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
(236, '2018', '\n\nQuestion\n\n(a) Define specific heat capacity.\n(b)(i) With the aid of a labelled diagram, describe an experiment to determine the specific heat capacity of copper using a copper ball.\n(ii) State two precautions necessary to obtain accurate results.\n(c) A piece of copper block of mass 24 g at 230°C is placed in a copper calorimeter of mass 60 g containing 54 g of water at 31°C. Assuming heat losses are negligible, calculate the final steady temperature of the mixture. [specific heat capacity of water = 4200 J kg\\(^{-1}\\) K\\(^{-1}\\)] [specific heat capacity of copper = 400 J kg\\(^{-1}\\) K\\(^{-1}\\)]\n\n', '', '(a) Define specific heat capacity.\n(b)(i) With the aid of a labelled diagram, describe an experiment to determine the specific heat capacity of copper using a copper ball.\n(ii) State two precautions necessary to obtain accurate results.\n(c) A piece of copper block of mass 24 g at 230°C is placed in a copper calorimeter of mass 60 g containing 54 g of water at 31°C. Assuming heat losses are negligible, calculate the final steady temperature of the mixture. [specific heat capacity of water = 4200 J kg\\(^{-1}\\) K\\(^{-1}\\)] [specific heat capacity of copper = 400 J kg\\(^{-1}\\) K\\(^{-1}\\)]', '(a) Define specific heat capacity.~(b)(i) With the aid of a labelled diagram, describe an experiment to determine the specific heat capacity of copper using a copper ball.\n(ii) State two precautions necessary to obtain accurate results.\n(c) A piece of copper block of mass 24 g at 230°C is placed in a copper calorimeter of mass 60 g containing 54 g of water at 31°C. Assuming heat losses are negligible, calculate the final steady temperature of the mixture. [specific heat capacity of water = 4200 J kg\\(^{-1}\\) K\\(^{-1}\\)] [specific heat capacity of copper = 400 J kg\\(^{-1}\\) K\\(^{-1}\\)]~(c) A piece of copper block of mass 24 g at 230°C is placed in a copper calorimeter of mass 60 g containing 54 g of water at 31°C. Assuming heat losses are negligible, calculate the final steady temperature of the mixture. [specific heat capacity of water = 4200 J kg\\(^{-1}\\) K\\(^{-1}\\)] [specific heat capacity of copper = 400 J kg\\(^{-1}\\) K\\(^{-1}\\)]', NULL, '\n\nExplanation\n(a) Specific heat capacity is the quantity of heat energy required to change the temperature of unit mass (1kg) of a substance by 1k.\n(b) Diagram\n\nMethod Weigh empty calorimeter plus stirrer and record the mass M\\(_1\\), weigh empty calorimeter plus stirrer plus water and record the mass m. Weigh copper ball and record the mass Mc. Take the initial temperature of the water T\\(_1\\). Heat copper ball in water to temperature T and record T. Quickly transfer the copper ball into the calorimeter and stir. Take the final temperature of the mixture T\\(_2\\). Heat lost by copper ball = Heat gained by calorimeter; water and stirrer.\nMcCc (T - T\\(_2\\)) = (M\\(_2\\) - M\\(_1\\))C\\(_w\\)\n(T\\(_2\\) - T\\(_1\\)) + M\\(_1\\) Cal(T\\(_2\\) - T\\(_1\\))\nWere Cc, Cw and Cz = specific heat capacity of copper, water and calorimeter respectively.\nCc = \\(\\frac{ (M - M_1C + M Cal (T - T)}{Mc(T - T_2)}\\)\nPrecautions:\n(i) Calorimeter should be lagged.\n(ii) Quickly transfer the hot ball to the calorimeter.\n(iii) Shake off water from hot calorimeter metal ball before transfer.\n(iv) Mix thoroughly the mixture with stirrer.\n(c) Heat lost = 0.024 x 400 (230 - T).\nHeat gained by water = 0.054 x 4200 (T - 31)\nHeat gained by calorimeter = 0.060 x 400 (T - 31)\nHeat lost = Heat gained = 0.024 x 400 (230 - T) = 0.054 x 4200\n(T - 31) + 0.060 x 400 (T - 31) T = 38.34°C\n\n', '', 'Explanation\n(a) Specific heat capacity is the quantity of heat energy required to change the temperature of unit mass (1kg) of a substance by 1k.\n(b) Diagram\n\nMethod Weigh empty calorimeter plus stirrer and record the mass M\\(_1\\), weigh empty calorimeter plus stirrer plus water and record the mass m. Weigh copper ball and record the mass Mc. Take the initial temperature of the water T\\(_1\\). Heat copper ball in water to temperature T and record T. Quickly transfer the copper ball into the calorimeter and stir. Take the final temperature of the mixture T\\(_2\\). Heat lost by copper ball = Heat gained by calorimeter; water and stirrer.\nMcCc (T - T\\(_2\\)) = (M\\(_2\\) - M\\(_1\\))C\\(_w\\)\n(T\\(_2\\) - T\\(_1\\)) + M\\(_1\\) Cal(T\\(_2\\) - T\\(_1\\))\nWere Cc, Cw and Cz = specific heat capacity of copper, water and calorimeter respectively.\nCc = \\(\\frac{ (M - M_1C + M Cal (T - T)}{Mc(T - T_2)}\\)\nPrecautions:\n(i) Calorimeter should be lagged.\n(ii) Quickly transfer the hot ball to the calorimeter.\n(iii) Shake off water from hot calorimeter metal ball before transfer.\n(iv) Mix thoroughly the mixture with stirrer.\n(c) Heat lost = 0.024 x 400 (230 - T).\nHeat gained by water = 0.054 x 4200 (T - 31)\nHeat gained by calorimeter = 0.060 x 400 (T - 31)\nHeat lost = Heat gained = 0.024 x 400 (230 - T) = 0.054 x 4200\n(T - 31) + 0.060 x 400 (T - 31) T = 38.34°C\n\n', '(a) Specific heat capacity is the quantity of heat energy required to change the temperature of unit mass (1kg) of a substance by 1k.~(b) Diagram\n\nMethod Weigh empty calorimeter plus stirrer and record the mass M\\(_1\\), weigh empty calorimeter plus stirrer plus water and record the mass m. Weigh copper ball and record the mass Mc. Take the initial temperature of the water T\\(_1\\). Heat copper ball in water to temperature T and record T. Quickly transfer the copper ball into the calorimeter and stir. Take the final temperature of the mixture T\\(_2\\). Heat lost by copper ball = Heat gained by calorimeter; water and stirrer.\nMcCc (T - T\\(_2\\)) = (M\\(_2\\) - M\\(_1\\))C\\(_w\\)\n(T\\(_2\\) - T\\(_1\\)) + M\\(_1\\) Cal(T\\(_2\\) - T\\(_1\\))\nWere Cc, Cw and Cz = specific heat capacity of copper, water and calorimeter respectively.\nCc = \\(\\frac{ (M - M_1C + M Cal (T - T)}{Mc(T - T_2)}\\)\nPrecautions:\n(i) Calorimeter should be lagged.\n(ii) Quickly transfer the hot ball to the calorimeter.\n(iii) Shake off water from hot calorimeter metal ball before transfer.\n(iv) Mix thoroughly the mixture with stirrer.\n(c) Heat lost = 0.024 x 400 (230 - T).\nHeat gained by water = 0.054 x 4200 (T - 31)\nHeat gained by calorimeter = 0.060 x 400 (T - 31)\nHeat lost = Heat gained = 0.024 x 400 (230 - T) = 0.054 x 4200\n(T - 31) + 0.060 x 400 (T - 31) T = 38.34°C\n~(c) Heat lost = 0.024 x 400 (230 - T).\nHeat gained by water = 0.054 x 4200 (T - 31)\nHeat gained by calorimeter = 0.060 x 400 (T - 31)\nHeat lost = Heat gained = 0.024 x 400 (230 - T) = 0.054 x 4200\n(T - 31) + 0.060 x 400 (T - 31) T = 38.34°C\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(237, '2018', '\n\nQuestion\n\n(a)(i) What is an echo? (ii) State two useful applications of echoes.\n(iii) Why are the walls, floors and ceilings of a recording studio heavily padded?\n(b)(i) Explain timbre and overtones.\n(ii) What is resonance?\n(c) As a ship approaches a cliff, its siren is sounded and the echo is heard in the ship after 12 seconds. 2.1 minutes later the siren was sounded again and the echo was heard 8 seconds later. If the speed of sound in air is 340 ms\\(^{-1}\\), calculate the velocity at which the ship was approaching the cliff.\n\n', '', '(a)(i) What is an echo? (ii) State two useful applications of echoes.\n(iii) Why are the walls, floors and ceilings of a recording studio heavily padded?\n(b)(i) Explain timbre and overtones.\n(ii) What is resonance?\n(c) As a ship approaches a cliff, its siren is sounded and the echo is heard in the ship after 12 seconds. 2.1 minutes later the siren was sounded again and the echo was heard 8 seconds later. If the speed of sound in air is 340 ms\\(^{-1}\\), calculate the velocity at which the ship was approaching the cliff.', '(a)(i) What is an echo? (ii) State two useful applications of echoes.\n(iii) Why are the walls, floors and ceilings of a recording studio heavily padded?~(b)(i) Explain timbre and overtones.\n(ii) What is resonance?\n(c) As a ship approaches a cliff, its siren is sounded and the echo is heard in the ship after 12 seconds. 2.1 minutes later the siren was sounded again and the echo was heard 8 seconds later. If the speed of sound in air is 340 ms\\(^{-1}\\), calculate the velocity at which the ship was approaching the cliff.~(c) As a ship approaches a cliff, its siren is sounded and the echo is heard in the ship after 12 seconds. 2.1 minutes later the siren was sounded again and the echo was heard 8 seconds later. If the speed of sound in air is 340 ms\\(^{-1}\\), calculate the velocity at which the ship was approaching the cliff.', NULL, '\n\nExplanation\n(a)(i) An echo is a reflected sound.\n(ii) Echoes can be used to:\n- Determine the speed of sound in air.\n- Determine the depth of a sea bed.\n- Locate the positions of shoals of fish.\n- Echo sounding is used in mineral prospecting.\n(iii) Padding absorbs sound and minimizes reflections at the walls, floors or ceilings. This helps to minimize the effects of echoes.\n(b)(i) Timbre is the characteristic quality of a sound produced by a particular voice or instrument. Timbre of the note is pure and it comes out unaccompanied by undesired notes. Overtones are sound notes which accompany the fundamental notes.\n(ii) Resonance is said to occur whenever a particular body or system is set in oscillation at its own natural frequency as a result of impulses received from some other system which is vibrating with the same frequency.\n(c) Let S\\(_1\\) = distance i between the ship and the cliff in the first instance\n\\(\\frac{2S}{12}\\) = 340;S\\(_1\\) = 2040m\nIn the second instance \\(\\frac{2S_2}{8}\\) = 340\nS\\(_2\\) = 1360m\nDistance travelled by ship in 2.1 minute S\\(_1\\) - S\\(_2\\)\n= 2040 - 1360 = 680m\nShip velocity V = \\(\\frac{680 }{2.1 \\times 60}\\) = 5.4ms\\(^{-1}\\)\n\n', '', 'Explanation\n(a)(i) An echo is a reflected sound.