reaction.\n\n(c)(i) A nuclear reaction is given \\(^2_1H + ^3_1H = ^4_0n\\) + energy\nWhat type of nuclear reaction is it?\n(ii) The isotope of a nuclide has a half life of 5.40 x 10\\(^3\\) s, Calculate its decay constant.\n\n', '', '(a) State two;\n(i) differences between nuclear fusion and nuclear fission;\n(ii) peaceful uses of atomic energy\n\n(b)(i) Explain chain reaction\n(ii) State (I) one condition necessary for chain reaction to occur.\n(II) two components in a nuclear reactor used to control chain reaction.\n\n(c)(i) A nuclear reaction is given \\(^2_1H + ^3_1H = ^4_0n\\) + energy\nWhat type of nuclear reaction is it?\n(ii) The isotope of a nuclide has a half life of 5.40 x 10\\(^3\\) s, Calculate its decay constant.', '(a) State two;\n(i) differences between nuclear fusion and nuclear fission;\n(ii) peaceful uses of atomic energy\n~(b)(i) Explain chain reaction\n(ii) State (I) one condition necessary for chain reaction to occur.\n(II) two components in a nuclear reactor used to control chain reaction.\n\n(c)(i) A nuclear reaction is given \\(^2_1H + ^3_1H = ^4_0n\\) + energy\nWhat type of nuclear reaction is it?\n(ii) The isotope of a nuclide has a half life of 5.40 x 10\\(^3\\) s, Calculate its decay constant.~(c)(i) A nuclear reaction is given \\(^2_1H + ^3_1H = ^4_0n\\) + energy\nWhat type of nuclear reaction is it?\n(ii) The isotope of a nuclide has a half life of 5.40 x 10\\(^3\\) s, Calculate its decay constant.', NULL, '\n\nExplanation\n(a)(i) In nuclear fission reaction results in splitting a heavy nucleus into two parts of nearly equal masses while in nuclear fusion process two light nuclei combine together to form a heavier nucleus.\n(ii) Nuclear fission releases less energy per nuclear while nuclear fusion releases more energy per nuclear.\n(iii) Nuclear fission can take place at room temperature while nuclear fusion takes place at very high temperature.\n(ii) Peaceful uses of atomic energy are:\n- Production of laser beams\n- Production of x-rays\n- Production of energy in discharge tubes.\n\n(b)(i) The neutrons produced as a result of fission of a nucleus can cause further fission if they strike other nuclei. This rapid multiplication of a fission process by fissile neutrons is called chain reaction.\n(ii) (i) Conditions necessary for chain reaction to occur are:\n- The mass of fissile material must exceed the critical mass\n- In addition to the two nuclei produced during the fission, one to three neutrons must be produced per parent nucleus.\n(ii) Components in a nuclear reactor used to control chain reactions are;\n- Boron steel\n- Heavy water or denterium oxide\n- Graphite or carbon rod.\n\n(c)(i) The type of nuclear reaction is fusion.\nT\\(_{\\frac{1}{2}} = \\frac{0.693}{\\lambda}\\); \\(\\lambda = \\frac{0.693}{T_{\\frac{1}{2}}}\\)\n= \\(\\frac{0.693}{5.40 \\times 10^{3}}\\)\n= 1.28 x 10\\(^{-4}\\)S\\(^{-1}\\)\n\n\n', '', 'Explanation\n(a)(i) In nuclear fission reaction results in splitting a heavy nucleus into two parts of nearly equal masses while in nuclear fusion process two light nuclei combine together to form a heavier nucleus.\n(ii) Nuclear fission releases less energy per nuclear while nuclear fusion releases more energy per nuclear.\n(iii) Nuclear fission can take place at room temperature while nuclear fusion takes place at very high temperature.\n(ii) Peaceful uses of atomic energy are:\n- Production of laser beams\n- Production of x-rays\n- Production of energy in discharge tubes.\n\n(b)(i) The neutrons produced as a result of fission of a nucleus can cause further fission if they strike other nuclei. This rapid multiplication of a fission process by fissile neutrons is called chain reaction.\n(ii) (i) Conditions necessary for chain reaction to occur are:\n- The mass of fissile material must exceed the critical mass\n- In addition to the two nuclei produced during the fission, one to three neutrons must be produced per parent nucleus.\n(ii) Components in a nuclear reactor used to control chain reactions are;\n- Boron steel\n- Heavy water or denterium oxide\n- Graphite or carbon rod.\n\n(c)(i) The type of nuclear reaction is fusion.\nT\\(_{\\frac{1}{2}} = \\frac{0.693}{\\lambda}\\); \\(\\lambda = \\frac{0.693}{T_{\\frac{1}{2}}}\\)\n= \\(\\frac{0.693}{5.40 \\times 10^{3}}\\)\n= 1.28 x 10\\(^{-4}\\)S\\(^{-1}\\)\n\n\n', '(a)(i) In nuclear fission reaction results in splitting a heavy nucleus into two parts of nearly equal masses while in nuclear fusion process two light nuclei combine together to form a heavier nucleus.\n(ii) Nuclear fission releases less energy per nuclear while nuclear fusion releases more energy per nuclear.\n(iii) Nuclear fission can take place at room temperature while nuclear fusion takes place at very high temperature.\n(ii) Peaceful uses of atomic energy are:\n- Production of laser beams\n- Production of x-rays\n- Production of energy in discharge tubes.\n~(b)(i) The neutrons produced as a result of fission of a nucleus can cause further fission if they strike other nuclei. This rapid multiplication of a fission process by fissile neutrons is called chain reaction.\n(ii) (i) Conditions necessary for chain reaction to occur are:\n- The mass of fissile material must exceed the critical mass\n- In addition to the two nuclei produced during the fission, one to three neutrons must be produced per parent nucleus.\n(ii) Components in a nuclear reactor used to control chain reactions are;\n- Boron steel\n- Heavy water or denterium oxide\n- Graphite or carbon rod.\n\n(c)(i) The type of nuclear reaction is fusion.\nT\\(_{\\frac{1}{2}} = \\frac{0.693}{\\lambda}\\); \\(\\lambda = \\frac{0.693}{T_{\\frac{1}{2}}}\\)\n= \\(\\frac{0.693}{5.40 \\times 10^{3}}\\)\n= 1.28 x 10\\(^{-4}\\)S\\(^{-1}\\)\n\n~(c)(i) The type of nuclear reaction is fusion.\nT\\(_{\\frac{1}{2}} = \\frac{0.693}{\\lambda}\\); \\(\\lambda = \\frac{0.693}{T_{\\frac{1}{2}}}\\)\n= \\(\\frac{0.693}{5.40 \\times 10^{3}}\\)\n= 1.28 x 10\\(^{-4}\\)S\\(^{-1}\\)\n\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(196, '2018', '\n\nQuestion\n\nIn his first attempt, a long jumper took off from the spring board with a speed of 8 ms\\(^{-1}\\) at 30° to the horizontal. He makes a second attempt with the same speed at 45° to the horizontal. Given that the expression for the horizontal range of a projectile is \\(\\frac{u^2 sin 2\\theta}{g}\\) where all the symbols have their usual meanings, show that he gains a distance of 0.8576 m in his second attempt. [g = 10ms\\(^{-2}\\)]\n\n', 'In his first attempt, a long jumper took off from the spring board with a speed of 8 ms\\(^{-1}\\) at 30° to the horizontal. He makes a second attempt with the same speed at 45° to the horizontal. Given that the expression for the horizontal range of a projectile is \\(\\frac{u^2 sin 2\\theta}{g}\\) where all the symbols have their usual meanings, show that he gains a distance of 0.8576 m in his second attempt. [g = 10ms\\(^{-2}\\)]', '', '', NULL, '\n\nExplanation\n1st Attempt R = \\(\\frac{U^2 sin 2 \\theta}{g} = \\frac{8^2 sin 60}{10} = 5.5424m\\)\n2nd Attempt R = \\(U^2sin 2 \\theta\\)\n= \\(\\frac{8^2sin 90}{10}\\) = 6.4m\nGain = 6.4 - 5.5425 = 0.8576m\n\n', '1st Attempt R = \\(\\frac{U^2 sin 2 \\theta}{g} = \\frac{8^2 sin 60}{10} = 5.5424m\\)\n2nd Attempt R = \\(U^2sin 2 \\theta\\)\n= \\(\\frac{8^2sin 90}{10}\\) = 6.4m\nGain = 6.4 - 5.5425 = 0.8576m\n\n', '', '', 'WAEC-PK-1', 'WAEC-PHY', 1),
(197, '2018', '\n\nQuestion\n\nThe diagram above illustrates the path ABC, in a vertical x z plane, of a bullet shot into the air at an angle above the horizontal. Copy the diagram, and, using arrows, indicate the relative magnitudes and directions of the vertical and horizontal components of the velocities of the bullet at the point A, B and C.\n\n\n', 'The diagram above illustrates the path ABC, in a vertical x z plane, of a bullet shot into the air at an angle above the horizontal. Copy the diagram, and, using arrows, indicate the relative magnitudes and directions of the vertical and horizontal components of the velocities of the bullet at the point A, B and C.\n', '', '', NULL, '\n\nExplanation\n', '', '', '', 'WAEC-PK-1', 'WAEC-PHY', 1),
(198, '2018', '\n\nQuestion\n\n(a) For a water voltameter, identify the;\n(i) electrolyte;\n(ii) electrodes;\n(iii) substances deposited on the electrodes\n(b)State the ratio of the volume of the substances deposited in (a) (iii) above\n\n', '', '(a) For a water voltameter, identify the;\n(i) electrolyte;\n(ii) electrodes;\n(iii) substances deposited on the electrodes\n(b)State the ratio of the volume of the substances deposited in (a) (iii) above', '(a) For a water voltameter, identify the;\n(i) electrolyte;\n(ii) electrodes;\n(iii) substances deposited on the electrodes~(b)State the ratio of the volume of the substances deposited in (a) (iii) above', NULL, '\n\nExplanation\n(a) (i) Electrolyte = dilute tetraoxosulphate (vi) acid\n(ii) Electrode = platinum\n(iii) Hydrogen and oxygen are deposited at the cathode and anode respectively.\n(b) Volume of hydrogen is twice that of oxygen.\n\n', '', 'Explanation\n(a) (i) Electrolyte = dilute tetraoxosulphate (vi) acid\n(ii) Electrode = platinum\n(iii) Hydrogen and oxygen are deposited at the cathode and anode respectively.\n(b) Volume of hydrogen is twice that of oxygen.\n\n', '(a) (i) Electrolyte = dilute tetraoxosulphate (vi) acid\n(ii) Electrode = platinum\n(iii) Hydrogen and oxygen are deposited at the cathode and anode respectively.~(b) Volume of hydrogen is twice that of oxygen.\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(199, '2018', '\n\nQuestion\n\nA parallel beam of unpolarised light is incident on a plane glass surface at an angle of 58\\(^{o}\\) to the normal. If the reflected beam is completely polarised, calculate the refractive index of the glass.\n\n', 'A parallel beam of unpolarised light is incident on a plane glass surface at an angle of 58\\(^{o}\\) to the normal. If the reflected beam is completely polarised, calculate the refractive index of the glass.', '', '', NULL, '\n\nExplanation\nWhen i = 58°, r = 90\' - 58° = 32°\nn = \\(\\frac{sin i}{sin r}\\) = \\(\\frac{sin 58^o}{sin 32^o}\\)\n= 1.60\n\n', 'When i = 58°, r = 90\' - 58° = 32°\nn = \\(\\frac{sin i}{sin r}\\) = \\(\\frac{sin 58^o}{sin 32^o}\\)\n= 1.60\n\n', '', '', 'WAEC-PK-1', 'WAEC-PHY', 1),
(200, '2018', '\n\nQuestion\n\nState three methods of polarising an unpolarised light\n\n', 'State three methods of polarising an unpolarised light', '', '', NULL, '\n\nExplanation\nUnpolarised light can be polarised.\n(i) By reflection\n(ii) By double refraction\n(iii) Polarising pile of plates\n(iv). Selective absorption by crystals.\n(v) Passing unpolarised light through a amid.\n\n', '', 'Explanation\nUnpolarised light can be polarised.\n(i) By reflection\n(ii) By double refraction\n(iii) Polarising pile of plates\n(iv). Selective absorption by crystals.\n(v) Passing unpolarised light through a amid.\n\n', '', 'WAEC-PK-1', 'WAEC-PHY', 1),
(201, '2018', '\n\nQuestion\n\n(a) Define diffusion\n(b) State Graham\'s law of diffusion\n\n', '', '(a) Define diffusion\n(b) State Graham\'s law of diffusion', '(a) Define diffusion~(b) State Graham\'s law of diffusion', NULL, '\n\nExplanation\n(a) Diffusion is the process by which substances mix with one another due to the random motion of their molecules.\n(b) Graham\'s law of diffusion states that the rate of diffusion of gases is inversely proportional to the square - root of their densities provided temperature remains constant.\n\n', '', 'Explanation\n(a) Diffusion is the process by which substances mix with one another due to the random motion of their molecules.\n(b) Graham\'s law of diffusion states that the rate of diffusion of gases is inversely proportional to the square - root of their densities provided temperature remains constant.\n\n', '(a) Diffusion is the process by which substances mix with one another due to the random motion of their molecules.~(b) Graham\'s law of diffusion states that the rate of diffusion of gases is inversely proportional to the square - root of their densities provided temperature remains constant.\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(202, '2018', '\n\nQuestion\n\nA Spiral spring of natural length 20.00cm has a scale hanging freely in its lower end. When an object ass 40 g is placed in the pan. its length becomes cm. When the object is replaced with another of 60g, the length becomes 22.05cm. Calculate the mass scale pan. [g = 10 ms\\(^{-2}\\)]\n\n', 'A Spiral spring of natural length 20.00cm has a scale hanging freely in its lower end. When an object ass 40 g is placed in the pan. its length becomes cm. When the object is replaced with another of 60g, the length becomes 22.05cm. Calculate the mass scale pan. [g = 10 ms\\(^{-2}\\)]', '', '', NULL, '\n\nExplanation\nMg = Ke ; (m + 40)g = 1.80k .................. (i)\n(m + 60)g = 2.05K ................ (ii)\n\\(\\frac{m + 40}{m + 60} = \\frac{1.80}{2.05}\\)\n1.80 (m + 60) = 2.05 (m + 40)\nm = \\(\\frac{26}{0.25} = 104g\\)\n\n', 'Mg = Ke ; (m + 40)g = 1.80k .................. (i)\n(m + 60)g = 2.05K ................ (ii)\n\\(\\frac{m + 40}{m + 60} = \\frac{1.80}{2.05}\\)\n1.80 (m + 60) = 2.05 (m + 40)\nm = \\(\\frac{26}{0.25} = 104g\\)\n\n', '', '', 'WAEC-PK-1', 'WAEC-PHY', 1),