\n(ii) Echoes can be used to:\n- Determine the speed of sound in air.\n- Determine the depth of a sea bed.\n- Locate the positions of shoals of fish.\n- Echo sounding is used in mineral prospecting.\n(iii) Padding absorbs sound and minimizes reflections at the walls, floors or ceilings. This helps to minimize the effects of echoes.\n(b)(i) Timbre is the characteristic quality of a sound produced by a particular voice or instrument. Timbre of the note is pure and it comes out unaccompanied by undesired notes. Overtones are sound notes which accompany the fundamental notes.\n(ii) Resonance is said to occur whenever a particular body or system is set in oscillation at its own natural frequency as a result of impulses received from some other system which is vibrating with the same frequency.\n(c) Let S\\(_1\\) = distance i between the ship and the cliff in the first instance\n\\(\\frac{2S}{12}\\) = 340;S\\(_1\\) = 2040m\nIn the second instance \\(\\frac{2S_2}{8}\\) = 340\nS\\(_2\\) = 1360m\nDistance travelled by ship in 2.1 minute S\\(_1\\) - S\\(_2\\)\n= 2040 - 1360 = 680m\nShip velocity V = \\(\\frac{680 }{2.1 \\times 60}\\) = 5.4ms\\(^{-1}\\)\n\n', '(a)(i) An echo is a reflected sound.\n(ii) Echoes can be used to:\n- Determine the speed of sound in air.\n- Determine the depth of a sea bed.\n- Locate the positions of shoals of fish.\n- Echo sounding is used in mineral prospecting.\n(iii) Padding absorbs sound and minimizes reflections at the walls, floors or ceilings. This helps to minimize the effects of echoes.~(b)(i) Timbre is the characteristic quality of a sound produced by a particular voice or instrument. Timbre of the note is pure and it comes out unaccompanied by undesired notes. Overtones are sound notes which accompany the fundamental notes.\n(ii) Resonance is said to occur whenever a particular body or system is set in oscillation at its own natural frequency as a result of impulses received from some other system which is vibrating with the same frequency.\n(c) Let S\\(_1\\) = distance i between the ship and the cliff in the first instance\n\\(\\frac{2S}{12}\\) = 340;S\\(_1\\) = 2040m\nIn the second instance \\(\\frac{2S_2}{8}\\) = 340\nS\\(_2\\) = 1360m\nDistance travelled by ship in 2.1 minute S\\(_1\\) - S\\(_2\\)\n= 2040 - 1360 = 680m\nShip velocity V = \\(\\frac{680 }{2.1 \\times 60}\\) = 5.4ms\\(^{-1}\\)\n~(c) Let S\\(_1\\) = distance i between the ship and the cliff in the first instance\n\\(\\frac{2S}{12}\\) = 340;S\\(_1\\) = 2040m\nIn the second instance \\(\\frac{2S_2}{8}\\) = 340\nS\\(_2\\) = 1360m\nDistance travelled by ship in 2.1 minute S\\(_1\\) - S\\(_2\\)\n= 2040 - 1360 = 680m\nShip velocity V = \\(\\frac{680 }{2.1 \\times 60}\\) = 5.4ms\\(^{-1}\\)\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(238, '2018', '\n\nQuestion\n\n(a) Explain the statement the capacitance of a capacitor is 5\\(\\mu\\)F.\n(b)(i) State the factors upon which the capacitance of a parallel plate capacitor depend.\n(ii) State how the capacitance depends on each of these factors stated in (b)(i).\n(c) A series arrangement of three capacitors of values 8uF, 12\\(\\mu\\)F, and 24\\(\\mu\\)F is connected in series with 90-V battery.\n(i) Draw an open-circuit diagram for this arrangement.\n(ii) Calculate the effective capacitance in the circuit.\n(iii) On closed circuit, calculate the charge on each capacitor when fully charged.\n(iv) Determine the p.d across the 8\\(\\mu\\)F capacitor.\n\n', '', '(a) Explain the statement the capacitance of a capacitor is 5\\(\\mu\\)F.\n(b)(i) State the factors upon which the capacitance of a parallel plate capacitor depend.\n(ii) State how the capacitance depends on each of these factors stated in (b)(i).\n(c) A series arrangement of three capacitors of values 8uF, 12\\(\\mu\\)F, and 24\\(\\mu\\)F is connected in series with 90-V battery.\n(i) Draw an open-circuit diagram for this arrangement.\n(ii) Calculate the effective capacitance in the circuit.\n(iii) On closed circuit, calculate the charge on each capacitor when fully charged.\n(iv) Determine the p.d across the 8\\(\\mu\\)F capacitor.', '(a) Explain the statement the capacitance of a capacitor is 5\\(\\mu\\)F.~(b)(i) State the factors upon which the capacitance of a parallel plate capacitor depend.\n(ii) State how the capacitance depends on each of these factors stated in (b)(i).\n(c) A series arrangement of three capacitors of values 8uF, 12\\(\\mu\\)F, and 24\\(\\mu\\)F is connected in series with 90-V battery.\n(i) Draw an open-circuit diagram for this arrangement.\n(ii) Calculate the effective capacitance in the circuit.\n(iii) On closed circuit, calculate the charge on each capacitor when fully charged.\n(iv) Determine the p.d across the 8\\(\\mu\\)F capacitor.~(c) A series arrangement of three capacitors of values 8uF, 12\\(\\mu\\)F, and 24\\(\\mu\\)F is connected in series with 90-V battery.\n(i) Draw an open-circuit diagram for this arrangement.\n(ii) Calculate the effective capacitance in the circuit.\n(iii) On closed circuit, calculate the charge on each capacitor when fully charged.\n(iv) Determine the p.d across the 8\\(\\mu\\)F capacitor.', NULL, '\n\nExplanation\n(a) The statement means that an electric charge of 5\\(\\mu\\)F is stored by the fully charged capacitor when connected to a d.c source of 1V.\n(b)(i)\n\n\n\n\n(i) Factors\n\n\n(ii) Form of dependence\n\n\n\n\n- Common area of plates\n- Distance between plates\n- Permitivity or nature of dielectric\n\n\n- Direct proportionality\n- Inverse proportionality\n- Direct proportionality\n\n\n\n\n\n(c)(i)\n\n\n(ii) \\(\\frac{1}{C} = \\frac{1}{C_1} + \\frac{1}{C_1} + \\frac{1}{C_3}\\)\n= \\(\\frac{1}{8} + \\frac{1}{12} + \\frac{1}{20} = \\frac{3 + 2 + 1}{24}\\)\nC = \\(\\frac{24}{6} = 4 \\mu F\\)\n(iii) Q = CV = 4 x 10\\(^{-6}\\) x 90 = 360\\(\\mu\\)C\ncharge on 8\\(\\mu\\) FV = \\(\\frac{Q}{C} = \\frac{360}{8}\\) = 45V\n\n', '', 'Explanation\n(a) The statement means that an electric charge of 5\\(\\mu\\)F is stored by the fully charged capacitor when connected to a d.c source of 1V.\n(b)(i)\n\n\n\n\n(i) Factors\n\n\n(ii) Form of dependence\n\n\n\n\n- Common area of plates\n- Distance between plates\n- Permitivity or nature of dielectric\n\n\n- Direct proportionality\n- Inverse proportionality\n- Direct proportionality\n\n\n\n\n\n(c)(i)\n\n\n(ii) \\(\\frac{1}{C} = \\frac{1}{C_1} + \\frac{1}{C_1} + \\frac{1}{C_3}\\)\n= \\(\\frac{1}{8} + \\frac{1}{12} + \\frac{1}{20} = \\frac{3 + 2 + 1}{24}\\)\nC = \\(\\frac{24}{6} = 4 \\mu F\\)\n(iii) Q = CV = 4 x 10\\(^{-6}\\) x 90 = 360\\(\\mu\\)C\ncharge on 8\\(\\mu\\) FV = \\(\\frac{Q}{C} = \\frac{360}{8}\\) = 45V\n\n', '(a) The statement means that an electric charge of 5\\(\\mu\\)F is stored by the fully charged capacitor when connected to a d.c source of 1V.~(b)(i)\n\n\n\n\n(i) Factors\n\n\n(ii) Form of dependence\n\n\n\n\n- Common area of plates\n- Distance between plates\n- Permitivity or nature of dielectric\n\n\n- Direct proportionality\n- Inverse proportionality\n- Direct proportionality\n\n\n\n\n\n(c)(i)\n\n\n(ii) \\(\\frac{1}{C} = \\frac{1}{C_1} + \\frac{1}{C_1} + \\frac{1}{C_3}\\)\n= \\(\\frac{1}{8} + \\frac{1}{12} + \\frac{1}{20} = \\frac{3 + 2 + 1}{24}\\)\nC = \\(\\frac{24}{6} = 4 \\mu F\\)\n(iii) Q = CV = 4 x 10\\(^{-6}\\) x 90 = 360\\(\\mu\\)C\ncharge on 8\\(\\mu\\) FV = \\(\\frac{Q}{C} = \\frac{360}{8}\\) = 45V\n~(c)(i)\n\n\n(ii) \\(\\frac{1}{C} = \\frac{1}{C_1} + \\frac{1}{C_1} + \\frac{1}{C_3}\\)\n= \\(\\frac{1}{8} + \\frac{1}{12} + \\frac{1}{20} = \\frac{3 + 2 + 1}{24}\\)\nC = \\(\\frac{24}{6} = 4 \\mu F\\)\n(iii) Q = CV = 4 x 10\\(^{-6}\\) x 90 = 360\\(\\mu\\)C\ncharge on 8\\(\\mu\\) FV = \\(\\frac{Q}{C} = \\frac{360}{8}\\) = 45V\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(239, '2018', '\n\nQuestion\n\n(a) When nitrogen (atomic mass = 14, atomic number = 7) is bombarded with neutrons, the collisions result in disintegrations in which alpha particles are produced. Represent this transmutation in a symbolic equation.\n(b)(i) How does a radioactive atom differ from a stable one?\n(ii) Explain \'half life\'.\n(iii) A sample of radioactive material has a haft life of 35 days. Calculate the fraction of the original quantity that will remain after 105 days.\n(c) Light of wavelength 5.00 x 10\\(^{-7}\\)m is incident on a material of work function 1.90 eV. Calculate\n(i) photon energy.\n(ii) kinetic energy of the most energetic photo electron.\n(iii) stopping potential [Plancks constant h =6.6 x 10\\(^{-34}\\)Js] [c= 3.0 x10\\(^8\\)ms\\(^{-2}\\), leV= 1.6 x 10\\(^{19}\\)J]\n\n', '', '(a) When nitrogen (atomic mass = 14, atomic number = 7) is bombarded with neutrons, the collisions result in disintegrations in which alpha particles are produced. Represent this transmutation in a symbolic equation.\n(b)(i) How does a radioactive atom differ from a stable one?\n(ii) Explain \'half life\'.\n(iii) A sample of radioactive material has a haft life of 35 days. Calculate the fraction of the original quantity that will remain after 105 days.\n(c) Light of wavelength 5.00 x 10\\(^{-7}\\)m is incident on a material of work function 1.90 eV. Calculate\n(i) photon energy.\n(ii) kinetic energy of the most energetic photo electron.\n(iii) stopping potential [Plancks constant h =6.6 x 10\\(^{-34}\\)Js] [c= 3.0 x10\\(^8\\)ms\\(^{-2}\\), leV= 1.6 x 10\\(^{19}\\)J]', '(a) When nitrogen (atomic mass = 14, atomic number = 7) is bombarded with neutrons, the collisions result in disintegrations in which alpha particles are produced. Represent this transmutation in a symbolic equation.~(b)(i) How does a radioactive atom differ from a stable one?\n(ii) Explain \'half life\'.\n(iii) A sample of radioactive material has a haft life of 35 days. Calculate the fraction of the original quantity that will remain after 105 days.\n(c) Light of wavelength 5.00 x 10\\(^{-7}\\)m is incident on a material of work function 1.90 eV. Calculate\n(i) photon energy.\n(ii) kinetic energy of the most energetic photo electron.\n(iii) stopping potential [Plancks constant h =6.6 x 10\\(^{-34}\\)Js] [c= 3.0 x10\\(^8\\)ms\\(^{-2}\\), leV= 1.6 x 10\\(^{19}\\)J]~(c) Light of wavelength 5.00 x 10\\(^{-7}\\)m is incident on a material of work function 1.90 eV. Calculate\n(i) photon energy.\n(ii) kinetic energy of the most energetic photo electron.\n(iii) stopping potential [Plancks constant h =6.6 x 10\\(^{-34}\\)Js] [c= 3.0 x10\\(^8\\)ms\\(^{-2}\\), leV= 1.6 x 10\\(^{19}\\)J]', NULL, '\n\nExplanation\n(a) \\(^{14}_7 + ^1_0n\\) -> \\(^4_2He + ^{11}_5X\\)\n\n(b)(i) The radioactive atom disintegrates spontaneously while a stable atom does not disintegrate on its own.\n(ii) The time at which the number of radioactive atoms remaining is just one-half of the original number is called half-life of the element. The half-life T\\(_{\\frac{1}{2}}\\) is related to the decay constant (\\(\\lambda\\)) of the isotope by the equation.\n\\(T \\frac{1}{2} = \\frac{0.693}{ \\lambda}\\)\n(iii) After 35days, 1/2 of it remains.\nAfter 70 days, 1/2 of 1/2 i.e. 1/4 remains. After 105 days, % of 1/4 i.e. 1/8 remains.\n\n(c)(i) The photon energy Ep = hf = \\(\\frac{hc}{f}\\)\n= \\(\\frac{6.6 \\times 10^{-34} x 3.0 \\times 10^8}{ 5.0 x 10^{-7}}\\)\n= 3.96x 10-19J\n(ii) The kinetic energy E\\(_k\\) = (Ep - W)\nE\\(_k\\) = hf - W = \\(\\frac{hc - W}{\\lambda}\\)\n W = 1.9eV = 1.6 x 10\\(^{19}\\) x 1.9\n= 3.044 x 10\\(^{-19}\\)J.\nE\\(_k\\) = 3.96 x 10\\(^{-19}\\) - 3.044 x 10\\(^{19}\\) = 9.2 x 10\\(^{-20}\\)J.\n(iii) Stopping potential Vs = energy needed to stop most energetic electron.\neVs =E\\(_{kmax}\\)\nVs = \\(\\frac{E_{max}}{e}\\)\n= \\(\\frac{9.20 \\times 10^{-20}}{1.6 \\times 10^{-19}}\\)\n= 0.575 V\n\n', '', 'Explanation\n(a) \\(^{14}_7 + ^1_0n\\) -> \\(^4_2He + ^{11}_5X\\)\n\n(b)(i) The radioactive atom disintegrates spontaneously while a stable atom does not disintegrate on its own.\n(ii) The time at which the number of radioactive atoms remaining is just one-half of the original number is called half-life of the element. The half-life T\\(_{\\frac{1}{2}}\\) is related to the decay constant (\\(\\lambda\\)) of the isotope by the equation.\n\\(T \\frac{1}{2} = \\frac{0.693}{ \\lambda}\\)\n(iii) After 35days, 1/2 of it remains.\nAfter 70 days, 1/2 of 1/2 i.e. 1/4 remains. After 105 days, % of 1/4 i.e. 1/8 remains.\n\n(c)(i) The photon energy Ep = hf = \\(\\frac{hc}{f}\\)\n= \\(\\frac{6.6 \\times 10^{-34} x 3.0 \\times 10^8}{ 5.0 x 10^{-7}}\\)\n= 3.96x 10-19J\n(ii) The kinetic energy E\\(_k\\) = (Ep - W)\nE\\(_k\\) = hf - W = \\(\\frac{hc - W}{\\lambda}\\)\n W = 1.9eV = 1.6 x 10\\(^{19}\\) x 1.9\n= 3.044 x 10\\(^{-19}\\)J.\nE\\(_k\\) = 3.96 x 10\\(^{-19}\\) - 3.044 x 10\\(^{19}\\) = 9.2 x 10\\(^{-20}\\)J.\n(iii) Stopping potential Vs = energy needed to stop most energetic electron.\neVs =E\\(_{kmax}\\)\nVs = \\(\\frac{E_{max}}{e}\\)\n= \\(\\frac{9.20 \\times 10^{-20}}{1.6 \\times 10^{-19}}\\)\n= 0.575 V\n\n', '(a) \\(^{14}_7 + ^1_0n\\) -> \\(^4_2He + ^{11}_5X\\)\n~(b)(i) The radioactive atom disintegrates spontaneously while a stable atom does not disintegrate on its own.\n(ii) The time at which the number of radioactive atoms remaining is just one-half of the original number is called half-life of the element. The half-life T\\(_{\\frac{1}{2}}\\) is related to the decay constant (\\(\\lambda\\)) of the isotope by the equation.\n\\(T \\frac{1}{2} = \\frac{0.693}{ \\lambda}\\)\n(iii) After 35days, 1/2 of it remains.\nAfter 70 days, 1/2 of 1/2 i.e. 1/4 remains. After 105 days, % of 1/4 i.e. 1/8 remains.\n\n(c)(i) The photon energy Ep = hf = \\(\\frac{hc}{f}\\)\n= \\(\\frac{6.6 \\times 10^{-34} x 3.0 \\times 10^8}{ 5.0 x 10^{-7}}\\)\n= 3.96x 10-19J\n(ii) The kinetic energy E\\(_k\\) = (Ep - W)\nE\\(_k\\) = hf - W = \\(\\frac{hc - W}{\\lambda}\\)\n W = 1.9eV = 1.6 x 10\\(^{19}\\) x 1.9\n= 3.044 x 10\\(^{-19}\\)J.\nE\\(_k\\) = 3.96 x 10\\(^{-19}\\) - 3.044 x 10\\(^{19}\\) = 9.2 x 10\\(^{-20}\\)J.\n(iii) Stopping potential Vs = energy needed to stop most energetic electron.\neVs =E\\(_{kmax}\\)\nVs = \\(\\frac{E_{max}}{e}\\)\n= \\(\\frac{9.20 \\times 10^{-20}}{1.6 \\times 10^{-19}}\\)\n= 0.575 V\n~(c)(i) The photon energy Ep = hf = \\(\\frac{hc}{f}\\)\n= \\(\\frac{6.6 \\times 10^{-34} x 3.0 \\times 10^8}{ 5.0 x 10^{-7}}\\)\n= 3.96x 10-19J\n(ii) The kinetic energy E\\(_k\\) = (Ep - W)\nE\\(_k\\) = hf - W = \\(\\frac{hc - W}{\\lambda}\\)\n W = 1.9eV = 1.6 x 10\\(^{19}\\) x 1.9\n= 3.044 x 10\\(^{-19}\\)J.\nE\\(_k\\) = 3.96 x 10\\(^{-19}\\) - 3.044 x 10\\(^{19}\\) = 9.2 x 10\\(^{-20}\\)J.\n(iii) Stopping potential Vs = energy needed to stop most energetic electron.\neVs =E\\(_{kmax}\\)\nVs = \\(\\frac{E_{max}}{e}\\)\n= \\(\\frac{9.20 \\times 10^{-20}}{1.6 \\times 10^{-19}}\\)\n= 0.575 V\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(240, '2018', '\n\nQuestion\n\nA body of mass 0.6kg is thrown vertically upward from the ground with a speed of 20ms\\(^{-2}\\). Calculate its;\n(i) potential energy at the maximum height reached.\n(ii) kinetic energy just before it hits the ground.\n\n', '', 'A body of mass 0.6kg is thrown vertically upward from the ground with a speed of 20ms\\(^{-2}\\). Calculate its;\n(i) potential energy at the maximum height reached.\n(ii) kinetic energy just before it hits the ground.', '', NULL, '\n\nExplanation\nKinetic energy at point of projection = \\(\\frac{1}{2} mv^2 \\times 0.6 \\times 20^2\\) = 120J\nor\nV\\(^2\\) = U\\(^2\\) - 2gs\nS = \\(\\frac{v^2 - U^2}{-2g}\\) = \\(\\frac{0^2 - 2062}{-2 \\times 10}\\)\n= 20cm\nP.E. = mgs or mgh 0.6 x 10 x 20 = 120J\nii) Kinetic energy in reaching ground = Potential energy at highest point = 120J.\n\n', 'Kinetic energy at point of projection = \\(\\frac{1}{2} mv^2 \\times 0.6 \\times 20^2\\) = 120J\nor\nV\\(^2\\) = U\\(^2\\) - 2gs\nS = \\(\\frac{v^2 - U^2}{-2g}\\) = \\(\\frac{0^2 - 2062}{-2 \\times 10}\\)\n= 20cm\nP.E. = mgs or mgh 0.6 x 10 x 20 = 120J\nii) Kinetic energy in reaching ground = Potential energy at highest point = 120J.\n\n', '', '', 'WAEC-PK-1', 'WAEC-PHY', 1),
(241, '2019', '\n\nQuestion\n\nWhat is a projectile? Give four examples of projectiles in everyday life\n\n', 'What is a projectile? Give four examples of projectiles in everyday life', '', '', NULL, '\n\nExplanation\nA projectile is an object or body launched into the air and allowed to move on its own or moving freely under gravity. Examples are - a shot arrow, a thrown javeline, a fired bullet, a kicked football etc.\n\n', 'A projectile is an object or body launched into the air and allowed to move on its own or moving freely under gravity. Examples are - a shot arrow, a thrown javeline, a fired bullet, a kicked football etc.\n\n', '', '', 'WAEC-PK-1', 'WAEC-PHY', 1),
(242, '2019', '\n\nQuestion\n\na) Define the term surface tension\n(b) Calculate the force required to lift a needle 4cm long off the surface of water if the surface tension of water is 7.3 x 10\\(^{-4}\\) NM\\(^{-1}\\)\n\n\n', '', 'a) Define the term surface tension\n(b) Calculate the force required to lift a needle 4cm long off the surface of water if the surface tension of water is 7.3 x 10\\(^{-4}\\) NM\\(^{-1}\\)\n', '~(b) Calculate the force required to lift a needle 4cm long off the surface of water if the surface tension of water is 7.3 x 10\\(^{-4}\\) NM\\(^{-1}\\)\n', NULL, '\n\nExplanation\na) Surface tension is defined as a force per unit length perpendicular to any line on the surface of the liquid OR Property of a liquid which makes the surface of liquid behave like an elastic skin.\n(b) Surface tension = F/21\n=7.3 x 1\\(^{-2}\\) x 2 x 0.04 = 5.8 x 10\\(^{-3}\\)N\n\n', '', 'Explanation\na) Surface tension is defined as a force per unit length perpendicular to any line on the surface of the liquid OR Property of a liquid which makes the surface of liquid behave like an elastic skin.\n(b) Surface tension = F/21\n=7.3 x 1\\(^{-2}\\) x 2 x 0.04 = 5.8 x 10\\(^{-3}\\)N\n\n', 'a) Surface tension is defined as a force per unit length perpendicular to any line on the surface of the liquid OR Property of a liquid which makes the surface of liquid behave like an elastic skin.~(b) Surface tension = F/21\n=7.3 x 1\\(^{-2}\\) x 2 x 0.04 = 5.8 x 10\\(^{-3}\\)N\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(243, '2019', '\n\nQuestion\n\nState any three properties of matter which are common to all substances.\n\n', 'State any three properties of matter which are common to all substances.', '', '', NULL, '\n\nExplanation\n(i) Matter is made up of tiny particles, molecules or atoms\n(ii) Matter occupies space\n(iii) Matter possesses inertia, mass or weight\n(iv) Matter exerts pressure.\n\n', '', 'Explanation\n(i) Matter is made up of tiny particles, molecules or atoms\n(ii) Matter occupies space\n(iii) Matter possesses inertia, mass or weight\n(iv) Matter exerts pressure.\n\n', '', 'WAEC-PK-1', 'WAEC-PHY', 1),
(244, '2019', '\n\nQuestion\n\nDefine (i) Elasticity; (ii) Young\'s modulus; (iii) Force constant.\n\n', 'Define (i) Elasticity; (ii) Young\'s modulus; (iii) Force constant.', '', '', NULL, '\n\nExplanation\n(i) Elasticity is the ability of a material to regain its original shape or size after deformations. OR after removal of stress force, OR after it has been 1 compressed.\n(ii) Young\'s Modulus is the ratio of tensile I or compressive stress to tensile or compressive strain\n(iii) Force constant = the amount of force that causes a unit extension of an elastic material. OR\n\\(\\frac{Force (F)}{ Extension(e)}\\) of an elastic material.\n\n', '', 'Explanation\n(i) Elasticity is the ability of a material to regain its original shape or size after deformations. OR after removal of stress force, OR after it has been 1 compressed.\n(ii) Young\'s Modulus is the ratio of tensile I or compressive stress to tensile or compressive strain\n(iii) Force constant = the amount of force that causes a unit extension of an elastic material. OR\n\\(\\frac{Force (F)}{ Extension(e)}\\) of an elastic material.\n\n', '~(e)}\\) of an elastic material.\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(245, '2019', '\n\nQuestion\n\nA force of 40 N is applied at the end of a wire 4m long and produces an extension of 0.24mm. If the diameter of the wire is 2.00mm, calculate the;\n(i) stress on the wire; (ii) strain in the wire.\n\n', '', 'A force of 40 N is applied at the end of a wire 4m long and produces an extension of 0.24mm. If the diameter of the wire is 2.00mm, calculate the;\n(i) stress on the wire; (ii) strain in the wire.', '', NULL, '\n\nExplanation\n(i) Stress = \\(\\frac{force}{\\text{cross sectiona area}}\\)\n= \\(\\frac{40}{3.142 (1 \\times 10^{-3})^2}\\) = 12.74 x 10\\(^{6}\\)Nm\\(^{-2}\\)\n(ii) Strain = \\(\\frac{e}{I}\\) = \\(\\frac{24 \\times 10^{-5}}{4}\\)\n= 6.0 x 10\\(^{-5}\\)\n\n', '', 'Explanation\n(i) Stress = \\(\\frac{force}{\\text{cross sectiona area}}\\)\n= \\(\\frac{40}{3.142 (1 \\times 10^{-3})^2}\\) = 12.74 x 10\\(^{6}\\)Nm\\(^{-2}\\)\n(ii) Strain = \\(\\frac{e}{I}\\) = \\(\\frac{24 \\times 10^{-5}}{4}\\)\n= 6.0 x 10\\(^{-5}\\)\n\n', '', 'WAEC-PK-1', 'WAEC-PHY', 1),