(203, '2018', '\n\nQuestion\n\n(a) Define Young\'s modulus\n(b) When a force of 50 N is applied to the free end of an elastic cord, an extension of 4 cm is produced in the cord. Calculate the work done on the cord.\n\n', '', '(a) Define Young\'s modulus\n(b) When a force of 50 N is applied to the free end of an elastic cord, an extension of 4 cm is produced in the cord. Calculate the work done on the cord.', '(a) Define Young\'s modulus~(b) When a force of 50 N is applied to the free end of an elastic cord, an extension of 4 cm is produced in the cord. Calculate the work done on the cord.', NULL, '\n\nExplanation\n(a) Young\'s modulus is the ratio of tensile or compressive stress to tensile or compressive strain.\n(b) W = \\(\\frac{1}{2}\\)K = \\(\\frac{1}{T}\\) x 50 x 0.04 = 1.0J\n\n', '', 'Explanation\n(a) Young\'s modulus is the ratio of tensile or compressive stress to tensile or compressive strain.\n(b) W = \\(\\frac{1}{2}\\)K = \\(\\frac{1}{T}\\) x 50 x 0.04 = 1.0J\n\n', '(a) Young\'s modulus is the ratio of tensile or compressive stress to tensile or compressive strain.~(b) W = \\(\\frac{1}{2}\\)K = \\(\\frac{1}{T}\\) x 50 x 0.04 = 1.0J\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(204, '2018', '\n\nQuestion\n\n(a) List two properties of cathode rays\n(b) Explain how the intensity and energy of cathode rays may be increased\n\n', '', '(a) List two properties of cathode rays\n(b) Explain how the intensity and energy of cathode rays may be increased', '(a) List two properties of cathode rays~(b) Explain how the intensity and energy of cathode rays may be increased', NULL, '\n\nExplanation\n(a) Properties of cathode rays: They are;\n(i) Negatively charged\n(ii) Possess energy\n(iii) Possess momentum\n(iv) Travel in straight lines in field free space.\n(v) Are deflected by electric and magnetic fields.\n\n(b) (i) The intensity of cathode rays can be increased by increasing the temperature of the cathode.\n(ii) The energy of cathode ray cen be increased by increasing the P.d between the anode and the cathode.\n\n', '', 'Explanation\n(a) Properties of cathode rays: They are;\n(i) Negatively charged\n(ii) Possess energy\n(iii) Possess momentum\n(iv) Travel in straight lines in field free space.\n(v) Are deflected by electric and magnetic fields.\n\n(b) (i) The intensity of cathode rays can be increased by increasing the temperature of the cathode.\n(ii) The energy of cathode ray cen be increased by increasing the P.d between the anode and the cathode.\n\n', '(a) Properties of cathode rays: They are;\n(i) Negatively charged\n(ii) Possess energy\n(iii) Possess momentum\n(iv) Travel in straight lines in field free space.\n(v) Are deflected by electric and magnetic fields.\n~(b) (i) The intensity of cathode rays can be increased by increasing the temperature of the cathode.\n(ii) The energy of cathode ray cen be increased by increasing the P.d between the anode and the cathode.\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(205, '2018', '\n\nQuestion\n\n Give three observations in support of de Broglie\'s assumption that moving particles behave like waves.\n\n', ' Give three observations in support of de Broglie\'s assumption that moving particles behave like waves.', '', '', NULL, '\n\nExplanation\nThe observations are:\n(i) Photoelectric effect\n(ii) Electrons show refraction or reflection\n(iii) They have wavelength or frequency\n(iv) Electrons show diffraction patterns.\n(v) Compton effect\n(vi) Determination of molecular velocities in gases.\n\n\n', '', 'Explanation\nThe observations are:\n(i) Photoelectric effect\n(ii) Electrons show refraction or reflection\n(iii) They have wavelength or frequency\n(iv) Electrons show diffraction patterns.\n(v) Compton effect\n(vi) Determination of molecular velocities in gases.\n\n\n', '', 'WAEC-PK-1', 'WAEC-PHY', 1);
INSERT INTO `exam_simulation2` (`id`, `examYear`, `questions`, `parent_que`, `child_que`, `gen_child_que_desc`, `imageDir`, `explanation`, `parent_que_expl`, `child_que_exp`, `gen_child_que_expl`, `package_id`, `subject_id`, `status`) VALUES
(206, '2018', '\n\nQuestion\n\n(a) State two;\n(i) properties of x-rays\n(ii) reasons to show that x-rays are waves\n(iii) uses of x-rays other than those in medicine;\n(iv) hazards of x-rays.\n\n(b) The potential difference between the cathode and target of an x-ray tube is 5.00 x 10\\(^4\\)V and the current in the tube is 2.00 x 10\\(^{-2}\\)A. Given that only one percent of the total energy supplied is emitted as x-radiation, determine the ospheric pressure.\n(i) maximum frequency of the emitted radiation\n(ii) rate at which heat is removed from the target in order to keep it at steady temperature. [Planck\'s constant, h = 6.63 x 10\\(^{-34}\\) Js, electronic charge e = 1.60 x 10\\(^{-19}\\) C]\n\n', '', '(a) State two;\n(i) properties of x-rays\n(ii) reasons to show that x-rays are waves\n(iii) uses of x-rays other than those in medicine;\n(iv) hazards of x-rays.\n\n(b) The potential difference between the cathode and target of an x-ray tube is 5.00 x 10\\(^4\\)V and the current in the tube is 2.00 x 10\\(^{-2}\\)A. Given that only one percent of the total energy supplied is emitted as x-radiation, determine the ospheric pressure.\n(i) maximum frequency of the emitted radiation\n(ii) rate at which heat is removed from the target in order to keep it at steady temperature. [Planck\'s constant, h = 6.63 x 10\\(^{-34}\\) Js, electronic charge e = 1.60 x 10\\(^{-19}\\) C]', '(a) State two;\n(i) properties of x-rays\n(ii) reasons to show that x-rays are waves\n(iii) uses of x-rays other than those in medicine;\n(iv) hazards of x-rays.\n~(b) The potential difference between the cathode and target of an x-ray tube is 5.00 x 10\\(^4\\)V and the current in the tube is 2.00 x 10\\(^{-2}\\)A. Given that only one percent of the total energy supplied is emitted as x-radiation, determine the ospheric pressure.\n(i) maximum frequency of the emitted radiation\n(ii) rate at which heat is removed from the target in order to keep it at steady temperature. [Planck\'s constant, h = 6.63 x 10\\(^{-34}\\) Js, electronic charge e = 1.60 x 10\\(^{-19}\\) C]', NULL, '\n\nExplanation\n(a)(i) Properties of X-rays are:;\nThey (a) penetrates substance\n(b) travel in straight lines\n(c) are able to ionise gases.\n(d) affect photographic plate or paper making them to fogge\n(e) undergo reflection, or refraction, or polarization or diffraction etc.\n\n(ii) X-rays are electromagnetic waves with wave properties such as;\n- reflection\n- refraction\n- diffraction\n- polarisation\n- travel in straight line\n- cannot be deflected by magnetic or electric field.\n\n(iii) X - rays (a) used to ionise gases causing them to become conducting.\n(b) used in industries to locate fractures in machines and metal castings.\n(c) used to analyse crystal structure i.e. crystallography.\n\n(iv) X - rays are hazardous because they\n- kill mall organisms\n- cause leukaemia can damage riving tissues\n- cause genetic mutation\n- cause damage to cells of large animals.\n(b)(i) hf = eV\nf = \\(\\frac{eV}{h}\\)\nf = 1% of eV = \\(\\frac{1 \\times 1.60 \\times 10^{-17} \\times 5.00 \\times 10^4}{100 \\times 6.63 \\times 10^{-34}}\\) = 1.207 x 10\\(^7\\) Hz.\n(ii) Times rate at which heat is removed = 99% x I x V\n= \\(\\frac{99}{100}\\) x 2.00 x 10\\(^{-2}\\) x 5.00 x 10\\(^4\\)\n= 990W\n\n', '', 'Explanation\n(a)(i) Properties of X-rays are:;\nThey (a) penetrates substance\n(b) travel in straight lines\n(c) are able to ionise gases.\n(d) affect photographic plate or paper making them to fogge\n(e) undergo reflection, or refraction, or polarization or diffraction etc.\n\n(ii) X-rays are electromagnetic waves with wave properties such as;\n- reflection\n- refraction\n- diffraction\n- polarisation\n- travel in straight line\n- cannot be deflected by magnetic or electric field.\n\n(iii) X - rays (a) used to ionise gases causing them to become conducting.\n(b) used in industries to locate fractures in machines and metal castings.\n(c) used to analyse crystal structure i.e. crystallography.\n\n(iv) X - rays are hazardous because they\n- kill mall organisms\n- cause leukaemia can damage riving tissues\n- cause genetic mutation\n- cause damage to cells of large animals.\n(b)(i) hf = eV\nf = \\(\\frac{eV}{h}\\)\nf = 1% of eV = \\(\\frac{1 \\times 1.60 \\times 10^{-17} \\times 5.00 \\times 10^4}{100 \\times 6.63 \\times 10^{-34}}\\) = 1.207 x 10\\(^7\\) Hz.\n(ii) Times rate at which heat is removed = 99% x I x V\n= \\(\\frac{99}{100}\\) x 2.00 x 10\\(^{-2}\\) x 5.00 x 10\\(^4\\)\n= 990W\n\n', '(a)(i) Properties of X-rays are:;\nThey (a) penetrates substance\n(b) travel in straight lines\n(c) are able to ionise gases.\n(d) affect photographic plate or paper making them to fogge\n(e) undergo reflection, or refraction, or polarization or diffraction etc.\n\n(ii) X-rays are electromagnetic waves with wave properties such as;\n- reflection\n- refraction\n- diffraction\n- polarisation\n- travel in straight line\n- cannot be deflected by magnetic or electric field.\n\n(iii) X - rays (a) used to ionise gases causing them to become conducting.\n(b) used in industries to locate fractures in machines and metal castings.\n(c) used to analyse crystal structure i.e. crystallography.\n\n(iv) X - rays are hazardous because they\n- kill mall organisms\n- cause leukaemia can damage riving tissues\n- cause genetic mutation\n- cause damage to cells of large animals.~(b) travel in straight lines\n(c) are able to ionise gases.\n(d) affect photographic plate or paper making them to fogge\n(e) undergo reflection, or refraction, or polarization or diffraction etc.\n\n(ii) X-rays are electromagnetic waves with wave properties such as;\n- reflection\n- refraction\n- diffraction\n- polarisation\n- travel in straight line\n- cannot be deflected by magnetic or electric field.\n\n(iii) X - rays (a) used to ionise gases causing them to become conducting.\n(b) used in industries to locate fractures in machines and metal castings.\n(c) used to analyse crystal structure i.e. crystallography.\n\n(iv) X - rays are hazardous because they\n- kill mall organisms\n- cause leukaemia can damage riving tissues\n- cause genetic mutation\n- cause damage to cells of large animals.\n(b)(i) hf = eV\nf = \\(\\frac{eV}{h}\\)\nf = 1% of eV = \\(\\frac{1 \\times 1.60 \\times 10^{-17} \\times 5.00 \\times 10^4}{100 \\times 6.63 \\times 10^{-34}}\\) = 1.207 x 10\\(^7\\) Hz.\n(ii) Times rate at which heat is removed = 99% x I x V\n= \\(\\frac{99}{100}\\) x 2.00 x 10\\(^{-2}\\) x 5.00 x 10\\(^4\\)\n= 990W\n~(c) are able to ionise gases.\n(d) affect photographic plate or paper making them to fogge\n(e) undergo reflection, or refraction, or polarization or diffraction etc.\n\n(ii) X-rays are electromagnetic waves with wave properties such as;\n- reflection\n- refraction\n- diffraction\n- polarisation\n- travel in straight line\n- cannot be deflected by magnetic or electric field.\n\n(iii) X - rays (a) used to ionise gases causing them to become conducting.\n(b) used in industries to locate fractures in machines and metal castings.\n(c) used to analyse crystal structure i.e. crystallography.\n\n(iv) X - rays are hazardous because they\n- kill mall organisms\n- cause leukaemia can damage riving tissues\n- cause genetic mutation\n- cause damage to cells of large animals.\n(b)(i) hf = eV\nf = \\(\\frac{eV}{h}\\)\nf = 1% of eV = \\(\\frac{1 \\times 1.60 \\times 10^{-17} \\times 5.00 \\times 10^4}{100 \\times 6.63 \\times 10^{-34}}\\) = 1.207 x 10\\(^7\\) Hz.\n(ii) Times rate at which heat is removed = 99% x I x V\n= \\(\\frac{99}{100}\\) x 2.00 x 10\\(^{-2}\\) x 5.00 x 10\\(^4\\)\n= 990W\n~(d) affect photographic plate or paper making them to fogge\n(e) undergo reflection, or refraction, or polarization or diffraction etc.\n\n(ii) X-rays are electromagnetic waves with wave properties such as;\n- reflection\n- refraction\n- diffraction\n- polarisation\n- travel in straight line\n- cannot be deflected by magnetic or electric field.\n\n(iii) X - rays (a) used to ionise gases causing them to become conducting.\n(b) used in industries to locate fractures in machines and metal castings.\n(c) used to analyse crystal structure i.e. crystallography.\n\n(iv) X - rays are hazardous because they\n- kill mall organisms\n- cause leukaemia can damage riving tissues\n- cause genetic mutation\n- cause damage to cells of large animals.\n(b)(i) hf = eV\nf = \\(\\frac{eV}{h}\\)\nf = 1% of eV = \\(\\frac{1 \\times 1.60 \\times 10^{-17} \\times 5.00 \\times 10^4}{100 \\times 6.63 \\times 10^{-34}}\\) = 1.207 x 10\\(^7\\) Hz.\n(ii) Times rate at which heat is removed = 99% x I x V\n= \\(\\frac{99}{100}\\) x 2.00 x 10\\(^{-2}\\) x 5.00 x 10\\(^4\\)\n= 990W\n~(e) undergo reflection, or refraction, or polarization or diffraction etc.\n\n(ii) X-rays are electromagnetic waves with wave properties such as;\n- reflection\n- refraction\n- diffraction\n- polarisation\n- travel in straight line\n- cannot be deflected by magnetic or electric field.\n\n(iii) X - rays (a) used to ionise gases causing them to become conducting.\n(b) used in industries to locate fractures in machines and metal castings.\n(c) used to analyse crystal structure i.e. crystallography.\n\n(iv) X - rays are hazardous because they\n- kill mall organisms\n- cause leukaemia can damage riving tissues\n- cause genetic mutation\n- cause damage to cells of large animals.\n(b)(i) hf = eV\nf = \\(\\frac{eV}{h}\\)\nf = 1% of eV = \\(\\frac{1 \\times 1.60 \\times 10^{-17} \\times 5.00 \\times 10^4}{100 \\times 6.63 \\times 10^{-34}}\\) = 1.207 x 10\\(^7\\) Hz.\n(ii) Times rate at which heat is removed = 99% x I x V\n= \\(\\frac{99}{100}\\) x 2.00 x 10\\(^{-2}\\) x 5.00 x 10\\(^4\\)\n= 990W\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(207, '2018', '\n\nQuestion\n\n(a) Define gravitational field intensity\n(b) In an experiment to determine the acceleration of free-fall due to gravity, g, using a simple pendulum of length I, six different values of I were used to obtain six corresponding values of period T. If a graph of I along the vertical axis is plotted against T\\(^2\\) on the horizontal axis;\n(i) make a sketch to show the nature of the graph,\n(ii) write down the equation that relates T, I and g hence obtain an expression for the slope of the graph\n(iii) given that the slope of the graph is 0.25, determine the value for g [Take \\(\\pi\\) = 3.142]\n(c) A stone, thrown horizontally from the top of a vertical wall with a velocity of 15 ms\\(^{-1}\\), hits the horizontal ground at a point 45m from the base of the wall. Calculate the\n(i) times of light of the stone\n(ii) height of the wall [g = 10ms\\(^{-2}\\)]\n\n', '', '(a) Define gravitational field intensity\n(b) In an experiment to determine the acceleration of free-fall due to gravity, g, using a simple pendulum of length I, six different values of I were used to obtain six corresponding values of period T. If a graph of I along the vertical axis is plotted against T\\(^2\\) on the horizontal axis;\n(i) make a sketch to show the nature of the graph,\n(ii) write down the equation that relates T, I and g hence obtain an expression for the slope of the graph\n(iii) given that the slope of the graph is 0.25, determine the value for g [Take \\(\\pi\\) = 3.142]\n(c) A stone, thrown horizontally from the top of a vertical wall with a velocity of 15 ms\\(^{-1}\\), hits the horizontal ground at a point 45m from the base of the wall. Calculate the\n(i) times of light of the stone\n(ii) height of the wall [g = 10ms\\(^{-2}\\)]', '(a) Define gravitational field intensity~(b) In an experiment to determine the acceleration of free-fall due to gravity, g, using a simple pendulum of length I, six different values of I were used to obtain six corresponding values of period T. If a graph of I along the vertical axis is plotted against T\\(^2\\) on the horizontal axis;\n(i) make a sketch to show the nature of the graph,\n(ii) write down the equation that relates T, I and g hence obtain an expression for the slope of the graph\n(iii) given that the slope of the graph is 0.25, determine the value for g [Take \\(\\pi\\) = 3.142]\n(c) A stone, thrown horizontally from the top of a vertical wall with a velocity of 15 ms\\(^{-1}\\), hits the horizontal ground at a point 45m from the base of the wall. Calculate the\n(i) times of light of the stone\n(ii) height of the wall [g = 10ms\\(^{-2}\\)]~(c) A stone, thrown horizontally from the top of a vertical wall with a velocity of 15 ms\\(^{-1}\\), hits the horizontal ground at a point 45m from the base of the wall. Calculate the\n(i) times of light of the stone\n(ii) height of the wall [g = 10ms\\(^{-2}\\)]', NULL, '\n\nExplanation\n(a) The gravitational field intensity at any point on the earth\'s field is the gravitational force acting on a unit mass at that point\n\n(b)(i)\n\n(ii) A simple Pendulum of length L, swinging in a vertical plane at a small angle of displacement has a period T given as\n\\(T = 2 \\pi \\sqrt{\\frac{L}{g}}\\)\n\\(T^{2} = \\frac{4 \\pi^2L}{g}\\)\nL = \\(\\frac{gT^2}{4 \\pi^2}\\)\nSlope S of the graph I against \\(T^2 = \\frac{g}{4 \\pi^2}\\)\n(iii) g = 4\\(\\pi^2\\)S = 4(3.142)\\(^2\\) x 0.25 = 9.87m/s\\(^{2}\\)\n\n(c)(i) Time t = \\(\\frac{distance}{speed} = \\frac{45}{15}\\) = 3(s)\n(ii) Height h = \\(\\frac{1}{2} gt^2 = \\frac{1}{2} \\times 10 \\times 3^3 = 45m\\)\n\n\n\n\n', '', 'Explanation\n(a) The gravitational field intensity at any point on the earth\'s field is the gravitational force acting on a unit mass at that point\n\n(b)(i)\n\n(ii) A simple Pendulum of length L, swinging in a vertical plane at a small angle of displacement has a period T given as\n\\(T = 2 \\pi \\sqrt{\\frac{L}{g}}\\)\n\\(T^{2} = \\frac{4 \\pi^2L}{g}\\)\nL = \\(\\frac{gT^2}{4 \\pi^2}\\)\nSlope S of the graph I against \\(T^2 = \\frac{g}{4 \\pi^2}\\)\n(iii) g = 4\\(\\pi^2\\)S = 4(3.142)\\(^2\\) x 0.25 = 9.87m/s\\(^{2}\\)\n\n(c)(i) Time t = \\(\\frac{distance}{speed} = \\frac{45}{15}\\) = 3(s)\n(ii) Height h = \\(\\frac{1}{2} gt^2 = \\frac{1}{2} \\times 10 \\times 3^3 = 45m\\)\n\n\n\n\n', '(a) The gravitational field intensity at any point on the earth\'s field is the gravitational force acting on a unit mass at that point\n~(b)(i)\n\n(ii) A simple Pendulum of length L, swinging in a vertical plane at a small angle of displacement has a period T given as\n\\(T = 2 \\pi \\sqrt{\\frac{L}{g}}\\)\n\\(T^{2} = \\frac{4 \\pi^2L}{g}\\)\nL = \\(\\frac{gT^2}{4 \\pi^2}\\)\nSlope S of the graph I against \\(T^2 = \\frac{g}{4 \\pi^2}\\)\n(iii) g = 4\\(\\pi^2\\)S = 4(3.142)\\(^2\\) x 0.25 = 9.87m/s\\(^{2}\\)\n\n(c)(i) Time t = \\(\\frac{distance}{speed} = \\frac{45}{15}\\) = 3(s)\n(ii) Height h = \\(\\frac{1}{2} gt^2 = \\frac{1}{2} \\times 10 \\times 3^3 = 45m\\)\n\n\n\n~(c)(i) Time t = \\(\\frac{distance}{speed} = \\frac{45}{15}\\) = 3(s)\n(ii) Height h = \\(\\frac{1}{2} gt^2 = \\frac{1}{2} \\times 10 \\times 3^3 = 45m\\)\n\n\n\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(208, '2018', '\n\nQuestion\n\nDefine upper fixed point and lower fixed point as used in thermometry\n(b) The electrical resistances of the element in a platinum restistance thermometer at 100\\(^o\\)C, 0\\(^o\\) and room temperature are 75.000, 63.000 and 64.992 \\(\\Omega\\) respectively. Use these data to determine the room temperature.\n(c) (i) State Boyle\'s law\n(ii) A uniform capillary tube, closed at one end contained dry air trapped by a thread of mecury 8.5 x 10\\(^{-2}\\)m long. When the tube was held horinzontally, the length of the air column was 5.0 x 10\\(^{-2}\\)m, when it was held vertically with the closed end downwards, the length was 4.5 x 10\\(^{-2}\\)m, Determine the value of the atmospheric pressure. [g = 10ms\\(^{-2}\\), density of mecury = 1.36 x 10\\(^4\\) kg m\\(^{-3}\\)]\n\n', '', 'Define upper fixed point and lower fixed point as used in thermometry\n(b) The electrical resistances of the element in a platinum restistance thermometer at 100\\(^o\\)C, 0\\(^o\\) and room temperature are 75.000, 63.000 and 64.992 \\(\\Omega\\) respectively. Use these data to determine the room temperature.\n(c) (i) State Boyle\'s law\n(ii) A uniform capillary tube, closed at one end contained dry air trapped by a thread of mecury 8.5 x 10\\(^{-2}\\)m long. When the tube was held horinzontally, the length of the air column was 5.0 x 10\\(^{-2}\\)m, when it was held vertically with the closed end downwards, the length was 4.5 x 10\\(^{-2}\\)m, Determine the value of the atmospheric pressure. [g = 10ms\\(^{-2}\\), density of mecury = 1.36 x 10\\(^4\\) kg m\\(^{-3}\\)]', '~(b) The electrical resistances of the element in a platinum restistance thermometer at 100\\(^o\\)C, 0\\(^o\\) and room temperature are 75.000, 63.000 and 64.992 \\(\\Omega\\) respectively. Use these data to determine the room temperature.\n(c) (i) State Boyle\'s law\n(ii) A uniform capillary tube, closed at one end contained dry air trapped by a thread of mecury 8.5 x 10\\(^{-2}\\)m long. When the tube was held horinzontally, the length of the air column was 5.0 x 10\\(^{-2}\\)m, when it was held vertically with the closed end downwards, the length was 4.5 x 10\\(^{-2}\\)m, Determine the value of the atmospheric pressure. [g = 10ms\\(^{-2}\\), density of mecury = 1.36 x 10\\(^4\\) kg m\\(^{-3}\\)]~(c) (i) State Boyle\'s law\n(ii) A uniform capillary tube, closed at one end contained dry air trapped by a thread of mecury 8.5 x 10\\(^{-2}\\)m long. When the tube was held horinzontally, the length of the air column was 5.0 x 10\\(^{-2}\\)m, when it was held vertically with the closed end downwards, the length was 4.5 x 10\\(^{-2}\\)m, Determine the value of the atmospheric pressure. [g = 10ms\\(^{-2}\\), density of mecury = 1.36 x 10\\(^4\\) kg m\\(^{-3}\\)]', NULL, '\n\nExplanation\n(a) Upper fixed point is the temperature of steam from boiling water at standard atmospheric pressure of 760 mmHg. Lower fixed point is the temperature of pure melting ice.\n\n(b) 75.000 \\(\\Omega\\) - 1000\\(o\\)C\n\\(\\frac{64.992 - 63.000}{75.000 - 63.000} = \\frac{R - 0}{100 - 0}; 64.992 \\Omega - R\\)\n\\(\\frac{1.992}{12.000 } = \\frac{R_1}{100}; 63.000 \\Omega - 0^oC\\)\n\\(R_1 = \\frac{1992 \\times 100}{12.000} = 16.6^oC\\)\n\n(c) (i) Boyle\'s law states that the volume of a fixed unit varies inversely as the pressure provided the temperature remains constant.\n(ii) At horizontal position pressure = A.P = H\nGas volume = 4.5 x 10\\(^{-2}\\) x cross sectional area of the tube A. But PV = Constant (Boyle\'s law)\nH x 5.0 x 10\\(^{-2}\\) x A = (H + 8.5 x 10\\(^{-2}\\)) x 4.5 x 10\\(^{-2}\\) x A\nH = \\(\\frac{8.5 \\times 10^{-2} \\times 4.5}{0.5} = 0.765mHg\\)\nA.P = hpg = 0.765 x 1.36 x 10\\(^4\\) x 10 = 1.04 x 10\\(^{5}\\)pa\n\n\n\n', '', 'Explanation\n(a) Upper fixed point is the temperature of steam from boiling water at standard atmospheric pressure of 760 mmHg. Lower fixed point is the temperature of pure melting ice.\n\n(b) 75.000 \\(\\Omega\\) - 1000\\(o\\)C\n\\(\\frac{64.992 - 63.000}{75.000 - 63.000} = \\frac{R - 0}{100 - 0}; 64.992 \\Omega - R\\)\n\\(\\frac{1.992}{12.000 } = \\frac{R_1}{100}; 63.000 \\Omega - 0^oC\\)\n\\(R_1 = \\frac{1992 \\times 100}{12.000} = 16.6^oC\\)\n\n(c) (i) Boyle\'s law states that the volume of a fixed unit varies inversely as the pressure provided the temperature remains constant.\n(ii) At horizontal position pressure = A.P = H\nGas volume = 4.5 x 10\\(^{-2}\\) x cross sectional area of the tube A. But PV = Constant (Boyle\'s law)\nH x 5.0 x 10\\(^{-2}\\) x A = (H + 8.5 x 10\\(^{-2}\\)) x 4.5 x 10\\(^{-2}\\) x A\nH = \\(\\frac{8.5 \\times 10^{-2} \\times 4.5}{0.5} = 0.765mHg\\)\nA.P = hpg = 0.765 x 1.36 x 10\\(^4\\) x 10 = 1.04 x 10\\(^{5}\\)pa\n\n\n\n', '(a) Upper fixed point is the temperature of steam from boiling water at standard atmospheric pressure of 760 mmHg. Lower fixed point is the temperature of pure melting ice.\n~(b) 75.000 \\(\\Omega\\) - 1000\\(o\\)C\n\\(\\frac{64.992 - 63.000}{75.000 - 63.000} = \\frac{R - 0}{100 - 0}; 64.992 \\Omega - R\\)\n\\(\\frac{1.992}{12.000 } = \\frac{R_1}{100}; 63.000 \\Omega - 0^oC\\)\n\\(R_1 = \\frac{1992 \\times 100}{12.000} = 16.6^oC\\)\n\n(c) (i) Boyle\'s law states that the volume of a fixed unit varies inversely as the pressure provided the temperature remains constant.\n(ii) At horizontal position pressure = A.P = H\nGas volume = 4.5 x 10\\(^{-2}\\) x cross sectional area of the tube A. But PV = Constant (Boyle\'s law)\nH x 5.0 x 10\\(^{-2}\\) x A = (H + 8.5 x 10\\(^{-2}\\)) x 4.5 x 10\\(^{-2}\\) x A\nH = \\(\\frac{8.5 \\times 10^{-2} \\times 4.5}{0.5} = 0.765mHg\\)\nA.P = hpg = 0.765 x 1.36 x 10\\(^4\\) x 10 = 1.04 x 10\\(^{5}\\)pa\n\n\n~(c) (i) Boyle\'s law states that the volume of a fixed unit varies inversely as the pressure provided the temperature remains constant.\n(ii) At horizontal position pressure = A.P = H\nGas volume = 4.5 x 10\\(^{-2}\\) x cross sectional area of the tube A. But PV = Constant (Boyle\'s law)\nH x 5.0 x 10\\(^{-2}\\) x A = (H + 8.5 x 10\\(^{-2}\\)) x 4.5 x 10\\(^{-2}\\) x A\nH = \\(\\frac{8.5 \\times 10^{-2} \\times 4.5}{0.5} = 0.765mHg\\)\nA.P = hpg = 0.765 x 1.36 x 10\\(^4\\) x 10 = 1.04 x 10\\(^{5}\\)pa\n\n\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(209, '2018', '\n\nQuestion\n\n(a) Explain;\n(i) wave motion\n(ii) stationary wave\n(b)(i) List four physical properties of a wave\n(ii) Define amplitude and use it to distinguish between the node and antinode of a stationary wave\n(iii) List the factors on which the frequency of vibration in a stretched string depends\n(c) The equation, y = 5 sin (3x - 4t), where y is in millimeters, x is in meres and t is in seconds represents a wave motion . Determine the;\n(i) frequency\n(ii) period\n(iii) speed of the wave\n\n', '', '(a) Explain;\n(i) wave motion\n(ii) stationary wave\n(b)(i) List four physical properties of a wave\n(ii) Define amplitude and use it to distinguish between the node and antinode of a stationary wave\n(iii) List the factors on which the frequency of vibration in a stretched string depends\n(c) The equation, y = 5 sin (3x - 4t), where y is in millimeters, x is in meres and t is in seconds represents a wave motion . Determine the;\n(i) frequency\n(ii) period\n(iii) speed of the wave', '(a) Explain;\n(i) wave motion\n(ii) stationary wave~(b)(i) List four physical properties of a wave\n(ii) Define amplitude and use it to distinguish between the node and antinode of a stationary wave\n(iii) List the factors on which the frequency of vibration in a stretched string depends\n(c) The equation, y = 5 sin (3x - 4t), where y is in millimeters, x is in meres and t is in seconds represents a wave motion . Determine the;\n(i) frequency\n(ii) period\n(iii) speed of the wave~(c) The equation, y = 5 sin (3x - 4t), where y is in millimeters, x is in meres and t is in seconds represents a wave motion . Determine the;\n(i) frequency\n(ii) period\n(iii) speed of the wave', NULL, '\n\nExplanation\na) (i) A wave motion is a disturbance which travels through a medium and transfers energy from one point to another without causing any permanent displacement of the medium itself.\n\n\n(ii) A stationary wave is a wave got when two progressive waves of equal amplitude and frequency or wavelength travelling in opposite direction are superimposed. In stationary wave nodes are pendulum of permanently at rest while antinodes are at maximum amplitude\n\nStationary wave along a string\n(b)(i) Physical properties of wave are: diffraction, refraction, reflection and.interference.\n(ii) Amplitude is the maximum displacement of a oscillating object from its equilibrium position. A node is a point on a stationary wave at zero amplitude while an antinode is a point at maximum amplitude.