(246, '2019', '\n\nQuestion\n\n(a) Differentiate between interference and polarisation as applied to waves.\n(b) Mention two uses of polaroids\n\n', '', '(a) Differentiate between interference and polarisation as applied to waves.\n(b) Mention two uses of polaroids', '(a) Differentiate between interference and polarisation as applied to waves.~(b) Mention two uses of polaroids', NULL, '\n\nExplanation\na)(i) Interference occurs in all kinds of waves i.e. longitudinal or transverse, while polarisation occurs only in transverse waves.\n(ii) - In sun glasses to reduce the intensity of light.\n- To produce polarised light.\n- To produce three dimensional pictures.\n\n\n\n', '', 'Explanation\na)(i) Interference occurs in all kinds of waves i.e. longitudinal or transverse, while polarisation occurs only in transverse waves.\n(ii) - In sun glasses to reduce the intensity of light.\n- To produce polarised light.\n- To produce three dimensional pictures.\n\n\n\n', '', 'WAEC-PK-1', 'WAEC-PHY', 1),
(247, '2019', '\n\nQuestion\n\n(a) Define angle of contact\n(b) Draw sketches to show angles of contact for a capillary tube dipped vertically in (i) water; (ii) mercury.\n\n', '', '(a) Define angle of contact\n(b) Draw sketches to show angles of contact for a capillary tube dipped vertically in (i) water; (ii) mercury.', '(a) Define angle of contact~(b) Draw sketches to show angles of contact for a capillary tube dipped vertically in (i) water; (ii) mercury.', NULL, '\n\nExplanation\n(a) Angle of contact - the angle, measured through the liquid between the wall of the container and the tangent of the meniscus.\n(b)\n\n\n\n\n', '', 'Explanation\n(a) Angle of contact - the angle, measured through the liquid between the wall of the container and the tangent of the meniscus.\n(b)\n\n\n\n\n', '(a) Angle of contact - the angle, measured through the liquid between the wall of the container and the tangent of the meniscus.~(b)\n\n\n\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(248, '2019', '\n\nQuestion\n\nDuring the electrolysis of copper (II) tetraoxosulphate (VI) solution, a steady current of 4.0 x 10\\(^2\\) A flowing for one hour liberated 0.48 g of copper. Calculate the mass of copper liberated by one coulomb of charge.\n\n', 'During the electrolysis of copper (II) tetraoxosulphate (VI) solution, a steady current of 4.0 x 10\\(^2\\) A flowing for one hour liberated 0.48 g of copper. Calculate the mass of copper liberated by one coulomb of charge.', '', '', NULL, '\n\nExplanation\nM = Zit\nZ = \\(\\frac{M}{It}\\)\n= \\(\\frac{0.48 \\times 10^{-3}}{4 \\times 10^2 \\times 3600}\\) = 3.3 x 10\\(^{-10}\\)kgc\\(^{-1}\\)\n\n', 'M = Zit\nZ = \\(\\frac{M}{It}\\)\n= \\(\\frac{0.48 \\times 10^{-3}}{4 \\times 10^2 \\times 3600}\\) = 3.3 x 10\\(^{-10}\\)kgc\\(^{-1}\\)\n\n', '', '', 'WAEC-PK-1', 'WAEC-PHY', 1),
(249, '2019', '\n\nQuestion\n\nAn electron of mass 9. 1 x 10\\(^{-31}\\) kg moves with a velocity of 4.2 x 10 ms\\(^{-1}\\) between the cathode and anode of an x-ray tube. Calculate the wavelength. (Take Planck\'s constant h = 6.6 x 10\\(^{-34}\\) J S)\n\n', 'An electron of mass 9. 1 x 10\\(^{-31}\\) kg moves with a velocity of 4.2 x 10 ms\\(^{-1}\\) between the cathode and anode of an x-ray tube. Calculate the wavelength. (Take Planck\'s constant h = 6.6 x 10\\(^{-34}\\) J S)', '', '', NULL, '\n\nExplanation\nP = mv = 9.1 x 10\\(^{31}\\) x 4.2 x 10\\(^{-7}\\)\n= 9.1 x 4.2 x 10\\(^{-24}\\) Kgms\\(^{-1}\\)\n\\(\\lambda = \\frac{h}{p} = \\frac{6.6 \\times 10^{-34}}{9.1 \\times 4.2 \\times 10^{-24}}\\)\n= 1.73 x 10\\(^{-11}\\)m\nor\n\\(\\lambda = \\frac{h}{mv} = \\frac{6.6 \\times 10^{-34}}{9.1 \\times 10^{-31} \\times 4.2 \\times 10^7}\\) = 1.73 x 10\\(^{-11}\\)\n\n', 'P = mv = 9.1 x 10\\(^{31}\\) x 4.2 x 10\\(^{-7}\\)\n= 9.1 x 4.2 x 10\\(^{-24}\\) Kgms\\(^{-1}\\)\n\\(\\lambda = \\frac{h}{p} = \\frac{6.6 \\times 10^{-34}}{9.1 \\times 4.2 \\times 10^{-24}}\\)\n= 1.73 x 10\\(^{-11}\\)m\nor\n\\(\\lambda = \\frac{h}{mv} = \\frac{6.6 \\times 10^{-34}}{9.1 \\times 10^{-31} \\times 4.2 \\times 10^7}\\) = 1.73 x 10\\(^{-11}\\)\n\n', '', '', 'WAEC-PK-1', 'WAEC-PHY', 1),
(250, '2019', '\n\nQuestion\n\n(a)(i) Define the term linear momentum.\n(ii) State the law of conservation of linear momentum.\n(b) A ball P of mass 0.25 kg, loses one-third of its velocity when it makes a head on collision with an identical ball Q at rest. After the collision, Q moves off with a speed of 2ms\\(^{-1}\\) in the original direction of P. Calculate the initial velocity of R\n(c)(i) State Newton\'s second law of motion.\n(ii) Show that F = ma where F is the magnitude of the force acting on a body of mass m to give it an acceleration of magnitude a.\n(iii) The engine of a vehicle moves it forward with a force of 9600 N against a resistive force of 2200 N. If the mass of the vehicle is 3400 kg, calculate the acceleration produced.\n\n', '', '(a)(i) Define the term linear momentum.\n(ii) State the law of conservation of linear momentum.\n(b) A ball P of mass 0.25 kg, loses one-third of its velocity when it makes a head on collision with an identical ball Q at rest. After the collision, Q moves off with a speed of 2ms\\(^{-1}\\) in the original direction of P. Calculate the initial velocity of R\n(c)(i) State Newton\'s second law of motion.\n(ii) Show that F = ma where F is the magnitude of the force acting on a body of mass m to give it an acceleration of magnitude a.\n(iii) The engine of a vehicle moves it forward with a force of 9600 N against a resistive force of 2200 N. If the mass of the vehicle is 3400 kg, calculate the acceleration produced.', '(a)(i) Define the term linear momentum.\n(ii) State the law of conservation of linear momentum.~(b) A ball P of mass 0.25 kg, loses one-third of its velocity when it makes a head on collision with an identical ball Q at rest. After the collision, Q moves off with a speed of 2ms\\(^{-1}\\) in the original direction of P. Calculate the initial velocity of R\n(c)(i) State Newton\'s second law of motion.\n(ii) Show that F = ma where F is the magnitude of the force acting on a body of mass m to give it an acceleration of magnitude a.\n(iii) The engine of a vehicle moves it forward with a force of 9600 N against a resistive force of 2200 N. If the mass of the vehicle is 3400 kg, calculate the acceleration produced.~(c)(i) State Newton\'s second law of motion.\n(ii) Show that F = ma where F is the magnitude of the force acting on a body of mass m to give it an acceleration of magnitude a.\n(iii) The engine of a vehicle moves it forward with a force of 9600 N against a resistive force of 2200 N. If the mass of the vehicle is 3400 kg, calculate the acceleration produced.', NULL, '\n\nExplanation\n(a)(i) The momentum of a body is the product of its mass and its velocity.\n(ii) The total momentum of a system before collision equals to the total momentum after collision provided no external force acts. OR In a system of colliding bodies the total momentum is constant provided no external force acts.\n(b) Mp Up + Mq Uq = Mp Vp + Mq Vq\n0.25 Up + 0 = 0.25 x \\(\\frac{2}{3}\\) Up + 0.25 x 2; 0.25\n(Up - 2/3 Up) = 0.50; 1/3\nUp = \\(\\frac{0.50}{0.25}\\) = 2m/s\nUp = 2 x 3 = 6m/s\n(c)(i) Newton\'s second law of motion states that the time rate of change of linear momentum is directly proportional to the force causing the change and the change takes place in the direction of the force\n(ii) Let the momentum before collision = mu and momentum after collision = my, change in momentum = my - mu, Change in momentum in unit time\n= \\(\\frac{mv - mu}{t}\\)\nFrom second law: \\(\\frac{mv - mu}{t}\\) = KF\n\\(\\frac{m(v - u)}{t}\\) = KF\nIf m = 1kg, a = 1ms\\(^{-2}\\) and F = 1N, K = 1\nma = F\n(iii) Resultant force on the vehicle = 9600 - 200 = 7400N.\nBut F = ma\na = \\(\\frac{F}{m} = \\frac{7400}{3400}\\) = 2.18ms\\(^{-2}\\) or 2.2ms\\(^{-2}\\)\n\n', '', 'Explanation\n(a)(i) The momentum of a body is the product of its mass and its velocity.\n(ii) The total momentum of a system before collision equals to the total momentum after collision provided no external force acts. OR In a system of colliding bodies the total momentum is constant provided no external force acts.\n(b) Mp Up + Mq Uq = Mp Vp + Mq Vq\n0.25 Up + 0 = 0.25 x \\(\\frac{2}{3}\\) Up + 0.25 x 2; 0.25\n(Up - 2/3 Up) = 0.50; 1/3\nUp = \\(\\frac{0.50}{0.25}\\) = 2m/s\nUp = 2 x 3 = 6m/s\n(c)(i) Newton\'s second law of motion states that the time rate of change of linear momentum is directly proportional to the force causing the change and the change takes place in the direction of the force\n(ii) Let the momentum before collision = mu and momentum after collision = my, change in momentum = my - mu, Change in momentum in unit time\n= \\(\\frac{mv - mu}{t}\\)\nFrom second law: \\(\\frac{mv - mu}{t}\\) = KF\n\\(\\frac{m(v - u)}{t}\\) = KF\nIf m = 1kg, a = 1ms\\(^{-2}\\) and F = 1N, K = 1\nma = F\n(iii) Resultant force on the vehicle = 9600 - 200 = 7400N.