\n(iii) Frequency of vibration on a stretch string depends on;\n- Tension on the string\n- Length of stretch string\n- Mass per unit length and density of the string\n(c) y = 5 sin (3x - 4t); y = A sin (\\(\\frac{2 \\pi x}{\\pi} \\times 2 \\pi ft\\))\n\\(\\frac{2 \\pi x}{\\pi}\\) = 3x\n\\(\\pi = \\frac{2 \\pi}{3}\\)m = 2\\(\\pi ft\\) = 4t\nf = \\(\\frac{2}{\\pi }\\)\n(i) = \\(\\frac{2}{\\pi}\\) = 0.64Hz\n(ii) T = \\(\\frac{1}{f}\\) = \\(\\frac{1}{0.64}\\) = 1.56(s)\n(iii) V = f\\(\\pi\\) = 0.64 x \\(\\frac{2 \\pi}{3} = 1.34\\)m/s\n\n', '', 'Explanation\na) (i) A wave motion is a disturbance which travels through a medium and transfers energy from one point to another without causing any permanent displacement of the medium itself.\n\n\n(ii) A stationary wave is a wave got when two progressive waves of equal amplitude and frequency or wavelength travelling in opposite direction are superimposed. In stationary wave nodes are pendulum of permanently at rest while antinodes are at maximum amplitude\n\nStationary wave along a string\n(b)(i) Physical properties of wave are: diffraction, refraction, reflection and.interference.\n(ii) Amplitude is the maximum displacement of a oscillating object from its equilibrium position. A node is a point on a stationary wave at zero amplitude while an antinode is a point at maximum amplitude.\n(iii) Frequency of vibration on a stretch string depends on;\n- Tension on the string\n- Length of stretch string\n- Mass per unit length and density of the string\n(c) y = 5 sin (3x - 4t); y = A sin (\\(\\frac{2 \\pi x}{\\pi} \\times 2 \\pi ft\\))\n\\(\\frac{2 \\pi x}{\\pi}\\) = 3x\n\\(\\pi = \\frac{2 \\pi}{3}\\)m = 2\\(\\pi ft\\) = 4t\nf = \\(\\frac{2}{\\pi }\\)\n(i) = \\(\\frac{2}{\\pi}\\) = 0.64Hz\n(ii) T = \\(\\frac{1}{f}\\) = \\(\\frac{1}{0.64}\\) = 1.56(s)\n(iii) V = f\\(\\pi\\) = 0.64 x \\(\\frac{2 \\pi}{3} = 1.34\\)m/s\n\n', 'a) (i) A wave motion is a disturbance which travels through a medium and transfers energy from one point to another without causing any permanent displacement of the medium itself.\n\n\n(ii) A stationary wave is a wave got when two progressive waves of equal amplitude and frequency or wavelength travelling in opposite direction are superimposed. In stationary wave nodes are pendulum of permanently at rest while antinodes are at maximum amplitude\n\nStationary wave along a string~(b)(i) Physical properties of wave are: diffraction, refraction, reflection and.interference.\n(ii) Amplitude is the maximum displacement of a oscillating object from its equilibrium position. A node is a point on a stationary wave at zero amplitude while an antinode is a point at maximum amplitude.\n(iii) Frequency of vibration on a stretch string depends on;\n- Tension on the string\n- Length of stretch string\n- Mass per unit length and density of the string\n(c) y = 5 sin (3x - 4t); y = A sin (\\(\\frac{2 \\pi x}{\\pi} \\times 2 \\pi ft\\))\n\\(\\frac{2 \\pi x}{\\pi}\\) = 3x\n\\(\\pi = \\frac{2 \\pi}{3}\\)m = 2\\(\\pi ft\\) = 4t\nf = \\(\\frac{2}{\\pi }\\)\n(i) = \\(\\frac{2}{\\pi}\\) = 0.64Hz\n(ii) T = \\(\\frac{1}{f}\\) = \\(\\frac{1}{0.64}\\) = 1.56(s)\n(iii) V = f\\(\\pi\\) = 0.64 x \\(\\frac{2 \\pi}{3} = 1.34\\)m/s\n~(c) y = 5 sin (3x - 4t); y = A sin (\\(\\frac{2 \\pi x}{\\pi} \\times 2 \\pi ft\\))\n\\(\\frac{2 \\pi x}{\\pi}\\) = 3x\n\\(\\pi = \\frac{2 \\pi}{3}\\)m = 2\\(\\pi ft\\) = 4t\nf = \\(\\frac{2}{\\pi }\\)\n(i) = \\(\\frac{2}{\\pi}\\) = 0.64Hz\n(ii) T = \\(\\frac{1}{f}\\) = \\(\\frac{1}{0.64}\\) = 1.56(s)\n(iii) V = f\\(\\pi\\) = 0.64 x \\(\\frac{2 \\pi}{3} = 1.34\\)m/s\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(210, '2018', '\n\nQuestion\n\n(a) (i) Explain electromotive force\n(ii) list two sources of electromotive force other than a chemical cell\n(b) A chemical cell of electromotive force, E, and internal resistance, r, is connected in series with an ammeter, a plug key a plug key and an external load of resistance R. A volumeter is connected across the cell. Draw a circuit diagram to illustrate the arrangement,\n(c) for the arrangement in (b) above, with the key opened and closed, the voltmeter readings are V\\(_o\\) and V respectively.\n(i) Explain the physical meanings of V\\(_o\\) and V\n(ii) Find an expression for the (I) current passing through the circuit (II) maximum power dissipated in the cell and external load respectively; (III) efficiency of the cell\n\n', '', '(a) (i) Explain electromotive force\n(ii) list two sources of electromotive force other than a chemical cell\n(b) A chemical cell of electromotive force, E, and internal resistance, r, is connected in series with an ammeter, a plug key a plug key and an external load of resistance R. A volumeter is connected across the cell. Draw a circuit diagram to illustrate the arrangement,\n(c) for the arrangement in (b) above, with the key opened and closed, the voltmeter readings are V\\(_o\\) and V respectively.\n(i) Explain the physical meanings of V\\(_o\\) and V\n(ii) Find an expression for the (I) current passing through the circuit (II) maximum power dissipated in the cell and external load respectively; (III) efficiency of the cell', '(a) (i) Explain electromotive force\n(ii) list two sources of electromotive force other than a chemical cell~(b) A chemical cell of electromotive force, E, and internal resistance, r, is connected in series with an ammeter, a plug key a plug key and an external load of resistance R. A volumeter is connected across the cell. Draw a circuit diagram to illustrate the arrangement,\n(c) for the arrangement in (b) above, with the key opened and closed, the voltmeter readings are V\\(_o\\) and V respectively.\n(i) Explain the physical meanings of V\\(_o\\) and V\n(ii) Find an expression for the (I) current passing through the circuit (II) maximum power dissipated in the cell and external load respectively; (III) efficiency of the cell~(c) for the arrangement in (b) above, with the key opened and closed, the voltmeter readings are V\\(_o\\) and V respectively.\n(i) Explain the physical meanings of V\\(_o\\) and V\n(ii) Find an expression for the (I) current passing through the circuit (II) maximum power dissipated in the cell and external load respectively; (III) efficiency of the cell', NULL, '\n\nExplanation\n(i) E.m.f is the work doone in taking a unit electric charge round a closed circuit. H is measured in volts.\n(ii) Other sources of e.m.f. are; solar cells, fuel cells, dynamo or electric generators, thermocouple etc.\n(b)\n\n(c) (i) In open circuit Vo e.m.f of the cell.\nIn closed circuit V terminal p.d of the cell.\n(ii) (a) I = \\(\\frac{E}{R + r}\\)\n(b) Maximum power dissipated in the cell occurs when the external resistance = the internal resistance of the cell i.e. R = r. Maximum power dissipated in the cell = IE\n= \\(\\frac{E}{R + r} \\times E = \\frac{E^2}{R + r} = \\frac{E^2}{2R}\\)\nMaximuim power dissipated in the external load\n= I\\(^2\\)R = \\(\\frac{E^2R}{(R + r)^2}\\)\n= \\(\\frac{E^2R}{4R^2} = \\frac{E^2}{4R}\\)\n(iii) Efficiency of the call = \\(\\frac{\\text{power output}}{\\text{power input}}\\) x 100%\n= \\(\\frac{2R}{E^2} \\times \\frac{E^2}{4R}\\) x 100% = 50%\n\n', '', 'Explanation\n(i) E.m.f is the work doone in taking a unit electric charge round a closed circuit. H is measured in volts.\n(ii) Other sources of e.m.f. are; solar cells, fuel cells, dynamo or electric generators, thermocouple etc.\n(b)\n\n(c) (i) In open circuit Vo e.m.f of the cell.\nIn closed circuit V terminal p.d of the cell.\n(ii) (a) I = \\(\\frac{E}{R + r}\\)\n(b) Maximum power dissipated in the cell occurs when the external resistance = the internal resistance of the cell i.e. R = r. Maximum power dissipated in the cell = IE\n= \\(\\frac{E}{R + r} \\times E = \\frac{E^2}{R + r} = \\frac{E^2}{2R}\\)\nMaximuim power dissipated in the external load\n= I\\(^2\\)R = \\(\\frac{E^2R}{(R + r)^2}\\)\n= \\(\\frac{E^2R}{4R^2} = \\frac{E^2}{4R}\\)\n(iii) Efficiency of the call = \\(\\frac{\\text{power output}}{\\text{power input}}\\) x 100%\n= \\(\\frac{2R}{E^2} \\times \\frac{E^2}{4R}\\) x 100% = 50%\n\n', '(i) E.m.f is the work doone in taking a unit electric charge round a closed circuit. H is measured in volts.\n(ii) Other sources of e.m.f. are; solar cells, fuel cells, dynamo or electric generators, thermocouple etc.\n(b)\n\n(c) (i) In open circuit Vo e.m.f of the cell.\nIn closed circuit V terminal p.d of the cell.\n(ii) (a) I = \\(\\frac{E}{R + r}\\)~(b)\n\n(c) (i) In open circuit Vo e.m.f of the cell.\nIn closed circuit V terminal p.d of the cell.\n(ii) (a) I = \\(\\frac{E}{R + r}\\)\n(b) Maximum power dissipated in the cell occurs when the external resistance = the internal resistance of the cell i.e. R = r. Maximum power dissipated in the cell = IE\n= \\(\\frac{E}{R + r} \\times E = \\frac{E^2}{R + r} = \\frac{E^2}{2R}\\)\nMaximuim power dissipated in the external load\n= I\\(^2\\)R = \\(\\frac{E^2R}{(R + r)^2}\\)\n= \\(\\frac{E^2R}{4R^2} = \\frac{E^2}{4R}\\)\n(iii) Efficiency of the call = \\(\\frac{\\text{power output}}{\\text{power input}}\\) x 100%\n= \\(\\frac{2R}{E^2} \\times \\frac{E^2}{4R}\\) x 100% = 50%\n~(c) (i) In open circuit Vo e.m.f of the cell.\nIn closed circuit V terminal p.d of the cell.\n(ii) (a) I = \\(\\frac{E}{R + r}\\)\n(b) Maximum power dissipated in the cell occurs when the external resistance = the internal resistance of the cell i.e. R = r. Maximum power dissipated in the cell = IE\n= \\(\\frac{E}{R + r} \\times E = \\frac{E^2}{R + r} = \\frac{E^2}{2R}\\)\nMaximuim power dissipated in the external load\n= I\\(^2\\)R = \\(\\frac{E^2R}{(R + r)^2}\\)\n= \\(\\frac{E^2R}{4R^2} = \\frac{E^2}{4R}\\)\n(iii) Efficiency of the call = \\(\\frac{\\text{power output}}{\\text{power input}}\\) x 100%\n= \\(\\frac{2R}{E^2} \\times \\frac{E^2}{4R}\\) x 100% = 50%\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(211, '2018', '\n\nQuestion\n\n(a) State two objects each used in sports and warfare which may be considered as projectiles.\n(b) the horizontal range R, of a projectile is given by the expression;\nR = \\(\\frac{u^2 sin 2\\theta}{g}\\) where \\(\\theta\\) is the angle of projection, g the acceleration of free fall due to gravity and u the initial velocity. At what value of \\(\\theta\\) will R be maximium?\n\n', '', '(a) State two objects each used in sports and warfare which may be considered as projectiles.\n(b) the horizontal range R, of a projectile is given by the expression;\nR = \\(\\frac{u^2 sin 2\\theta}{g}\\) where \\(\\theta\\) is the angle of projection, g the acceleration of free fall due to gravity and u the initial velocity. At what value of \\(\\theta\\) will R be maximium?', '(a) State two objects each used in sports and warfare which may be considered as projectiles.~(b) the horizontal range R, of a projectile is given by the expression;\nR = \\(\\frac{u^2 sin 2\\theta}{g}\\) where \\(\\theta\\) is the angle of projection, g the acceleration of free fall due to gravity and u the initial velocity. At what value of \\(\\theta\\) will R be maximium?', NULL, '\n\nExplanation\n(a) Projectiles are used in;\n(i)Sports as – Shortput, football, javelin etc\n(ii) Warfare a bullet, spear,arrow etc.\n(b) A R maximum \\(\\theta\\) = 45°\n\n', '', 'Explanation\n(a) Projectiles are used in;\n(i)Sports as – Shortput, football, javelin etc\n(ii) Warfare a bullet, spear,arrow etc.\n(b) A R maximum \\(\\theta\\) = 45°\n\n', '(a) Projectiles are used in;\n(i)Sports as – Shortput, football, javelin etc\n(ii) Warfare a bullet, spear,arrow etc.~(b) A R maximum \\(\\theta\\) = 45°\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(212, '2018', '\n\nQuestion\n\n(a) What is surface tension?\n(b) State two methods by which the surface tension of a liquid can be reduced.\n\n', '', '(a) What is surface tension?\n(b) State two methods by which the surface tension of a liquid can be reduced.', '(a) What is surface tension?~(b) State two methods by which the surface tension of a liquid can be reduced.', NULL, '\n\nExplanation\n(a) Surface tension is the property of a liquid which makes its surface behave as though covered by an elastic skin.\n(b) Surface tension can be reduce by;\n(i) Addition of impurities e.g. soap, oil, alcohol etc.\n(ii) Heating the liquid.\n\n', '', 'Explanation\n(a) Surface tension is the property of a liquid which makes its surface behave as though covered by an elastic skin.\n(b) Surface tension can be reduce by;\n(i) Addition of impurities e.g. soap, oil, alcohol etc.\n(ii) Heating the liquid.\n\n', '(a) Surface tension is the property of a liquid which makes its surface behave as though covered by an elastic skin.~(b) Surface tension can be reduce by;\n(i) Addition of impurities e.g. soap, oil, alcohol etc.\n(ii) Heating the liquid.\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(213, '2018', '\n\nQuestion\n\n(a) Explain diffusion\n(b) State one factor that can affect the rate of diffusion.\n\n', '', '(a) Explain diffusion\n(b) State one factor that can affect the rate of diffusion.', '(a) Explain diffusion~(b) State one factor that can affect the rate of diffusion.', NULL, '\n\nExplanation\nDiffusion is the tendency of a gas to mix with another and fill an empty space as a result of the constant random motion of the molecules\n(b) Factors affecting rate of diffusiom are;\n(i) mass\n(ii) temperature\n(iii) density\n\n', '', 'Explanation\nDiffusion is the tendency of a gas to mix with another and fill an empty space as a result of the constant random motion of the molecules\n(b) Factors affecting rate of diffusiom are;\n(i) mass\n(ii) temperature\n(iii) density\n\n', '~(b) Factors affecting rate of diffusiom are;\n(i) mass\n(ii) temperature\n(iii) density\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(214, '2018', '\n\nQuestion\n\n(a) List two types of waves, apart from light, that can be plane polarized.