\nBut F = ma\na = \\(\\frac{F}{m} = \\frac{7400}{3400}\\) = 2.18ms\\(^{-2}\\) or 2.2ms\\(^{-2}\\)\n\n', '(a)(i) The momentum of a body is the product of its mass and its velocity.\n(ii) The total momentum of a system before collision equals to the total momentum after collision provided no external force acts. OR In a system of colliding bodies the total momentum is constant provided no external force acts.~(b) Mp Up + Mq Uq = Mp Vp + Mq Vq\n0.25 Up + 0 = 0.25 x \\(\\frac{2}{3}\\) Up + 0.25 x 2; 0.25\n(Up - 2/3 Up) = 0.50; 1/3\nUp = \\(\\frac{0.50}{0.25}\\) = 2m/s\nUp = 2 x 3 = 6m/s\n(c)(i) Newton\'s second law of motion states that the time rate of change of linear momentum is directly proportional to the force causing the change and the change takes place in the direction of the force\n(ii) Let the momentum before collision = mu and momentum after collision = my, change in momentum = my - mu, Change in momentum in unit time\n= \\(\\frac{mv - mu}{t}\\)\nFrom second law: \\(\\frac{mv - mu}{t}\\) = KF\n\\(\\frac{m(v - u)}{t}\\) = KF\nIf m = 1kg, a = 1ms\\(^{-2}\\) and F = 1N, K = 1\nma = F\n(iii) Resultant force on the vehicle = 9600 - 200 = 7400N.\nBut F = ma\na = \\(\\frac{F}{m} = \\frac{7400}{3400}\\) = 2.18ms\\(^{-2}\\) or 2.2ms\\(^{-2}\\)\n~(c)(i) Newton\'s second law of motion states that the time rate of change of linear momentum is directly proportional to the force causing the change and the change takes place in the direction of the force\n(ii) Let the momentum before collision = mu and momentum after collision = my, change in momentum = my - mu, Change in momentum in unit time\n= \\(\\frac{mv - mu}{t}\\)\nFrom second law: \\(\\frac{mv - mu}{t}\\) = KF\n\\(\\frac{m(v - u)}{t}\\) = KF\nIf m = 1kg, a = 1ms\\(^{-2}\\) and F = 1N, K = 1\nma = F\n(iii) Resultant force on the vehicle = 9600 - 200 = 7400N.\nBut F = ma\na = \\(\\frac{F}{m} = \\frac{7400}{3400}\\) = 2.18ms\\(^{-2}\\) or 2.2ms\\(^{-2}\\)\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(251, '2019', '\n\nQuestion\n\n(a)(i) Mention two modes of heat transfer other than convection.\n(ii) Explain land and sea breezes.\n(b) An iron rod of length 30cm is heated through 50 kelvin. Calculate its increase in length. (linear expansivity of iron = 1.2 x 10\\(^{-5}\\)K\\(^{-1}\\)\n(c) An electric heater immersed in some water raises the temperature of the water from 40°C to 100°C in 6 minutes. After another 25 minutes, it is noticed that half the water has boiled away. Neglecting heat losses to the surrounding, calculate the specific latent heat of vaporisation of water.\n\n', '', '(a)(i) Mention two modes of heat transfer other than convection.\n(ii) Explain land and sea breezes.\n(b) An iron rod of length 30cm is heated through 50 kelvin. Calculate its increase in length. (linear expansivity of iron = 1.2 x 10\\(^{-5}\\)K\\(^{-1}\\)\n(c) An electric heater immersed in some water raises the temperature of the water from 40°C to 100°C in 6 minutes. After another 25 minutes, it is noticed that half the water has boiled away. Neglecting heat losses to the surrounding, calculate the specific latent heat of vaporisation of water.', '(a)(i) Mention two modes of heat transfer other than convection.\n(ii) Explain land and sea breezes.~(b) An iron rod of length 30cm is heated through 50 kelvin. Calculate its increase in length. (linear expansivity of iron = 1.2 x 10\\(^{-5}\\)K\\(^{-1}\\)\n(c) An electric heater immersed in some water raises the temperature of the water from 40°C to 100°C in 6 minutes. After another 25 minutes, it is noticed that half the water has boiled away. Neglecting heat losses to the surrounding, calculate the specific latent heat of vaporisation of water.~(c) An electric heater immersed in some water raises the temperature of the water from 40°C to 100°C in 6 minutes. After another 25 minutes, it is noticed that half the water has boiled away. Neglecting heat losses to the surrounding, calculate the specific latent heat of vaporisation of water.', NULL, '\n\nExplanation\n(a)(i) Conduction and Radiation.\n(ii)\n\n\nThe specific heat capacity of land is less than that of sea. Land Breeze: At night land cools faster than sea, warm air from above the sea rises upward. It\'s place is taken by cooler air from above the land. Sea Breeze: During the day the land is hotter than the sea. Warm air above land rises, and its place is taken by cooler air from the sea.\n(b) Increase in length (L\\(_2\\) - L\\(_1\\) = L\\(_1 \\alpha \\Delta \\theta = 30 \\times 1.2 \\times 10^{-5} \\times 50 = 1.8 \\times 10^{-2}\\) cm\n\n(c) Ivt = mc\\(\\Delta \\theta\\)\nPower x time = mc\\(\\Delta \\theta\\)\n(i) Iv x 360 = m x c 60 ----(i) Ivt = \\(\\frac{1}{2}\\)mL\n(ii) Ivx 25 x 60 = \\(\\frac{1}{2}\\)mL ------(ii)\n\\(\\frac{Iv \\times 25 \\times 60}{Iv \\times 360} = \\frac{\\frac{1}{2}mL}{mc \\times 60}\\)\nL = \\(\\frac{25 \\times 60 \\times C \\times 60 \\times 2}{360}\\)\n= \\(\\frac{25 \\times 60 \\times 4200 \\times 60 \\times 2}{360}\\)\n= 2.1 x 10\\(^5\\) Jkg\\(^{-1}\\)\n\n', '', 'Explanation\n(a)(i) Conduction and Radiation.\n(ii)\n\n\nThe specific heat capacity of land is less than that of sea. Land Breeze: At night land cools faster than sea, warm air from above the sea rises upward. It\'s place is taken by cooler air from above the land. Sea Breeze: During the day the land is hotter than the sea. Warm air above land rises, and its place is taken by cooler air from the sea.\n(b) Increase in length (L\\(_2\\) - L\\(_1\\) = L\\(_1 \\alpha \\Delta \\theta = 30 \\times 1.2 \\times 10^{-5} \\times 50 = 1.8 \\times 10^{-2}\\) cm\n\n(c) Ivt = mc\\(\\Delta \\theta\\)\nPower x time = mc\\(\\Delta \\theta\\)\n(i) Iv x 360 = m x c 60 ----(i) Ivt = \\(\\frac{1}{2}\\)mL\n(ii) Ivx 25 x 60 = \\(\\frac{1}{2}\\)mL ------(ii)\n\\(\\frac{Iv \\times 25 \\times 60}{Iv \\times 360} = \\frac{\\frac{1}{2}mL}{mc \\times 60}\\)\nL = \\(\\frac{25 \\times 60 \\times C \\times 60 \\times 2}{360}\\)\n= \\(\\frac{25 \\times 60 \\times 4200 \\times 60 \\times 2}{360}\\)\n= 2.1 x 10\\(^5\\) Jkg\\(^{-1}\\)\n\n', '(a)(i) Conduction and Radiation.\n(ii)\n\n\nThe specific heat capacity of land is less than that of sea. Land Breeze: At night land cools faster than sea, warm air from above the sea rises upward. It\'s place is taken by cooler air from above the land. Sea Breeze: During the day the land is hotter than the sea. Warm air above land rises, and its place is taken by cooler air from the sea.~(b) Increase in length (L\\(_2\\) - L\\(_1\\) = L\\(_1 \\alpha \\Delta \\theta = 30 \\times 1.2 \\times 10^{-5} \\times 50 = 1.8 \\times 10^{-2}\\) cm\n\n(c) Ivt = mc\\(\\Delta \\theta\\)\nPower x time = mc\\(\\Delta \\theta\\)\n(i) Iv x 360 = m x c 60 ----(i) Ivt = \\(\\frac{1}{2}\\)mL\n(ii) Ivx 25 x 60 = \\(\\frac{1}{2}\\)mL ------(ii)\n\\(\\frac{Iv \\times 25 \\times 60}{Iv \\times 360} = \\frac{\\frac{1}{2}mL}{mc \\times 60}\\)\nL = \\(\\frac{25 \\times 60 \\times C \\times 60 \\times 2}{360}\\)\n= \\(\\frac{25 \\times 60 \\times 4200 \\times 60 \\times 2}{360}\\)\n= 2.1 x 10\\(^5\\) Jkg\\(^{-1}\\)\n~(c) Ivt = mc\\(\\Delta \\theta\\)\nPower x time = mc\\(\\Delta \\theta\\)\n(i) Iv x 360 = m x c 60 ----(i) Ivt = \\(\\frac{1}{2}\\)mL\n(ii) Ivx 25 x 60 = \\(\\frac{1}{2}\\)mL ------(ii)\n\\(\\frac{Iv \\times 25 \\times 60}{Iv \\times 360} = \\frac{\\frac{1}{2}mL}{mc \\times 60}\\)\nL = \\(\\frac{25 \\times 60 \\times C \\times 60 \\times 2}{360}\\)\n= \\(\\frac{25 \\times 60 \\times 4200 \\times 60 \\times 2}{360}\\)\n= 2.1 x 10\\(^5\\) Jkg\\(^{-1}\\)\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
(252, '2019', '\n\nQuestion\n\n(a)(i) State the laws of refraction of light.\n(ii) Describe an experiment to determine the refractive index, n, of the material of an equilaleral triangular glass prism using the minimum deviation method.\n(b) A rectangular glass prism of thickness 12cm is placed on a mark on a piece of paper resting on a horizontal bench:\n(i) Draw a ray diagram to show the apparent position of the mark in the glass prism.\n(ii) If the refractive-index of the material of the prism is 1.5, calculate the apparent displacement of the mark.\n\n', '', '(a)(i) State the laws of refraction of light.\n(ii) Describe an experiment to determine the refractive index, n, of the material of an equilaleral triangular glass prism using the minimum deviation method.\n(b) A rectangular glass prism of thickness 12cm is placed on a mark on a piece of paper resting on a horizontal bench:\n(i) Draw a ray diagram to show the apparent position of the mark in the glass prism.\n(ii) If the refractive-index of the material of the prism is 1.5, calculate the apparent displacement of the mark.', '(a)(i) State the laws of refraction of light.\n(ii) Describe an experiment to determine the refractive index, n, of the material of an equilaleral triangular glass prism using the minimum deviation method.~(b) A rectangular glass prism of thickness 12cm is placed on a mark on a piece of paper resting on a horizontal bench:\n(i) Draw a ray diagram to show the apparent position of the mark in the glass prism.\n(ii) If the refractive-index of the material of the prism is 1.5, calculate the apparent displacement of the mark.', NULL, '\n\nExplanation\n(a)(i) The incident ray, the refracted ray, and the normal at the point of incidence, all lie in the same plane. OR The ratio of the sine of angle of incidence to the sine of angle of refraction is a constant \\(\\frac{\\text{Sin i}}{\\text{Sin r}}\\) = Constant\n\nPlace the triangular prism on the drawing sheet and trace the outline. Arrange two pins on one side vertically and using two other pins, determine the emergent ray on the other side. Remove the prism and. join the position of pins to meet the outline of the prism. Draw the normal at the point of incidence. Produce both the incident and the emergent rays to intersect and measure the angles of incidence I° and the deviation d°. Repeat the procedure for at least four other values of i. Plot a graph of d on the vertical axis against i on the horizontal axis. Obtain minimum deviation D as below.