\n(b) State two applications of plane polarized light.\n\n', '', '(a) List two types of waves, apart from light, that can be plane polarized.\n(b) State two applications of plane polarized light.', '(a) List two types of waves, apart from light, that can be plane polarized.~(b) State two applications of plane polarized light.', NULL, '\n\nExplanation\n(a) Plane polarized waves are:\n(i) radio waves\n(ii) x-rays.\n(iii) infrared radiation etc\n(b) Application of plane polarized light\n– To reduce glare from reflecting surfaces.\n– In sunglasses to reduce the intensity of incident sunlight.\n– To determine sugar solution concentration etc.\n\n', '', 'Explanation\n(a) Plane polarized waves are:\n(i) radio waves\n(ii) x-rays.\n(iii) infrared radiation etc\n(b) Application of plane polarized light\n– To reduce glare from reflecting surfaces.\n– In sunglasses to reduce the intensity of incident sunlight.\n– To determine sugar solution concentration etc.\n\n', '(a) Plane polarized waves are:\n(i) radio waves\n(ii) x-rays.\n(iii) infrared radiation etc~(b) Application of plane polarized light\n– To reduce glare from reflecting surfaces.\n– In sunglasses to reduce the intensity of incident sunlight.\n– To determine sugar solution concentration etc.\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(215, '2018', '\n\nQuestion\n\nExplain the rise of water in a glass capillary tube using the kinetic theory.\n\n', 'Explain the rise of water in a glass capillary tube using the kinetic theory.', '', '', NULL, '\n\nExplanation\nGlass and water are made up of molecules. The force of attraction between glass molecules and water molecules is stronger than that among water molecules, hence the resultant force acts upwards leading to the rise of water in the glass tube.\n\n', 'Glass and water are made up of molecules. The force of attraction between glass molecules and water molecules is stronger than that among water molecules, hence the resultant force acts upwards leading to the rise of water in the glass tube.\n\n', '', '', 'WAEC-PK-1', 'WAEC-PHY', 1),
(216, '2018', '\n\nQuestion\n\n(a) Explain the term electrolyte and give two examples.\n(b) State the relationship between the mass of a substance liberated during electrolysis and the charge passed.\n\n', '', '(a) Explain the term electrolyte and give two examples.\n(b) State the relationship between the mass of a substance liberated during electrolysis and the charge passed.', '(a) Explain the term electrolyte and give two examples.~(b) State the relationship between the mass of a substance liberated during electrolysis and the charge passed.', NULL, '\n\nExplanation\n(a) Electrolyte is a substance in liquid or molten state which conducts electricity and undergoes decomposition e.g. metallic salts, acids and alkalis.\n(b) M \\(\\alpha\\) Q\nM = kQ. Mass m is proportional to charge Q where K is a constant of proportionality.\n\n', '', 'Explanation\n(a) Electrolyte is a substance in liquid or molten state which conducts electricity and undergoes decomposition e.g. metallic salts, acids and alkalis.\n(b) M \\(\\alpha\\) Q\nM = kQ. Mass m is proportional to charge Q where K is a constant of proportionality.\n\n', '(a) Electrolyte is a substance in liquid or molten state which conducts electricity and undergoes decomposition e.g. metallic salts, acids and alkalis.~(b) M \\(\\alpha\\) Q\nM = kQ. Mass m is proportional to charge Q where K is a constant of proportionality.\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(217, '2018', '\n\nQuestion\n\n(a) What is the principle upon which the lighting in fluorescent tubes operate?\n(b) State two factors which determine the colour of projection, g the acceleration of free fall due to gravity light from a fluorescent tube\n\n', '', '(a) What is the principle upon which the lighting in fluorescent tubes operate?\n(b) State two factors which determine the colour of projection, g the acceleration of free fall due to gravity light from a fluorescent tube', '(a) What is the principle upon which the lighting in fluorescent tubes operate?~(b) State two factors which determine the colour of projection, g the acceleration of free fall due to gravity light from a fluorescent tube', NULL, '\n\nExplanation\n(a)The principles is electric discharge through low pressure gas.\n(b) Colour of light is determined by\n– Nature of gas in the tube\n– Nature of coating inside the tube\n\n', '', 'Explanation\n(a)The principles is electric discharge through low pressure gas.\n(b) Colour of light is determined by\n– Nature of gas in the tube\n– Nature of coating inside the tube\n\n', '(a)The principles is electric discharge through low pressure gas.~(b) Colour of light is determined by\n– Nature of gas in the tube\n– Nature of coating inside the tube\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(218, '2018', '\n\nQuestion\n\nDefine;\n(i) Elasticity\n(ii) Young\'s modulus\n(iii) Force constant\n\n', '', 'Define;\n(i) Elasticity\n(ii) Young\'s modulus\n(iii) Force constant', '', NULL, '\n\nExplanation\n(i) Elasticity is the ability of a material to regain its original shape and size after undergoing a stretching or compression.\n(ii) Young\'s modulus is the ratio of tensile stress to tensile strain.\n(iii) Force constant is the amount of force that causes unit extension.\n\n', '', 'Explanation\n(i) Elasticity is the ability of a material to regain its original shape and size after undergoing a stretching or compression.\n(ii) Young\'s modulus is the ratio of tensile stress to tensile strain.\n(iii) Force constant is the amount of force that causes unit extension.\n\n', '', 'WAEC-PK-1', 'WAEC-PHY', 1),
(219, '2018', '\n\nQuestion\n\nA force of 40 N applied at the end of a wire of length 4m and diameter 2.00 mm produces an extension of 0.24 mm. Calculate the;\n(a) stress on the wire;\n(b) strain in the wire (\\(\\pi = 3.142\\))\n\n\n', '', 'A force of 40 N applied at the end of a wire of length 4m and diameter 2.00 mm produces an extension of 0.24 mm. Calculate the;\n(a) stress on the wire;\n(b) strain in the wire (\\(\\pi = 3.142\\))\n', '(a) stress on the wire;~(b) strain in the wire (\\(\\pi = 3.142\\))\n', NULL, '\n\nExplanation\nStress = \\(\\frac{F}{A} = \\frac{40}{3.14(1 \\times 10^{-3})^2} = 12.74 \\times 10^{6} Nm^{-2}\\)\n(b) Strain = \\(\\frac{e}{I} = \\frac{24 \\times 10^{-5}}{4} = 6 \\times 10^{-5}\\)\n\n', '', 'Explanation\nStress = \\(\\frac{F}{A} = \\frac{40}{3.14(1 \\times 10^{-3})^2} = 12.74 \\times 10^{6} Nm^{-2}\\)\n(b) Strain = \\(\\frac{e}{I} = \\frac{24 \\times 10^{-5}}{4} = 6 \\times 10^{-5}\\)\n\n', '~(b) Strain = \\(\\frac{e}{I} = \\frac{24 \\times 10^{-5}}{4} = 6 \\times 10^{-5}\\)\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(220, '2018', '\n\nQuestion\n\n(a) What is meant by the wave-particles duality of matter?\n(b) mention one physical phenomenon, in each case, that can be explained in terms of the wave nature and particle nature of light.\n\n', '', '(a) What is meant by the wave-particles duality of matter?\n(b) mention one physical phenomenon, in each case, that can be explained in terms of the wave nature and particle nature of light.', '(a) What is meant by the wave-particles duality of matter?~(b) mention one physical phenomenon, in each case, that can be explained in terms of the wave nature and particle nature of light.', NULL, '\n\nExplanation\n(a) It means that light behaves either as a wave or as a particle but not as both simultaneously.\n(b) As wave nature e.g. diffraction, interference etc. As particle nature e.g. heated body radiation, photoelectric effect etc.\n\n', '', 'Explanation\n(a) It means that light behaves either as a wave or as a particle but not as both simultaneously.\n(b) As wave nature e.g. diffraction, interference etc. As particle nature e.g. heated body radiation, photoelectric effect etc.\n\n', '(a) It means that light behaves either as a wave or as a particle but not as both simultaneously.~(b) As wave nature e.g. diffraction, interference etc. As particle nature e.g. heated body radiation, photoelectric effect etc.\n', 'WAEC-PK-1', 'WAEC-PHY', 1);
INSERT INTO `exam_simulation2` (`id`, `examYear`, `questions`, `parent_que`, `child_que`, `gen_child_que_desc`, `imageDir`, `explanation`, `parent_que_expl`, `child_que_exp`, `gen_child_que_expl`, `package_id`, `subject_id`, `status`) VALUES
(221, '2018', '\n\nQuestion\n\n(a) Define (i) Linear momentum; (ii) Impulse\n(b) State the principle of conservation of linear momentum.\n(c) A tractor of mass 5.0 x 10\\(^{3}\\)kg is used a tow a car of mass 2.5 x 103 kg. The tractor moved with a speed of 3.0 ms\\(^{-1}\\) just before the towing rope becomes taut. Calculate the:\n(i) Speed of the tractor immediately the rope becomes taut\n(ii) Loss in kinetic energy of the system just after the car has started Moving;\n(iii) Impulse in the rope when it jerks the car into motion.\n\n', '', '(a) Define (i) Linear momentum; (ii) Impulse\n(b) State the principle of conservation of linear momentum.\n(c) A tractor of mass 5.0 x 10\\(^{3}\\)kg is used a tow a car of mass 2.5 x 103 kg. The tractor moved with a speed of 3.0 ms\\(^{-1}\\) just before the towing rope becomes taut. Calculate the:\n(i) Speed of the tractor immediately the rope becomes taut\n(ii) Loss in kinetic energy of the system just after the car has started Moving;\n(iii) Impulse in the rope when it jerks the car into motion.', '(a) Define (i) Linear momentum; (ii) Impulse~(b) State the principle of conservation of linear momentum.\n(c) A tractor of mass 5.0 x 10\\(^{3}\\)kg is used a tow a car of mass 2.5 x 103 kg. The tractor moved with a speed of 3.0 ms\\(^{-1}\\) just before the towing rope becomes taut. Calculate the:\n(i) Speed of the tractor immediately the rope becomes taut\n(ii) Loss in kinetic energy of the system just after the car has started Moving;\n(iii) Impulse in the rope when it jerks the car into motion.~(c) A tractor of mass 5.0 x 10\\(^{3}\\)kg is used a tow a car of mass 2.5 x 103 kg. The tractor moved with a speed of 3.0 ms\\(^{-1}\\) just before the towing rope becomes taut. Calculate the:\n(i) Speed of the tractor immediately the rope becomes taut\n(ii) Loss in kinetic energy of the system just after the car has started Moving;\n(iii) Impulse in the rope when it jerks the car into motion.', NULL, '\n\nExplanation\n(a)(i) Linear momentum is the product of the mass of a body and its velocity.\n(ii) Impulse is the product of a large force and short time for which the force acts.\n(b) The principle of conservation of linear momentum states that in system of colliding bodies the total momentum of the system is constant provided no external forces act on the system.\n(c) (i) \\(m_1u_1 + m_2U_2 = v(m_1 + m_2)\\) 5 x 10\\(^{3}\\) x 3 + 0\n = v(5 x 10\\(^{3}\\) + 2.5 x 10\\(^{3}\\)) v = \\(\\frac{15 \\times 10^3}{7.5 \\times 10^3}\\) = 2m/s\n(ii) Initial K.E = \\(\\frac{1}{2}mv^2 = \\frac{1}{2} MV^2 = \\frac{1}{2} times 5 \\times 10^{3} \\times 3^2\\)\n= 2.25 x 10\\(^4\\)\nFinal K.E = \\(\\frac{1}{2}mv^2 = \\frac{1}{2} (5 + 2.5) \\times 10^3 \\times 2^2\\)\n= 1.5 x 10\\(^{4}\\)J\nLoss in K.E of the couple = (2.25 - 1.5) x 10\\(^4\\)\n= 75 x 10\\(^3\\)J\n(iii) Impulse = (\\(mu_1U_1 + m_2U_2\\)) - v\n(\\(mu_1 +m_2) (0 + 5 \\times 10^3) - 2 (2.25 + 5) 10^3\\) = 0Ns\n\n', '', 'Explanation\n(a)(i) Linear momentum is the product of the mass of a body and its velocity.\n(ii) Impulse is the product of a large force and short time for which the force acts.\n(b) The principle of conservation of linear momentum states that in system of colliding bodies the total momentum of the system is constant provided no external forces act on the system.\n(c) (i) \\(m_1u_1 + m_2U_2 = v(m_1 + m_2)\\) 5 x 10\\(^{3}\\) x 3 + 0\n = v(5 x 10\\(^{3}\\) + 2.5 x 10\\(^{3}\\)) v = \\(\\frac{15 \\times 10^3}{7.5 \\times 10^3}\\) = 2m/s\n(ii) Initial K.E = \\(\\frac{1}{2}mv^2 = \\frac{1}{2} MV^2 = \\frac{1}{2} times 5 \\times 10^{3} \\times 3^2\\)\n= 2.25 x 10\\(^4\\)\nFinal K.E = \\(\\frac{1}{2}mv^2 = \\frac{1}{2} (5 + 2.5) \\times 10^3 \\times 2^2\\)\n= 1.5 x 10\\(^{4}\\)J\nLoss in K.E of the couple = (2.25 - 1.5) x 10\\(^4\\)\n= 75 x 10\\(^3\\)J\n(iii) Impulse = (\\(mu_1U_1 + m_2U_2\\)) - v\n(\\(mu_1 +m_2) (0 + 5 \\times 10^3) - 2 (2.25 + 5) 10^3\\) = 0Ns\n\n', '(a)(i) Linear momentum is the product of the mass of a body and its velocity.\n(ii) Impulse is the product of a large force and short time for which the force acts.~(b) The principle of conservation of linear momentum states that in system of colliding bodies the total momentum of the system is constant provided no external forces act on the system.\n(c) (i) \\(m_1u_1 + m_2U_2 = v(m_1 + m_2)\\) 5 x 10\\(^{3}\\) x 3 + 0\n = v(5 x 10\\(^{3}\\) + 2.5 x 10\\(^{3}\\)) v = \\(\\frac{15 \\times 10^3}{7.5 \\times 10^3}\\) = 2m/s\n(ii) Initial K.E = \\(\\frac{1}{2}mv^2 = \\frac{1}{2} MV^2 = \\frac{1}{2} times 5 \\times 10^{3} \\times 3^2\\)\n= 2.25 x 10\\(^4\\)\nFinal K.E = \\(\\frac{1}{2}mv^2 = \\frac{1}{2} (5 + 2.5) \\times 10^3 \\times 2^2\\)\n= 1.5 x 10\\(^{4}\\)J\nLoss in K.E of the couple = (2.25 - 1.5) x 10\\(^4\\)\n= 75 x 10\\(^3\\)J\n(iii) Impulse = (\\(mu_1U_1 + m_2U_2\\)) - v\n(\\(mu_1 +m_2) (0 + 5 \\times 10^3) - 2 (2.25 + 5) 10^3\\) = 0Ns\n~(c) (i) \\(m_1u_1 + m_2U_2 = v(m_1 + m_2)\\) 5 x 10\\(^{3}\\) x 3 + 0\n = v(5 x 10\\(^{3}\\) + 2.5 x 10\\(^{3}\\)) v = \\(\\frac{15 \\times 10^3}{7.5 \\times 10^3}\\) = 2m/s\n(ii) Initial K.E = \\(\\frac{1}{2}mv^2 = \\frac{1}{2} MV^2 = \\frac{1}{2} times 5 \\times 10^{3} \\times 3^2\\)\n= 2.25 x 10\\(^4\\)\nFinal K.E = \\(\\frac{1}{2}mv^2 = \\frac{1}{2} (5 + 2.5) \\times 10^3 \\times 2^2\\)\n= 1.5 x 10\\(^{4}\\)J\nLoss in K.E of the couple = (2.25 - 1.5) x 10\\(^4\\)\n= 75 x 10\\(^3\\)J\n(iii) Impulse = (\\(mu_1U_1 + m_2U_2\\)) - v\n(\\(mu_1 +m_2) (0 + 5 \\times 10^3) - 2 (2.25 + 5) 10^3\\) = 0Ns\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(222, '2018', '\n\nQuestion\n\n(a) Using the kinetic theory of matter, explain why;\n(i) Evaporation causes cooling\n(ii) Boiling water changes to steam without any change in temperature, although heat is being supplied to the water.