\n\nn = \\(\\frac{sin \\frac{1}{2}(A +D)}{sin \\frac{1}{2} A}\\)\n= \\(\\frac{sin \\frac{1}{2} (60 + D)}{sin 30}\\)\nPrecautions\n- neat traces\n- pins fixed vertically\n- Avoid parallax in reading protector\n\n(b)(i)\n\n(ii) n = \\(\\frac{\\text{real depth}}{\\text{apparent depth}}\\)\napparent depth = \\(\\frac{\\text{real depth}}{n}\\)\n= \\(\\frac{12}{1.5}\\)\n= 8cm\nApparent displacement = real depth - apparent depth\n= 12 - 8 = 4cm\n\n\n\n', '', 'Explanation\n(a)(i) The incident ray, the refracted ray, and the normal at the point of incidence, all lie in the same plane. OR The ratio of the sine of angle of incidence to the sine of angle of refraction is a constant \\(\\frac{\\text{Sin i}}{\\text{Sin r}}\\) = Constant\n\nPlace the triangular prism on the drawing sheet and trace the outline. Arrange two pins on one side vertically and using two other pins, determine the emergent ray on the other side. Remove the prism and. join the position of pins to meet the outline of the prism. Draw the normal at the point of incidence. Produce both the incident and the emergent rays to intersect and measure the angles of incidence I° and the deviation d°. Repeat the procedure for at least four other values of i. Plot a graph of d on the vertical axis against i on the horizontal axis. Obtain minimum deviation D as below.\n\nn = \\(\\frac{sin \\frac{1}{2}(A +D)}{sin \\frac{1}{2} A}\\)\n= \\(\\frac{sin \\frac{1}{2} (60 + D)}{sin 30}\\)\nPrecautions\n- neat traces\n- pins fixed vertically\n- Avoid parallax in reading protector\n\n(b)(i)\n\n(ii) n = \\(\\frac{\\text{real depth}}{\\text{apparent depth}}\\)\napparent depth = \\(\\frac{\\text{real depth}}{n}\\)\n= \\(\\frac{12}{1.5}\\)\n= 8cm\nApparent displacement = real depth - apparent depth\n= 12 - 8 = 4cm\n\n\n\n', '(a)(i) The incident ray, the refracted ray, and the normal at the point of incidence, all lie in the same plane. OR The ratio of the sine of angle of incidence to the sine of angle of refraction is a constant \\(\\frac{\\text{Sin i}}{\\text{Sin r}}\\) = Constant\n\nPlace the triangular prism on the drawing sheet and trace the outline. Arrange two pins on one side vertically and using two other pins, determine the emergent ray on the other side. Remove the prism and. join the position of pins to meet the outline of the prism. Draw the normal at the point of incidence. Produce both the incident and the emergent rays to intersect and measure the angles of incidence I° and the deviation d°. Repeat the procedure for at least four other values of i. Plot a graph of d on the vertical axis against i on the horizontal axis. Obtain minimum deviation D as below.\n\nn = \\(\\frac{sin \\frac{1}{2}(A +D)}{sin \\frac{1}{2} A}\\)\n= \\(\\frac{sin \\frac{1}{2} (60 + D)}{sin 30}\\)\nPrecautions\n- neat traces\n- pins fixed vertically\n- Avoid parallax in reading protector\n~(b)(i)\n\n(ii) n = \\(\\frac{\\text{real depth}}{\\text{apparent depth}}\\)\napparent depth = \\(\\frac{\\text{real depth}}{n}\\)\n= \\(\\frac{12}{1.5}\\)\n= 8cm\nApparent displacement = real depth - apparent depth\n= 12 - 8 = 4cm\n\n\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(253, '2019', '\n\nQuestion\n\n(a) Define the following terms:\n(i) Electric field intensity\n(ii) Electric potential\n(b) The diagram below illustrates two collinear electric charges of magnitudes + Q and -Q. The charges are equidistant from a point P at which a rest charge is placed.\n\nCopy the diagram and use arrows to indicate, from the point P, the direction of the;\n(i) electric force F\\(_1\\) due to + Q.\n(ii) electric force F\\(_2\\) due to -Q.\n(iii) electric field intensity E.\n\n(c) What is meant by dielectric substance?\n(ii) List the factors which determine the capacitance of a parallel plate capacitor and state the effect each of them has on the capacitance.The diagram above represents a section of a circuit. Calculate the effective capacitance in the section.\n(iii)\n\nThe diagram above represents a section of a circuit. Calculate the effective capacitance in the section.\n\n\n', '', '(a) Define the following terms:\n(i) Electric field intensity\n(ii) Electric potential\n(b) The diagram below illustrates two collinear electric charges of magnitudes + Q and -Q. The charges are equidistant from a point P at which a rest charge is placed.\n\nCopy the diagram and use arrows to indicate, from the point P, the direction of the;\n(i) electric force F\\(_1\\) due to + Q.\n(ii) electric force F\\(_2\\) due to -Q.\n(iii) electric field intensity E.\n\n(c) What is meant by dielectric substance?\n(ii) List the factors which determine the capacitance of a parallel plate capacitor and state the effect each of them has on the capacitance.The diagram above represents a section of a circuit. Calculate the effective capacitance in the section.\n(iii)\n\nThe diagram above represents a section of a circuit. Calculate the effective capacitance in the section.\n', '(a) Define the following terms:\n(i) Electric field intensity\n(ii) Electric potential~(b) The diagram below illustrates two collinear electric charges of magnitudes + Q and -Q. The charges are equidistant from a point P at which a rest charge is placed.\n\nCopy the diagram and use arrows to indicate, from the point P, the direction of the;\n(i) electric force F\\(_1\\) due to + Q.\n(ii) electric force F\\(_2\\) due to -Q.\n(iii) electric field intensity E.\n\n(c) What is meant by dielectric substance?\n(ii) List the factors which determine the capacitance of a parallel plate capacitor and state the effect each of them has on the capacitance.The diagram above represents a section of a circuit. Calculate the effective capacitance in the section.\n(iii)\n\nThe diagram above represents a section of a circuit. Calculate the effective capacitance in the section.\n~(c) What is meant by dielectric substance?\n(ii) List the factors which determine the capacitance of a parallel plate capacitor and state the effect each of them has on the capacitance.The diagram above represents a section of a circuit. Calculate the effective capacitance in the section.\n(iii)\n\nThe diagram above represents a section of a circuit. Calculate the effective capacitance in the section.\n', NULL, '\n\nExplanation\n(a)(i) Electric field intensity = Electric force per unit charge OR As negative potential i.e.E = \\(\\frac{-V}{d}\\)\nOR E = \\(\\frac{F}{Q}\\). where F is the force,\n\nQ is the charge and E electric field intensity.\n(ii) Electric potential =- work done per unit charge to bring a (positive) charge from infinity to a point in an electric field.\n\n(b)\n\n\n(c) (i) A dielectric substance is any insulating material placed between the plates of a capacitor\n\n\n\n(ii) Factors\nEffect on capacitor\n\n\n\ncommon area of plates\nDistance between plates\nNature of dielectric \n\n\ncapacitance increases with area\ncapacitance decreased with increasing distance\ncapacitance increases with dielectric constant\n\n\n\n\n\nC = \\(\\frac{KE_oA}{A}\\)\n(iii) C\\(_{2.3}\\) = C\\(_2\\) + C\\(_3\\) = 20 + 20 = 40\\(\\mu\\) F\n\\(\\frac{1}{C_T} = \\frac{1}{C_{2.3}} + \\frac{1}{C_1}\\)\n\\(C_T = \\frac{\\frac{C_{2.3} \\times C_1}{C_2.3 + C_1} = \\frac{40 \\times 40}{40 \\times 40}\\)\n= 20\\(\\mu F\\)\n\n', '', 'Explanation\n(a)(i) Electric field intensity = Electric force per unit charge OR As negative potential i.e.E = \\(\\frac{-V}{d}\\)\nOR E = \\(\\frac{F}{Q}\\). where F is the force,\n\nQ is the charge and E electric field intensity.\n(ii) Electric potential =- work done per unit charge to bring a (positive) charge from infinity to a point in an electric field.\n\n(b)\n\n\n(c) (i) A dielectric substance is any insulating material placed between the plates of a capacitor\n\n\n\n(ii) Factors\nEffect on capacitor\n\n\n\ncommon area of plates\nDistance between plates\nNature of dielectric \n\n\ncapacitance increases with area\ncapacitance decreased with increasing distance\ncapacitance increases with dielectric constant\n\n\n\n\n\nC = \\(\\frac{KE_oA}{A}\\)\n(iii) C\\(_{2.3}\\) = C\\(_2\\) + C\\(_3\\) = 20 + 20 = 40\\(\\mu\\) F\n\\(\\frac{1}{C_T} = \\frac{1}{C_{2.3}} + \\frac{1}{C_1}\\)\n\\(C_T = \\frac{\\frac{C_{2.3} \\times C_1}{C_2.3 + C_1} = \\frac{40 \\times 40}{40 \\times 40}\\)\n= 20\\(\\mu F\\)\n\n', '(a)(i) Electric field intensity = Electric force per unit charge OR As negative potential i.e.E = \\(\\frac{-V}{d}\\)\nOR E = \\(\\frac{F}{Q}\\). where F is the force,\n\nQ is the charge and E electric field intensity.\n(ii) Electric potential =- work done per unit charge to bring a (positive) charge from infinity to a point in an electric field.\n~(b)\n\n\n(c) (i) A dielectric substance is any insulating material placed between the plates of a capacitor\n\n\n\n(ii) Factors\nEffect on capacitor\n\n\n\ncommon area of plates\nDistance between plates\nNature of dielectric \n\n\ncapacitance increases with area\ncapacitance decreased with increasing distance\ncapacitance increases with dielectric constant\n\n\n\n\n\nC = \\(\\frac{KE_oA}{A}\\)\n(iii) C\\(_{2.3}\\) = C\\(_2\\) + C\\(_3\\) = 20 + 20 = 40\\(\\mu\\) F\n\\(\\frac{1}{C_T} = \\frac{1}{C_{2.3}} + \\frac{1}{C_1}\\)\n\\(C_T = \\frac{\\frac{C_{2.3} \\times C_1}{C_2.3 + C_1} = \\frac{40 \\times 40}{40 \\times 40}\\)\n= 20\\(\\mu F\\)\n~(c) (i) A dielectric substance is any insulating material placed between the plates of a capacitor\n\n\n\n(ii) Factors\nEffect on capacitor\n\n\n\ncommon area of plates\nDistance between plates\nNature of dielectric \n\n\ncapacitance increases with area\ncapacitance decreased with increasing distance\ncapacitance increases with dielectric constant\n\n\n\n\n\nC = \\(\\frac{KE_oA}{A}\\)\n(iii) C\\(_{2.3}\\) = C\\(_2\\) + C\\(_3\\) = 20 + 20 = 40\\(\\mu\\) F\n\\(\\frac{1}{C_T} = \\frac{1}{C_{2.3}} + \\frac{1}{C_1}\\)\n\\(C_T = \\frac{\\frac{C_{2.3} \\times C_1}{C_2.3 + C_1} = \\frac{40 \\times 40}{40 \\times 40}\\)\n= 20\\(\\mu F\\)\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(254, '2019', '\n\nQuestion\n\n(a)(1) State the energy transformations which take place during the operation of a modern x-ray tube.