\n\n(b) (i) State Boyle\'s law.\n(ii) With the aid of a labelled diagram, describe an experiment to illustrate the relationship between the volume and pressure of a given mass of gas at constant temperature. (iii) State two precautions necessary to obtain accurate results.\n\n\n', '', '(a) Using the kinetic theory of matter, explain why;\n(i) Evaporation causes cooling\n(ii) Boiling water changes to steam without any change in temperature, although heat is being supplied to the water.\n\n(b) (i) State Boyle\'s law.\n(ii) With the aid of a labelled diagram, describe an experiment to illustrate the relationship between the volume and pressure of a given mass of gas at constant temperature. (iii) State two precautions necessary to obtain accurate results.\n', '(a) Using the kinetic theory of matter, explain why;\n(i) Evaporation causes cooling\n(ii) Boiling water changes to steam without any change in temperature, although heat is being supplied to the water.\n~(b) (i) State Boyle\'s law.\n(ii) With the aid of a labelled diagram, describe an experiment to illustrate the relationship between the volume and pressure of a given mass of gas at constant temperature. (iii) State two precautions necessary to obtain accurate results.\n', NULL, '\n\nExplanation\n(a)(i) Liquids consist of molecules in random motion having different velocities, hence near the surface molecules with higher velocities and kinetic energy escape from the surface. Therefore the average kinetic energy of the remaining liquid is reduced. Since the average kinetic energy is a measure of temperature, therefore temperature is lowered. Hence evaporation causes cooling.\n(ii) The heat supplied to boiling water increases the kinetic energy of the molecules and also weaken the bonds between the molecules. Hence at boiling the heat supplied is used in breaking the bonds, therefore molecules escaped as steam without change in temperature.\n(b) (i) Boyle\'s law states that the volume of a fixed mass of gas at constant temperature varies inversely as the pressure upon it.\n(ii)\n\nIntroduce dry air into burette (B) and close the tap. Read and record the barometer reading H i.e. a.p., keeping B steady, raise the reservoir R, read the mercury head h from the scale, at the same time the corresponding volume V of air from the Repeat burette. Repeat the experiment with various values of H, and determine the corresponding values of h and v. Evaluate P = H ± h and V. Graph of P against V\\(^{-1}\\)\n\nThe graph P against V\\(^{-1}\\) is a straight line graph passing through the origin, showing that pressure is inversely proportional to volume.\n(iii) Precautions;\n– I ensured that dry air is used.\n– I avoided parallax error when reading metre rule.\n\n\n', '', 'Explanation\n(a)(i) Liquids consist of molecules in random motion having different velocities, hence near the surface molecules with higher velocities and kinetic energy escape from the surface. Therefore the average kinetic energy of the remaining liquid is reduced. Since the average kinetic energy is a measure of temperature, therefore temperature is lowered. Hence evaporation causes cooling.\n(ii) The heat supplied to boiling water increases the kinetic energy of the molecules and also weaken the bonds between the molecules. Hence at boiling the heat supplied is used in breaking the bonds, therefore molecules escaped as steam without change in temperature.\n(b) (i) Boyle\'s law states that the volume of a fixed mass of gas at constant temperature varies inversely as the pressure upon it.\n(ii)\n\nIntroduce dry air into burette (B) and close the tap. Read and record the barometer reading H i.e. a.p., keeping B steady, raise the reservoir R, read the mercury head h from the scale, at the same time the corresponding volume V of air from the Repeat burette. Repeat the experiment with various values of H, and determine the corresponding values of h and v. Evaluate P = H ± h and V. Graph of P against V\\(^{-1}\\)\n\nThe graph P against V\\(^{-1}\\) is a straight line graph passing through the origin, showing that pressure is inversely proportional to volume.\n(iii) Precautions;\n– I ensured that dry air is used.\n– I avoided parallax error when reading metre rule.\n\n\n', '(a)(i) Liquids consist of molecules in random motion having different velocities, hence near the surface molecules with higher velocities and kinetic energy escape from the surface. Therefore the average kinetic energy of the remaining liquid is reduced. Since the average kinetic energy is a measure of temperature, therefore temperature is lowered. Hence evaporation causes cooling.\n(ii) The heat supplied to boiling water increases the kinetic energy of the molecules and also weaken the bonds between the molecules. Hence at boiling the heat supplied is used in breaking the bonds, therefore molecules escaped as steam without change in temperature.~(b) (i) Boyle\'s law states that the volume of a fixed mass of gas at constant temperature varies inversely as the pressure upon it.\n(ii)\n\nIntroduce dry air into burette (B) and close the tap. Read and record the barometer reading H i.e. a.p., keeping B steady, raise the reservoir R, read the mercury head h from the scale, at the same time the corresponding volume V of air from the Repeat burette. Repeat the experiment with various values of H, and determine the corresponding values of h and v. Evaluate P = H ± h and V. Graph of P against V\\(^{-1}\\)\n\nThe graph P against V\\(^{-1}\\) is a straight line graph passing through the origin, showing that pressure is inversely proportional to volume.\n(iii) Precautions;\n– I ensured that dry air is used.\n– I avoided parallax error when reading metre rule.\n\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(223, '2018', '\n\nQuestion\n\na) (i) Illustrate, using a ray diagram, how an image can be formed by a convex mirror.\n(ii) State one advantage rid one disadvantage of using a convex mirror as a iving mirror.\n(iii) Explain the action of a compound microscope.\n(b) Illustrate using labelled diagrams only, sonometer wire of length I, vibrating at its fundamental (ii) first overtone (iii) second overtone\n(c) A tuning fork vibrating at a frequency of 512 Hz is held over the top of a jar filled with water and fitted with a tap at the buttom. If the jar is 60 cm tall and the speed of sound is 350 ms\\(^{-1}\\), determine the possible resonance position(s).\n\n', '', 'a) (i) Illustrate, using a ray diagram, how an image can be formed by a convex mirror.\n(ii) State one advantage rid one disadvantage of using a convex mirror as a iving mirror.\n(iii) Explain the action of a compound microscope.\n(b) Illustrate using labelled diagrams only, sonometer wire of length I, vibrating at its fundamental (ii) first overtone (iii) second overtone\n(c) A tuning fork vibrating at a frequency of 512 Hz is held over the top of a jar filled with water and fitted with a tap at the buttom. If the jar is 60 cm tall and the speed of sound is 350 ms\\(^{-1}\\), determine the possible resonance position(s).', '(i) Illustrate, using a ray diagram, how an image can be formed by a convex mirror.\n(ii) State one advantage rid one disadvantage of using a convex mirror as a iving mirror.\n(iii) Explain the action of a compound microscope.~(b) Illustrate using labelled diagrams only, sonometer wire of length I, vibrating at its fundamental (ii) first overtone (iii) second overtone\n(c) A tuning fork vibrating at a frequency of 512 Hz is held over the top of a jar filled with water and fitted with a tap at the buttom. If the jar is 60 cm tall and the speed of sound is 350 ms\\(^{-1}\\), determine the possible resonance position(s).~(c) A tuning fork vibrating at a frequency of 512 Hz is held over the top of a jar filled with water and fitted with a tap at the buttom. If the jar is 60 cm tall and the speed of sound is 350 ms\\(^{-1}\\), determine the possible resonance position(s).', NULL, '\n\nExplanation\n(a)(i)\n\n(ii) Advantages of using a convex mirror as driving mirror.\n– It has wide field of view\n– Image is always erect.\nDisadvantages;\n– Image is always smaller than the object\n– It gives a false impression of the distance as image seems further away.\n(iii) The objective lens forms the image of an object placed beyond its focal length within the focal length of eye piece. This image is real inverted and magnified. This image serves as object to the eye piece which functions as a simple microscope to form a final virtual and magnified image.\n\n\n(c) \\(\\lambda = \\frac{v}{f} = \\frac{350}{512} = 0.68m\\)\n1st resonance position = \\(\\frac{\\lambda}{f} = \\frac{0.68}{4} = 0.17m\\)\n2nd resonance position = \\frac{3\\lambda}{4} \\frac{3 \\times 0.68}{4} = 0.51m\\)\n3rd resonance position = \\(\\frac{5\\lambda}{4} = \\frac{5 \\times 0.85}{4} = 0.85m\\)\nSince 0.85m is greater than the length of the jar (0.6m), therefore only the first and second resonance positions are possible.\n\n\n', '', 'Explanation\n(a)(i)\n\n(ii) Advantages of using a convex mirror as driving mirror.\n– It has wide field of view\n– Image is always erect.\nDisadvantages;\n– Image is always smaller than the object\n– It gives a false impression of the distance as image seems further away.\n(iii) The objective lens forms the image of an object placed beyond its focal length within the focal length of eye piece. This image is real inverted and magnified. This image serves as object to the eye piece which functions as a simple microscope to form a final virtual and magnified image.\n\n\n(c) \\(\\lambda = \\frac{v}{f} = \\frac{350}{512} = 0.68m\\)\n1st resonance position = \\(\\frac{\\lambda}{f} = \\frac{0.68}{4} = 0.17m\\)\n2nd resonance position = \\frac{3\\lambda}{4} \\frac{3 \\times 0.68}{4} = 0.51m\\)\n3rd resonance position = \\(\\frac{5\\lambda}{4} = \\frac{5 \\times 0.85}{4} = 0.85m\\)\nSince 0.85m is greater than the length of the jar (0.6m), therefore only the first and second resonance positions are possible.\n\n\n', '~(c) \\(\\lambda = \\frac{v}{f} = \\frac{350}{512} = 0.68m\\)\n1st resonance position = \\(\\frac{\\lambda}{f} = \\frac{0.68}{4} = 0.17m\\)\n2nd resonance position = \\frac{3\\lambda}{4} \\frac{3 \\times 0.68}{4} = 0.51m\\)\n3rd resonance position = \\(\\frac{5\\lambda}{4} = \\frac{5 \\times 0.85}{4} = 0.85m\\)\nSince 0.85m is greater than the length of the jar (0.6m), therefore only the first and second resonance positions are possible.\n\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(224, '2018', '\n\nQuestion\n\n(a)(i) What is meant by neutral point in a magnetic field?\n(ii) Draw and label a diagram to show the pattern and direction of the magnetic field produced around a straight current-carrying wire.\n(b) When is an ammeter said to be (i) Sensitive (ii) accurate?\n(c)(i) Explain, using a labelled diagram, how a delicate magnetic material could be protected, from the Earth\'s magnetic field.\n(ii) A charge of 1.6 x 10\\(^{-19}\\) C enters a magnetic field of flux density 2.0 T with a velocity of 2.5 x 10\\(^7\\) ms\\(^{-1}\\) at an angle of 30° with the field. Calculate the magnitude of the force exerted on the charge by the field.\n(d) State the laws of electro-magnetic induction.\n\n', '', '(a)(i) What is meant by neutral point in a magnetic field?\n(ii) Draw and label a diagram to show the pattern and direction of the magnetic field produced around a straight current-carrying wire.\n(b) When is an ammeter said to be (i) Sensitive (ii) accurate?\n(c)(i) Explain, using a labelled diagram, how a delicate magnetic material could be protected, from the Earth\'s magnetic field.\n(ii) A charge of 1.6 x 10\\(^{-19}\\) C enters a magnetic field of flux density 2.0 T with a velocity of 2.5 x 10\\(^7\\) ms\\(^{-1}\\) at an angle of 30° with the field. Calculate the magnitude of the force exerted on the charge by the field.\n(d) State the laws of electro-magnetic induction.', '(a)(i) What is meant by neutral point in a magnetic field?\n(ii) Draw and label a diagram to show the pattern and direction of the magnetic field produced around a straight current-carrying wire.~(b) When is an ammeter said to be (i) Sensitive (ii) accurate?\n(c)(i) Explain, using a labelled diagram, how a delicate magnetic material could be protected, from the Earth\'s magnetic field.\n(ii) A charge of 1.6 x 10\\(^{-19}\\) C enters a magnetic field of flux density 2.0 T with a velocity of 2.5 x 10\\(^7\\) ms\\(^{-1}\\) at an angle of 30° with the field. Calculate the magnitude of the force exerted on the charge by the field.\n(d) State the laws of electro-magnetic induction.~(c)(i) Explain, using a labelled diagram, how a delicate magnetic material could be protected, from the Earth\'s magnetic field.\n(ii) A charge of 1.6 x 10\\(^{-19}\\) C enters a magnetic field of flux density 2.0 T with a velocity of 2.5 x 10\\(^7\\) ms\\(^{-1}\\) at an angle of 30° with the field. Calculate the magnitude of the force exerted on the charge by the field.\n(d) State the laws of electro-magnetic induction.~(d) State the laws of electro-magnetic induction.', NULL, '\n\nExplanation\n(a) (i) Neutral point is a point at which the resultant magnetic flux density is zero.\n(ii)\n\n(b)(i) (ii) An ammeter is accurate if the current it mea-sures is close to the true value of the current flowing through the instrument.\n\n(c)\n\nThe material is surrounded by a thick wall of ferro-magnetic material which does not allow magnetism to pass through them. This shields or protects the material from the earth\'s magnetic field.\n(ii) F = qvBsin\\(\\theta\\) = 1.6 x 10\\(^{-19}\\) x 2.5 x 10\\(^{7}\\) x 2 x 0.5 = 4 x 10\\(^{2}\\) N\n(d) Electromagnetic induction laws are\n- Whenever there is a change in the magnetic flux linked with a circuit, an e.m.f is induced, the magnitude of which is proportional to the rate of change of the flux linked with the circuit.\n- The direction of the induced current is always such as to oppose the change producing it.\n\n', '', 'Explanation\n(a) (i) Neutral point is a point at which the resultant magnetic flux density is zero.\n(ii)\n\n(b)(i) (ii) An ammeter is accurate if the current it mea-sures is close to the true value of the current flowing through the instrument.\n\n(c)\n\nThe material is surrounded by a thick wall of ferro-magnetic material which does not allow magnetism to pass through them. This shields or protects the material from the earth\'s magnetic field.\n(ii) F = qvBsin\\(\\theta\\) = 1.6 x 10\\(^{-19}\\) x 2.5 x 10\\(^{7}\\) x 2 x 0.5 = 4 x 10\\(^{2}\\) N\n(d) Electromagnetic induction laws are\n- Whenever there is a change in the magnetic flux linked with a circuit, an e.m.f is induced, the magnitude of which is proportional to the rate of change of the flux linked with the circuit.\n- The direction of the induced current is always such as to oppose the change producing it.\n\n', '(a) (i) Neutral point is a point at which the resultant magnetic flux density is zero.\n(ii)\n~(b)(i) (ii) An ammeter is accurate if the current it mea-sures is close to the true value of the current flowing through the instrument.\n\n(c)\n\nThe material is surrounded by a thick wall of ferro-magnetic material which does not allow magnetism to pass through them. This shields or protects the material from the earth\'s magnetic field.\n(ii) F = qvBsin\\(\\theta\\) = 1.6 x 10\\(^{-19}\\) x 2.5 x 10\\(^{7}\\) x 2 x 0.5 = 4 x 10\\(^{2}\\) N\n(d) Electromagnetic induction laws are\n- Whenever there is a change in the magnetic flux linked with a circuit, an e.m.f is induced, the magnitude of which is proportional to the rate of change of the flux linked with the circuit.\n- The direction of the induced current is always such as to oppose the change producing it.\n~(c)\n\nThe material is surrounded by a thick wall of ferro-magnetic material which does not allow magnetism to pass through them. This shields or protects the material from the earth\'s magnetic field.\n(ii) F = qvBsin\\(\\theta\\) = 1.6 x 10\\(^{-19}\\) x 2.5 x 10\\(^{7}\\) x 2 x 0.5 = 4 x 10\\(^{2}\\) N\n(d) Electromagnetic induction laws are\n- Whenever there is a change in the magnetic flux linked with a circuit, an e.m.f is induced, the magnitude of which is proportional to the rate of change of the flux linked with the circuit.\n- The direction of the induced current is always such as to oppose the change producing it.\n~(d) Electromagnetic induction laws are\n- Whenever there is a change in the magnetic flux linked with a circuit, an e.m.f is induced, the magnitude of which is proportional to the rate of change of the flux linked with the circuit.\n- The direction of the induced current is always such as to oppose the change producing it.\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(225, '2018', '\n\nQuestion\n\n(a)(i) Explain photoelectric emission.\n(ii) State four applications of photoelectric emission.\n(b) Draw and label a diagram showing the structure of a simple type of a photocell and explain its mode of operation.\n(c) In a photocell, no electrons are emitted until the threshold frequency of light is reached.\n(i) Explain what happens to the energy of the light before emission of electrons begin.\n(ii) State one factor that may affect the number of emitted electrons.\n\n\n', '', '(a)(i) Explain photoelectric emission.\n(ii) State four applications of photoelectric emission.\n(b) Draw and label a diagram showing the structure of a simple type of a photocell and explain its mode of operation.\n(c) In a photocell, no electrons are emitted until the threshold frequency of light is reached.\n(i) Explain what happens to the energy of the light before emission of electrons begin.\n(ii) State one factor that may affect the number of emitted electrons.\n', '(a)(i) Explain photoelectric emission.\n(ii) State four applications of photoelectric emission.~(b) Draw and label a diagram showing the structure of a simple type of a photocell and explain its mode of operation.\n(c) In a photocell, no electrons are emitted until the threshold frequency of light is reached.\n(i) Explain what happens to the energy of the light before emission of electrons begin.\n(ii) State one factor that may affect the number of emitted electrons.\n~(c) In a photocell, no electrons are emitted until the threshold frequency of light is reached.\n(i) Explain what happens to the energy of the light before emission of electrons begin.\n(ii) State one factor that may affect the number of emitted electrons.\n', NULL, '\n\nExplanation\n(a)(i) Photoelectric emission is the ejection of electrons from the surface of a metal when light of appropriate wavelength and frequency is incident on the metal.\n(ii) Applications of photoelectric emission i.e. photo cells:\n- Burglar alarms\n- Photo meters\n- Automatic doors\n- Sound production from film tracks\n- T.V. camera etc\n(b)\n\nWhen light of a set wavelength and frequency falls on the cathode, electrons are emitted. The electrons flow toward the anode. The flow of these electrons constitute current.\n\n(c) (i) For energy of light with frequency less than threshold frequency, complete absorption takes place. The energy absorbed is used to raise the energy of electrons or overcome the potential barrier in the photoelectric material.\n(ii) The number of emitted electrons are affected by:\n- intensity\n- Number of photons.\n\n', '', 'Explanation\n(a)(i) Photoelectric emission is the ejection of electrons from the surface of a metal when light of appropriate wavelength and frequency is incident on the metal.\n(ii) Applications of photoelectric emission i.e. photo cells:\n- Burglar alarms\n- Photo meters\n- Automatic doors\n- Sound production from film tracks\n- T.V. camera etc\n(b)\n\nWhen light of a set wavelength and frequency falls on the cathode, electrons are emitted. The electrons flow toward the anode. The flow of these electrons constitute current.\n\n(c) (i) For energy of light with frequency less than threshold frequency, complete absorption takes place. The energy absorbed is used to raise the energy of electrons or overcome the potential barrier in the photoelectric material.\n(ii) The number of emitted electrons are affected by:\n- intensity\n- Number of photons.\n\n', '(a)(i) Photoelectric emission is the ejection of electrons from the surface of a metal when light of appropriate wavelength and frequency is incident on the metal.\n(ii) Applications of photoelectric emission i.e. photo cells:\n- Burglar alarms\n- Photo meters\n- Automatic doors\n- Sound production from film tracks\n- T.V. camera etc~(b)\n\nWhen light of a set wavelength and frequency falls on the cathode, electrons are emitted. The electrons flow toward the anode. The flow of these electrons constitute current.\n\n(c) (i) For energy of light with frequency less than threshold frequency, complete absorption takes place. The energy absorbed is used to raise the energy of electrons or overcome the potential barrier in the photoelectric material.\n(ii) The number of emitted electrons are affected by:\n- intensity\n- Number of photons.\n~(c) (i) For energy of light with frequency less than threshold frequency, complete absorption takes place. The energy absorbed is used to raise the energy of electrons or overcome the potential barrier in the photoelectric material.\n(ii) The number of emitted electrons are affected by:\n- intensity\n- Number of photons.\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(226, '2018', '\n\nQuestion\n\nA ball thrown vertically upward reaches a maximum height of 50 m above the level of projection. Calculate the;\n(i) time taken to reach the maximum height,\n(ii) speed of the throw. [g = 10 ms\\(^{-2}\\)]\n\n', '', 'A ball thrown vertically upward reaches a maximum height of 50 m above the level of projection. Calculate the;\n(i) time taken to reach the maximum height,\n(ii) speed of the throw. [g = 10 ms\\(^{-2}\\)]', '', NULL, '\n\nExplanation\nV\\(_1\\) = U - gt\nU = V + gt = 0 + 10t\nh = ut\\(\\frac{1}{2}\\) - gt\\(^2\\),\n50 = 10t\\(^2\\) \\(\\frac{1}{2}\\) x 10 t\\(^2\\)\n50 = 10t\\(^2\\) - 5t\\(^2\\)\n50 = 5t\\(^2\\) = 10\nt\\(^2\\) = 10\nt = \\(\\sqrt{10}\\) = 3.16s\n(ii) V = U - gt, O = U - 10 x 3.16\nU = 31.16\n\n\n', '', 'Explanation\nV\\(_1\\) = U - gt\nU = V + gt = 0 + 10t\nh = ut\\(\\frac{1}{2}\\) - gt\\(^2\\),\n50 = 10t\\(^2\\) \\(\\frac{1}{2}\\) x 10 t\\(^2\\)\n50 = 10t\\(^2\\) - 5t\\(^2\\)\n50 = 5t\\(^2\\) = 10\nt\\(^2\\) = 10\nt = \\(\\sqrt{10}\\) = 3.16s\n(ii) V = U - gt, O = U - 10 x 3.16\nU = 31.16\n\n\n', '', 'WAEC-PK-1', 'WAEC-PHY', 1),
(227, '2018', '\n\nQuestion\n\nA lead shot is projected from tha ground level with a velocity u at an angle \\(\\theta\\) to the horizontal. Given the time, t for the lead shot to reach its maximum height as; t = \\(\\frac{u \\sin \\theta}{g}\\) where \"g\" is the acceleration of free fall due to gravity, show that the greatest height reached by the body is h\\(_{\\text{max}} = \\frac{u^2 \\sin^2 \\theta}{2g}\\)\n\n\n', 'A lead shot is projected from tha ground level with a velocity u at an angle \\(\\theta\\) to the horizontal. Given the time, t for the lead shot to reach its maximum height as; t = \\(\\frac{u \\sin \\theta}{g}\\) where \"g\" is the acceleration of free fall due to gravity, show that the greatest height reached by the body is h\\(_{\\text{max}} = \\frac{u^2 \\sin^2 \\theta}{2g}\\)\n', '', '', NULL, '\n\nExplanation\nFor verticsl motion of projectile h = (U sin \\(\\theta\\)) t \\(\\frac{1}{2}\\)gt\\(^2\\)\n= U sin \\(\\theta\\) (\\(\\frac{Usin \\theta}{g}\\))\\(\\frac{1}{2}\\)g (\\(\\frac{Usin \\theta}{g}\\)) = (\\(\\frac{U^2 \\sin^2 \\theta}{g}\\))\n\n\n\n', 'For verticsl motion of projectile h = (U sin \\(\\theta\\)) t \\(\\frac{1}{2}\\)gt\\(^2\\)\n= U sin \\(\\theta\\) (\\(\\frac{Usin \\theta}{g}\\))\\(\\frac{1}{2}\\)g (\\(\\frac{Usin \\theta}{g}\\)) = (\\(\\frac{U^2 \\sin^2 \\theta}{g}\\))\n\n\n\n', '', '', 'WAEC-PK-1', 'WAEC-PHY', 1),
(228, '2018', '\n\nQuestion\n\n(a) What is electrolysis?\n(b) A current of 2A is passed through a copper voltameter for 5 minutes. If the electrochemical equivalent of copper is 3.27 x 10\\(^{-7}\\) kg. Determine the mass of the corper deposited.\n\n', '', '(a) What is electrolysis?\n(b) A current of 2A is passed through a copper voltameter for 5 minutes. If the electrochemical equivalent of copper is 3.27 x 10\\(^{-7}\\) kg. Determine the mass of the corper deposited.', '(a) What is electrolysis?~(b) A current of 2A is passed through a copper voltameter for 5 minutes. If the electrochemical equivalent of copper is 3.27 x 10\\(^{-7}\\) kg. Determine the mass of the corper deposited.', NULL, '\n\nExplanation\n(a) Electrolysis is the decomposition of an electrolyte into ions by passing current through the electrolyte.\n(b) Mass of copper deposited M = zlt\n= 3.27 x 10\\(^{7}\\) x 2 x 300 = 1.962 x 10\\(^{-4}\\) kg\n\n', '', 'Explanation\n(a) Electrolysis is the decomposition of an electrolyte into ions by passing current through the electrolyte.\n(b) Mass of copper deposited M = zlt\n= 3.27 x 10\\(^{7}\\) x 2 x 300 = 1.962 x 10\\(^{-4}\\) kg\n\n', '(a) Electrolysis is the decomposition of an electrolyte into ions by passing current through the electrolyte.~(b) Mass of copper deposited M = zlt\n= 3.27 x 10\\(^{7}\\) x 2 x 300 = 1.962 x 10\\(^{-4}\\) kg\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(229, '2018', '\n\nQuestion\n\nThe value of the e.m.f of a voltaic cell, which has dilute tetraoxosulphate (VI) acid as its electrolyte and copper and zinc as its electrodes becomes less with use. Explain this observation and state how it can be corrected.\n\n', 'The value of the e.m.f of a voltaic cell, which has dilute tetraoxosulphate (VI) acid as its electrolyte and copper and zinc as its electrodes becomes less with use. Explain this observation and state how it can be corrected.', '', '', NULL, '\n\nExplanation\nExplanation: This is due to the accumulation of hydrogen bubbles in the positive copper plate, thereby insulating the electrodes. The hydrogen-zinc pair sets up opposing e.m.f. that reduces the e.m.f. originally established. Correction: It can be corrected by introducing a suitable depolarises.\n\n', 'Explanation: This is due to the accumulation of hydrogen bubbles in the positive copper plate, thereby insulating the electrodes. The hydrogen-zinc pair sets up opposing e.m.f. that reduces the e.m.f. originally established. Correction: It can be corrected by introducing a suitable depolarises.\n\n', '', '', 'WAEC-PK-1', 'WAEC-PHY', 1),