\n(ii) Distinguish between hard and soft x-rays.\n(iii) State three uses of x-rays.\n(iv) Mention one hazard of over-exposure to x-rays in a radiological laboratory, and indicate any two safety precautions.\n(b) A possible fusion reaction is \\(^2_1 H + ^2_1 H \\to ^3_1H + ^1_1H + Q\\)\nwhere Q is the energy released as a result of the reaction. If Q = 4.03 MeV, calculate the atomic mass of \\(^3_1H\\) in atomic mass units. (\\(^2_1 H = 2.01410 U; ^1_1H = 1.00783 U; 1U = 931 MeV\\))\n\n', '', '(a)(1) State the energy transformations which take place during the operation of a modern x-ray tube.\n(ii) Distinguish between hard and soft x-rays.\n(iii) State three uses of x-rays.\n(iv) Mention one hazard of over-exposure to x-rays in a radiological laboratory, and indicate any two safety precautions.\n(b) A possible fusion reaction is \\(^2_1 H + ^2_1 H \\to ^3_1H + ^1_1H + Q\\)\nwhere Q is the energy released as a result of the reaction. If Q = 4.03 MeV, calculate the atomic mass of \\(^3_1H\\) in atomic mass units. (\\(^2_1 H = 2.01410 U; ^1_1H = 1.00783 U; 1U = 931 MeV\\))', '(a)(1) State the energy transformations which take place during the operation of a modern x-ray tube.\n(ii) Distinguish between hard and soft x-rays.\n(iii) State three uses of x-rays.\n(iv) Mention one hazard of over-exposure to x-rays in a radiological laboratory, and indicate any two safety precautions.~(b) A possible fusion reaction is \\(^2_1 H + ^2_1 H \\to ^3_1H + ^1_1H + Q\\)\nwhere Q is the energy released as a result of the reaction. If Q = 4.03 MeV, calculate the atomic mass of \\(^3_1H\\) in atomic mass units. (\\(^2_1 H = 2.01410 U; ^1_1H = 1.00783 U; 1U = 931 MeV\\))', NULL, '\n\nExplanation\n(i) Electrical \\(to\\) thermal \\(to\\) mechanical (kinetic energy) \\(to\\) thermal \\(to\\) electromagnetic\n(ii) Distinction between hard and soft X-rays\n\n\n\nHard X-rays\nSoft x-rays\n\n\n\n(i) High (target) voltage\n(ii) short wavelength\n(iii) High penetration\n(iv) produced by high voltage\n\n\nLow (target) voltage\nLong wavelength\nLow penetration\nProduced by low voltage\n\n\n\n\n(iii) Uses:\n- Used to detect hidden cracks in materials\n- Used to detect alteration to work of art\n- To show broken bones in human body\n- used in crystallography i.e. to study crystals\n- used in diagnosis of ulcer\n(iv) Hazards: genetic mutations\n- skin burn\n- leukemia\n- Tissue damage\n- Tissue damage\n- baldness\n- graying\n- cataracts\n(v) Safety precautions:\nuse of dosimeter, badges of GM counters\n- wearing of lead coats\n- Periodic medical counters\n- wearing of lead coats\n- periodic medical check up\n\n(b)\n\\(^2_1H + ^2_1H \\to ^3_1H + ^1_1H + Q\\)\nLet atomic mass unit of tritium \\(^3_1H\\) be X\n2.01410 x 2 x 931 = (x + 1.00783) 931 + 4.03\n\\(\\frac{3750.254 - 4.03}{931}\\) = x + 1.00783\n4.0238 - 1.0078 = x\n3.106 = x\n\n\n', '', 'Explanation\n(i) Electrical \\(to\\) thermal \\(to\\) mechanical (kinetic energy) \\(to\\) thermal \\(to\\) electromagnetic\n(ii) Distinction between hard and soft X-rays\n\n\n\nHard X-rays\nSoft x-rays\n\n\n\n(i) High (target) voltage\n(ii) short wavelength\n(iii) High penetration\n(iv) produced by high voltage\n\n\nLow (target) voltage\nLong wavelength\nLow penetration\nProduced by low voltage\n\n\n\n\n(iii) Uses:\n- Used to detect hidden cracks in materials\n- Used to detect alteration to work of art\n- To show broken bones in human body\n- used in crystallography i.e. to study crystals\n- used in diagnosis of ulcer\n(iv) Hazards: genetic mutations\n- skin burn\n- leukemia\n- Tissue damage\n- Tissue damage\n- baldness\n- graying\n- cataracts\n(v) Safety precautions:\nuse of dosimeter, badges of GM counters\n- wearing of lead coats\n- Periodic medical counters\n- wearing of lead coats\n- periodic medical check up\n\n(b)\n\\(^2_1H + ^2_1H \\to ^3_1H + ^1_1H + Q\\)\nLet atomic mass unit of tritium \\(^3_1H\\) be X\n2.01410 x 2 x 931 = (x + 1.00783) 931 + 4.03\n\\(\\frac{3750.254 - 4.03}{931}\\) = x + 1.00783\n4.0238 - 1.0078 = x\n3.106 = x\n\n\n', '(i) Electrical \\(to\\) thermal \\(to\\) mechanical (kinetic energy) \\(to\\) thermal \\(to\\) electromagnetic\n(ii) Distinction between hard and soft X-rays\n\n\n\nHard X-rays\nSoft x-rays\n\n\n\n(i) High (target) voltage\n(ii) short wavelength\n(iii) High penetration\n(iv) produced by high voltage\n\n\nLow (target) voltage\nLong wavelength\nLow penetration\nProduced by low voltage\n\n\n\n\n(iii) Uses:\n- Used to detect hidden cracks in materials\n- Used to detect alteration to work of art\n- To show broken bones in human body\n- used in crystallography i.e. to study crystals\n- used in diagnosis of ulcer\n(iv) Hazards: genetic mutations\n- skin burn\n- leukemia\n- Tissue damage\n- Tissue damage\n- baldness\n- graying\n- cataracts\n(v) Safety precautions:\nuse of dosimeter, badges of GM counters\n- wearing of lead coats\n- Periodic medical counters\n- wearing of lead coats\n- periodic medical check up\n~(b)\n\\(^2_1H + ^2_1H \\to ^3_1H + ^1_1H + Q\\)\nLet atomic mass unit of tritium \\(^3_1H\\) be X\n2.01410 x 2 x 931 = (x + 1.00783) 931 + 4.03\n\\(\\frac{3750.254 - 4.03}{931}\\) = x + 1.00783\n4.0238 - 1.0078 = x\n3.106 = x\n\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(255, '2019', '\n\nQuestion\n\nExplain plane polarisation of light\n\n', 'Explain plane polarisation of light', '', '', NULL, '\n\nExplanation\nIn plane polarisation of light, the vibrations are made to vibrate only in one plane perpendicular to the direction of the light. It is used in polaroids.\n\n', 'In plane polarisation of light, the vibrations are made to vibrate only in one plane perpendicular to the direction of the light. It is used in polaroids.\n\n', '', '', 'WAEC-PK-1', 'WAEC-PHY', 1),
(256, '2019', '\n\nQuestion\n\nA stone is projected horizontally from the top of a tower with a speed of 5ms-1. It lands on the ground level at a horizontal distance of 20m from the foot of the tower. Calculate the height of the tower. (g = 10ms\\(^{-2}\\))\n\n', 'A stone is projected horizontally from the top of a tower with a speed of 5ms-1. It lands on the ground level at a horizontal distance of 20m from the foot of the tower. Calculate the height of the tower. (g = 10ms\\(^{-2}\\))', '', '', NULL, '\n\nExplanation\nH = \\(\\frac{1}{2} gt^2\\) S = ut\nt = \\(\\frac{5}{u} = \\frac{20}{5} = 45\\)\nH = \\(\\frac{1}{2}gt^2 = \\frac{1}{2} \\times 10 x 4^2\\)\n= 80m\n\n', 'H = \\(\\frac{1}{2} gt^2\\) S = ut\nt = \\(\\frac{5}{u} = \\frac{20}{5} = 45\\)\nH = \\(\\frac{1}{2}gt^2 = \\frac{1}{2} \\times 10 x 4^2\\)\n= 80m\n\n', '', '', 'WAEC-PK-1', 'WAEC-PHY', 1),
(257, '2019', '\n\nQuestion\n\nIn an electrolysis experiment, the ammeter records a steady current of 1A. The mass of copper deposited is 0.66g in 30 minutes. Calculate the error in the ammeter reading (Electrochemical equivalent of copper = 0.00033g C\\(^{-1}\\)).\n\n', 'In an electrolysis experiment, the ammeter records a steady current of 1A. The mass of copper deposited is 0.66g in 30 minutes. Calculate the error in the ammeter reading (Electrochemical equivalent of copper = 0.00033g C\\(^{-1}\\)).', '', '', NULL, '\n\nExplanation\nMass deposited = 21t\nI = \\(\\frac{m}{2t} = \\frac{0.66}{33 \\times 10^{-5} \\times 30 \\times 60}\\)\n= 1.11A\nerror = 1.11 - 1\n= 0.11A\n\n', 'Mass deposited = 21t\nI = \\(\\frac{m}{2t} = \\frac{0.66}{33 \\times 10^{-5} \\times 30 \\times 60}\\)\n= 1.11A\nerror = 1.11 - 1\n= 0.11A\n\n', '', '', 'WAEC-PK-1', 'WAEC-PHY', 1),
(258, '2019', '\n\nQuestion\n\n(a) Define Young\'s modulus.\n(b) State the physical quantities one has to measure in order to determine the Young\'s modulus of a wire.\n\n', '', '(a) Define Young\'s modulus.\n(b) State the physical quantities one has to measure in order to determine the Young\'s modulus of a wire.', '(a) Define Young\'s modulus.~(b) State the physical quantities one has to measure in order to determine the Young\'s modulus of a wire.', NULL, '\n\nExplanation\n(a) Young\'s Modulus is the ratio of tensile stress to tensile strain\n(b) Length of wire, extension of wire, load used, diameter of wire.\n\n', '', 'Explanation\n(a) Young\'s Modulus is the ratio of tensile stress to tensile strain\n(b) Length of wire, extension of wire, load used, diameter of wire.\n\n', '(a) Young\'s Modulus is the ratio of tensile stress to tensile strain~(b) Length of wire, extension of wire, load used, diameter of wire.\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(259, '2019', '\n\nQuestion\n\n(a) What is the principle upon which the lighting in fluorescent tubes operate?\n(b) State two factors on which the colour of light from a fluorescent tube depends.\n\n', '', '(a) What is the principle upon which the lighting in fluorescent tubes operate?\n(b) State two factors on which the colour of light from a fluorescent tube depends.', '(a) What is the principle upon which the lighting in fluorescent tubes operate?~(b) State two factors on which the colour of light from a fluorescent tube depends.', NULL, '\n\nExplanation\n(a) Electrical discharge through low pressure gas\n(b)\n- Nature of the gas in the tube\n- Nature of the coating d inside the tube.\n\n', '', 'Explanation\n(a) Electrical discharge through low pressure gas\n(b)\n- Nature of the gas in the tube\n- Nature of the coating d inside the tube.\n\n', '(a) Electrical discharge through low pressure gas~(b)\n- Nature of the gas in the tube\n- Nature of the coating d inside the tube.\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(260, '2019', '\n\nQuestion\n\n(a) List two types of waves, apart from light, that can be plane polarised\n(b) State two applications of plane polarised light.\n\n', '', '(a) List two types of waves, apart from light, that can be plane polarised\n(b) State two applications of plane polarised light.', '(a) List two types of waves, apart from light, that can be plane polarised~(b) State two applications of plane polarised light.', NULL, '\n\nExplanation\n(a) \\(_gamma\\) - rays. X - rays, radio waves, infra-red.\n- To determine the concentration of sugar solution\n- In polaroids used in sun glasses to reduce the intensity of incident sunlight glare from reflecting surfaces.\n\n', '(a) \\(_gamma\\) - rays. X - rays, radio waves, infra-red.\n- To determine the concentration of sugar solution\n- In polaroids used in sun glasses to reduce the intensity of incident sunlight glare from reflecting surfaces.\n\n', '', '', 'WAEC-PK-1', 'WAEC-PHY', 1),