(230, '2018', '\n\nQuestion\n\n(a) What is meant by \'a beam of polarised light?\n(b) With the aid of well labelled diagrams, illustrate the action in of a polaroid spectacle on a beam of sunlight.\n\n', '', '(a) What is meant by \'a beam of polarised light?\n(b) With the aid of well labelled diagrams, illustrate the action in of a polaroid spectacle on a beam of sunlight.', '(a) What is meant by \'a beam of polarised light?~(b) With the aid of well labelled diagrams, illustrate the action in of a polaroid spectacle on a beam of sunlight.', NULL, '\n\nExplanation\n(a) A beam of polarised light is on constrained in a single plane perpendicular to the beam.\n\n\n', '(a) A beam of polarised light is on constrained in a single plane perpendicular to the beam.\n\n\n', '', '', 'WAEC-PK-1', 'WAEC-PHY', 1),
(231, '2018', '\n\nQuestion\n\n(a) State two properties of cathode rays.\n(b) Explain how the intensity and energy of cathode rays may be increased.\n\n', '', '(a) State two properties of cathode rays.\n(b) Explain how the intensity and energy of cathode rays may be increased.', '(a) State two properties of cathode rays.~(b) Explain how the intensity and energy of cathode rays may be increased.', NULL, '\n\nExplanation\n(a) Cathode rays:\n(i) Are negatively charged particles\n(ii) In field free space travel in straight line.\n(iii) Are deflected by electric and magnetic fields.\n(b) The intensity of the cathode rays may be increased by increasing the current through the filament. The energy of the cathode rays may be increased by increasing the p.d. between the anode and the cathode i.e. the anode potential.\n\n', '', 'Explanation\n(a) Cathode rays:\n(i) Are negatively charged particles\n(ii) In field free space travel in straight line.\n(iii) Are deflected by electric and magnetic fields.\n(b) The intensity of the cathode rays may be increased by increasing the current through the filament. The energy of the cathode rays may be increased by increasing the p.d. between the anode and the cathode i.e. the anode potential.\n\n', '(a) Cathode rays:\n(i) Are negatively charged particles\n(ii) In field free space travel in straight line.\n(iii) Are deflected by electric and magnetic fields.~(b) The intensity of the cathode rays may be increased by increasing the current through the filament. The energy of the cathode rays may be increased by increasing the p.d. between the anode and the cathode i.e. the anode potential.\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(232, '2018', '\n\nQuestion\n\n(a) The mass and wavelength of a moving electron are 9.0 x 10\\(^{-31}\\) kg and 1.0 x 10\\(^{-10}\\)m respectively. Calculate the kinetic energy of the electron. [ h = 6.6 x 10\\(^{-34}\\) Js]\n\n', '(a) The mass and wavelength of a moving electron are 9.0 x 10\\(^{-31}\\) kg and 1.0 x 10\\(^{-10}\\)m respectively. Calculate the kinetic energy of the electron. [ h = 6.6 x 10\\(^{-34}\\) Js]', '', '', NULL, '\n\nExplanation\n\\(\\lambda = \\frac{h}{p} = \\frac{h}{mv}\\)\nV = \\(\\frac{h}{m_2}\\)\nElectron kitenic energy = \\(\\frac{1}{2} MV^2 = \\frac{h^2}{2m \\lambda^2}\\)\n= \\(\\frac{(6.6 \\times 10^{-34})^2}{2 \\times 9.0 \\times 10^{-37} \\times 1.0 \\times 10^{-20}}\\) = 2.39 x 10\\(^{-17}\\)J\n\n', '\\(\\lambda = \\frac{h}{p} = \\frac{h}{mv}\\)\nV = \\(\\frac{h}{m_2}\\)\nElectron kitenic energy = \\(\\frac{1}{2} MV^2 = \\frac{h^2}{2m \\lambda^2}\\)\n= \\(\\frac{(6.6 \\times 10^{-34})^2}{2 \\times 9.0 \\times 10^{-37} \\times 1.0 \\times 10^{-20}}\\) = 2.39 x 10\\(^{-17}\\)J\n\n', '', '', 'WAEC-PK-1', 'WAEC-PHY', 1),
(233, '2018', '\n\nQuestion\n\nThe above reprt the graph electron energy against the frequency of the radiation incident on a metal surface. lnterprete the;\n(i) slope of the graph;\n(ii) intercept, OC;\n(iii) intercept, OK.\n\n\n', '', 'The above reprt the graph electron energy against the frequency of the radiation incident on a metal surface. lnterprete the;\n(i) slope of the graph;\n(ii) intercept, OC;\n(iii) intercept, OK.\n', '', NULL, '\n\nExplanation\n(i) Slope represents the Planck\'s constant, h.\n(ii) Intercept OC represents the work function of the metal.\n(iii) The intercept OK represents the threshold frequency of the radiation.\n\n', '', 'Explanation\n(i) Slope represents the Planck\'s constant, h.\n(ii) Intercept OC represents the work function of the metal.\n(iii) The intercept OK represents the threshold frequency of the radiation.\n\n', '', 'WAEC-PK-1', 'WAEC-PHY', 1),
(234, '2018', '\n\nQuestion\n\n(a) Distinguish between stress and strain as used in elasticity.\n(b) When a force of 40 N is applied to the free end of an elastic cord, an extension of 5cm is produced in the cord. Calculate the work done on the cord.\n\n', '', '(a) Distinguish between stress and strain as used in elasticity.\n(b) When a force of 40 N is applied to the free end of an elastic cord, an extension of 5cm is produced in the cord. Calculate the work done on the cord.', '(a) Distinguish between stress and strain as used in elasticity.~(b) When a force of 40 N is applied to the free end of an elastic cord, an extension of 5cm is produced in the cord. Calculate the work done on the cord.', NULL, '\n\nExplanation\n(a)(i) Stress is force per unit cross sectional area, while strain is extension over original length\n(ii) Stress has unit of Nm\\(^{-2}\\) while strain has no unit.\n(b) Work done on the cord = \\(\\frac{1}{2}\\) Fe\n= \\(\\frac{1}{2}\\) x 40 x 0.05 = 1.0J\n\n', '', 'Explanation\n(a)(i) Stress is force per unit cross sectional area, while strain is extension over original length\n(ii) Stress has unit of Nm\\(^{-2}\\) while strain has no unit.\n(b) Work done on the cord = \\(\\frac{1}{2}\\) Fe\n= \\(\\frac{1}{2}\\) x 40 x 0.05 = 1.0J\n\n', '(a)(i) Stress is force per unit cross sectional area, while strain is extension over original length\n(ii) Stress has unit of Nm\\(^{-2}\\) while strain has no unit.~(b) Work done on the cord = \\(\\frac{1}{2}\\) Fe\n= \\(\\frac{1}{2}\\) x 40 x 0.05 = 1.0J\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(235, '2018', '\n\nQuestion\n\n(a) In his first attempt, a long jumper took off from the springboard with a speed of 8 ms\\(^{-1}\\) at 30° to the horizontal. He makes a second attempt with the same speed at 45° to the horizontal. Given that the expression for the horizontal range of a projectile is \\(\\frac{v^2 sin \\theta}{g}\\) where all the symbols have their usual meanings, show that he gains a distance of 0.8576 m in his second attempt.\n(b)(i) State Hooke\'s law of elasticity.\n(ii) Describe an experiment to verify Hooke\'s law.\n(iii) State two precautions you would take if you were to perform this experiment in the laboratory.\n(c) A spiral spring of natural length 20.00 cm has a scale pan hanging freely in its lower end. When an object of mass 40 g is placed in the pan, its length becomes 21.80 cm. When another object of mass 60g Is placed in the pan, the length becomes 22.05cm. Calculate the mass of the scale pan. [g = 10 ms\\(^{-2}\\)]\n\n', '', '(a) In his first attempt, a long jumper took off from the springboard with a speed of 8 ms\\(^{-1}\\) at 30° to the horizontal. He makes a second attempt with the same speed at 45° to the horizontal. Given that the expression for the horizontal range of a projectile is \\(\\frac{v^2 sin \\theta}{g}\\) where all the symbols have their usual meanings, show that he gains a distance of 0.8576 m in his second attempt.\n(b)(i) State Hooke\'s law of elasticity.\n(ii) Describe an experiment to verify Hooke\'s law.\n(iii) State two precautions you would take if you were to perform this experiment in the laboratory.\n(c) A spiral spring of natural length 20.00 cm has a scale pan hanging freely in its lower end. When an object of mass 40 g is placed in the pan, its length becomes 21.80 cm. When another object of mass 60g Is placed in the pan, the length becomes 22.05cm. Calculate the mass of the scale pan. [g = 10 ms\\(^{-2}\\)]', '(a) In his first attempt, a long jumper took off from the springboard with a speed of 8 ms\\(^{-1}\\) at 30° to the horizontal. He makes a second attempt with the same speed at 45° to the horizontal. Given that the expression for the horizontal range of a projectile is \\(\\frac{v^2 sin \\theta}{g}\\) where all the symbols have their usual meanings, show that he gains a distance of 0.8576 m in his second attempt.~(b)(i) State Hooke\'s law of elasticity.\n(ii) Describe an experiment to verify Hooke\'s law.\n(iii) State two precautions you would take if you were to perform this experiment in the laboratory.\n(c) A spiral spring of natural length 20.00 cm has a scale pan hanging freely in its lower end. When an object of mass 40 g is placed in the pan, its length becomes 21.80 cm. When another object of mass 60g Is placed in the pan, the length becomes 22.05cm. Calculate the mass of the scale pan. [g = 10 ms\\(^{-2}\\)]~(c) A spiral spring of natural length 20.00 cm has a scale pan hanging freely in its lower end. When an object of mass 40 g is placed in the pan, its length becomes 21.80 cm. When another object of mass 60g Is placed in the pan, the length becomes 22.05cm. Calculate the mass of the scale pan. [g = 10 ms\\(^{-2}\\)]', NULL, '\n\nExplanation\n(a) Range = \\(\\frac{U^2 sin 2\\theta}{g}\\)\nfor first attempy, Range = \\(\\frac{8^2 \\sin 60}{10} = \\frac{64 x\\times 0.8660}{10}\\)\n= 5.5424m\nfor second attemt; Range = \\(\\frac{8^2 \\sin 90}{10} = \\frac{64 \\times 1}{10}\\)\n= 6.4m\ngain 6.40 - 5.54 = 0.8576m\n(b)(i) Hooke\'s law states that provided the limit of proportionality is not exceeded, the force is proportional to the extension in an elastic material.\n(ii) Description: Suspend the spiral spring from a rigid support with a pointer attached at the free end. Suspend a stretching weight from the free end, and note the position of the pointer. Suspend a known mass, note the new position and determine the extension. Repeat the experiment using various known masses. Repeat the above procedure while unloading and determine mean extension. Plot a graph of load against mean extension. A straight line graph through the origin verifies Hooke\'s law.\nPrecautions: (i) Loading beyond elastic limit should be avoided\n(ii) Pointer to be rigidly attached to the free end and made perpendicular to the spring.\n(iii) Avoid parallax error while taking readings.\n(c) mg = Ke, m = total mass If m = mass of scale pan then;\n(m + 40)g = 1.8k-----------(i)\n(m + 60) = 2.05k --------- (ii)\nDividing (i) by (ii) \\(\\frac{m + 40}{m + 60} = \\frac{1.8}{2.05}\\)\n2.05(m + 40) = 1.8 (m + 60)\nM = \\(\\frac{26}{0.25}\\) = 104gm\n\n', '', 'Explanation\n(a) Range = \\(\\frac{U^2 sin 2\\theta}{g}\\)\nfor first attempy, Range = \\(\\frac{8^2 \\sin 60}{10} = \\frac{64 x\\times 0.8660}{10}\\)\n= 5.5424m\nfor second attemt; Range = \\(\\frac{8^2 \\sin 90}{10} = \\frac{64 \\times 1}{10}\\)\n= 6.4m\ngain 6.40 - 5.54 = 0.8576m\n(b)(i) Hooke\'s law states that provided the limit of proportionality is not exceeded, the force is proportional to the extension in an elastic material.\n(ii) Description: Suspend the spiral spring from a rigid support with a pointer attached at the free end. Suspend a stretching weight from the free end, and note the position of the pointer. Suspend a known mass, note the new position and determine the extension. Repeat the experiment using various known masses. Repeat the above procedure while unloading and determine mean extension. Plot a graph of load against mean extension. A straight line graph through the origin verifies Hooke\'s law.\nPrecautions: (i) Loading beyond elastic limit should be avoided\n(ii) Pointer to be rigidly attached to the free end and made perpendicular to the spring.\n(iii) Avoid parallax error while taking readings.\n(c) mg = Ke, m = total mass If m = mass of scale pan then;\n(m + 40)g = 1.8k-----------(i)\n(m + 60) = 2.05k --------- (ii)\nDividing (i) by (ii) \\(\\frac{m + 40}{m + 60} = \\frac{1.8}{2.05}\\)\n2.05(m + 40) = 1.8 (m + 60)\nM = \\(\\frac{26}{0.25}\\) = 104gm\n\n', '(a) Range = \\(\\frac{U^2 sin 2\\theta}{g}\\)\nfor first attempy, Range = \\(\\frac{8^2 \\sin 60}{10} = \\frac{64 x\\times 0.8660}{10}\\)\n= 5.5424m\nfor second attemt; Range = \\(\\frac{8^2 \\sin 90}{10} = \\frac{64 \\times 1}{10}\\)\n= 6.4m\ngain 6.40 - 5.54 = 0.8576m~(b)(i) Hooke\'s law states that provided the limit of proportionality is not exceeded, the force is proportional to the extension in an elastic material.\n(ii) Description: Suspend the spiral spring from a rigid support with a pointer attached at the free end. Suspend a stretching weight from the free end, and note the position of the pointer. Suspend a known mass, note the new position and determine the extension. Repeat the experiment using various known masses. Repeat the above procedure while unloading and determine mean extension. Plot a graph of load against mean extension. A straight line graph through the origin verifies Hooke\'s law.\nPrecautions: (i) Loading beyond elastic limit should be avoided\n(ii) Pointer to be rigidly attached to the free end and made perpendicular to the spring.\n(iii) Avoid parallax error while taking readings.\n(c) mg = Ke, m = total mass If m = mass of scale pan then;\n(m + 40)g = 1.8k-----------(i)\n(m + 60) = 2.05k --------- (ii)\nDividing (i) by (ii) \\(\\frac{m + 40}{m + 60} = \\frac{1.8}{2.05}\\)\n2.05(m + 40) = 1.8 (m + 60)\nM = \\(\\frac{26}{0.25}\\) = 104gm\n~(c) mg = Ke, m = total mass If m = mass of scale pan then;\n(m + 40)g = 1.8k-----------(i)\n(m + 60) = 2.05k --------- (ii)\nDividing (i) by (ii) \\(\\frac{m + 40}{m + 60} = \\frac{1.8}{2.05}\\)\n2.05(m + 40) = 1.8 (m + 60)\nM = \\(\\frac{26}{0.25}\\) = 104gm\n', 'WAEC-PK-1', 'WAEC-PHY', 1);