(261, '2019', '\n\nQuestion\n\n(a) State Hooke\'s law of elasticity\n(b) A spiral spring, loaded with a piece of metal, extends by 10.5cm hair. When the metal is fully submerged in water, the spring extends by 6.8cm. Calculate the relative density of the metal. (Assume Hooke\'s law is obeyed).\n\n', '', '(a) State Hooke\'s law of elasticity\n(b) A spiral spring, loaded with a piece of metal, extends by 10.5cm hair. When the metal is fully submerged in water, the spring extends by 6.8cm. Calculate the relative density of the metal. (Assume Hooke\'s law is obeyed).', '(a) State Hooke\'s law of elasticity~(b) A spiral spring, loaded with a piece of metal, extends by 10.5cm hair. When the metal is fully submerged in water, the spring extends by 6.8cm. Calculate the relative density of the metal. (Assume Hooke\'s law is obeyed).', NULL, '\n\nExplanation\n(a) Hooke\'s law states that the force applied on an elastic materal is proportional to its extension, provided the elastic limit is not exceeded\n(b) R.D = \\(\\frac{\\text{weight in air}}{\\text{upthrust}} = \\frac{10}{10.5 - 6.8} = \\frac{10.5}{3.7} = 2.84\\)\n\n\n', '', 'Explanation\n(a) Hooke\'s law states that the force applied on an elastic materal is proportional to its extension, provided the elastic limit is not exceeded\n(b) R.D = \\(\\frac{\\text{weight in air}}{\\text{upthrust}} = \\frac{10}{10.5 - 6.8} = \\frac{10.5}{3.7} = 2.84\\)\n\n\n', '(a) Hooke\'s law states that the force applied on an elastic materal is proportional to its extension, provided the elastic limit is not exceeded~(b) R.D = \\(\\frac{\\text{weight in air}}{\\text{upthrust}} = \\frac{10}{10.5 - 6.8} = \\frac{10.5}{3.7} = 2.84\\)\n\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(262, '2019', '\n\nQuestion\n\n(a) Explain electrolysis\n(b) Classify the following substances as electrolytes and non-electrolytes: Sugar solution: paraffin, suit solution and grape fruit juice.\n\n', '', '(a) Explain electrolysis\n(b) Classify the following substances as electrolytes and non-electrolytes: Sugar solution: paraffin, suit solution and grape fruit juice.', '(a) Explain electrolysis~(b) Classify the following substances as electrolytes and non-electrolytes: Sugar solution: paraffin, suit solution and grape fruit juice.', NULL, '\n\nExplanation\n(a) This is the process by which a solution or a molten substance is broken down into the component parts by the passage of an electric current through it.\n(b) \\(\\frac{Electrolytes}{\\text{salt solution}}\\) = \\(\\frac{\\text{non - electrolytes}}{\\text{Sugar solution}}\\)\ngrape fruit juice paraffin\n\n\n\n', '', 'Explanation\n(a) This is the process by which a solution or a molten substance is broken down into the component parts by the passage of an electric current through it.\n(b) \\(\\frac{Electrolytes}{\\text{salt solution}}\\) = \\(\\frac{\\text{non - electrolytes}}{\\text{Sugar solution}}\\)\ngrape fruit juice paraffin\n\n\n\n', '(a) This is the process by which a solution or a molten substance is broken down into the component parts by the passage of an electric current through it.~(b) \\(\\frac{Electrolytes}{\\text{salt solution}}\\) = \\(\\frac{\\text{non - electrolytes}}{\\text{Sugar solution}}\\)\ngrape fruit juice paraffin\n\n\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(263, '2019', '\n\nQuestion\n\n(a) State Heisenberg\'s uncertainty principle.\n(b) Mention two phenomena that can only be explained in terms of the particulate nature of light.\n\n', '', '(a) State Heisenberg\'s uncertainty principle.\n(b) Mention two phenomena that can only be explained in terms of the particulate nature of light.', '(a) State Heisenberg\'s uncertainty principle.~(b) Mention two phenomena that can only be explained in terms of the particulate nature of light.', NULL, '\n\nExplanation\n(a) Heinsenberg\'s uncertainty principle states that it is impossible to determine simultaneously the exact position and exact momentum of a particle\n(b) Black body radiation - Photo electricity.\n\n', '', 'Explanation\n(a) Heinsenberg\'s uncertainty principle states that it is impossible to determine simultaneously the exact position and exact momentum of a particle\n(b) Black body radiation - Photo electricity.\n\n', '(a) Heinsenberg\'s uncertainty principle states that it is impossible to determine simultaneously the exact position and exact momentum of a particle~(b) Black body radiation - Photo electricity.\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(264, '2019', '\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, using the principle of moments, to determine the mass of a metre rule\n(ii) State two precautions necessary to ensure accurate results.\n(c) A bullet of mass 120g is fired horizontally into a fixed wooden block with a speed of 20ms\\(^{-1}\\). The bullet is brought to rest in the block in 0.1s by a constant resistance. Calculate the:\n(i) magnitude of the resistance;\n(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, using the principle of moments, to determine the mass of a metre rule\n(ii) State two precautions necessary to ensure accurate results.\n(c) A bullet of mass 120g is fired horizontally into a fixed wooden block with a speed of 20ms\\(^{-1}\\). The bullet is brought to rest in the block in 0.1s by a constant resistance. Calculate the:\n(i) magnitude of the resistance;\n(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, using the principle of moments, to determine the mass of a metre rule\n(ii) State two precautions necessary to ensure accurate results.\n(c) A bullet of mass 120g is fired horizontally into a fixed wooden block with a speed of 20ms\\(^{-1}\\). The bullet is brought to rest in the block in 0.1s by a constant resistance. Calculate the:\n(i) magnitude of the resistance;\n(ii) distance moved by the bullet in the wood.~(c) A bullet of mass 120g is fired horizontally into a fixed wooden block with a speed of 20ms\\(^{-1}\\). The bullet is brought to rest in the block in 0.1s by a constant resistance. Calculate the:\n(i) magnitude of the resistance;\n(ii) distance moved by the bullet in the wood.', NULL, '\n\nExplanation\n(a) - Sum of forces in one direction is equal to the sum of forces in opposite direction. - Sum of anti-clockwise moments about a point is equal to the sum of clockwise moment about the same point.\n(b) Diagram:\n\nMethod: Suspend the metre rule on a knife edge. Adjust its position until it balances horizontally. Note the balance point G. Then suspend a mass M on one side of the rule. Adjust the rule until it balances horizontally again. Read and record X and Y as in the diagram above. According to the principle of moment:\nMgx = Mgy\nm = \\(\\frac{Mgy}{gx} = \\frac{My}{x}\\)\nm = mass of rule\nPrecautions\n- Avoid drought\n- Avoid parallax in reading the metre rile\n\n(c)(i) Deceleration a = \\(\\frac{u - v}{t} = \\frac{20 - 0}{0.1}\\) = 200ms\n Resistance = ma = 0.12 x 200 = 24N\n(ii) Distance, s \\(\\frac{u^2 - v^2}{2a} = \\frac{20^2 - O^2}{2 \\times 200} = \\frac{400}{400}\\)\n= 1.0m\n\n', '', 'Explanation\n(a) - Sum of forces in one direction is equal to the sum of forces in opposite direction. - Sum of anti-clockwise moments about a point is equal to the sum of clockwise moment about the same point.\n(b) Diagram:\n\nMethod: Suspend the metre rule on a knife edge. Adjust its position until it balances horizontally. Note the balance point G. Then suspend a mass M on one side of the rule. Adjust the rule until it balances horizontally again. Read and record X and Y as in the diagram above. According to the principle of moment:\nMgx = Mgy\nm = \\(\\frac{Mgy}{gx} = \\frac{My}{x}\\)\nm = mass of rule\nPrecautions\n- Avoid drought\n- Avoid parallax in reading the metre rile\n\n(c)(i) Deceleration a = \\(\\frac{u - v}{t} = \\frac{20 - 0}{0.1}\\) = 200ms\n Resistance = ma = 0.12 x 200 = 24N\n(ii) Distance, s \\(\\frac{u^2 - v^2}{2a} = \\frac{20^2 - O^2}{2 \\times 200} = \\frac{400}{400}\\)\n= 1.0m\n\n', '(a) - Sum of forces in one direction is equal to the sum of forces in opposite direction. - Sum of anti-clockwise moments about a point is equal to the sum of clockwise moment about the same point.~(b) Diagram:\n\nMethod: Suspend the metre rule on a knife edge. Adjust its position until it balances horizontally. Note the balance point G. Then suspend a mass M on one side of the rule. Adjust the rule until it balances horizontally again. Read and record X and Y as in the diagram above. According to the principle of moment:\nMgx = Mgy\nm = \\(\\frac{Mgy}{gx} = \\frac{My}{x}\\)\nm = mass of rule\nPrecautions\n- Avoid drought\n- Avoid parallax in reading the metre rile\n\n(c)(i) Deceleration a = \\(\\frac{u - v}{t} = \\frac{20 - 0}{0.1}\\) = 200ms\n Resistance = ma = 0.12 x 200 = 24N\n(ii) Distance, s \\(\\frac{u^2 - v^2}{2a} = \\frac{20^2 - O^2}{2 \\times 200} = \\frac{400}{400}\\)\n= 1.0m\n~(c)(i) Deceleration a = \\(\\frac{u - v}{t} = \\frac{20 - 0}{0.1}\\) = 200ms\n Resistance = ma = 0.12 x 200 = 24N\n(ii) Distance, s \\(\\frac{u^2 - v^2}{2a} = \\frac{20^2 - O^2}{2 \\times 200} = \\frac{400}{400}\\)\n= 1.0m\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(265, '2019', '\n\nQuestion\n\n(a) Explain (i) work: (ii) power\n(b) Show that the efficiency E, the force ratio M.A. and the velocity ratio V.R. of a machine are related by the equation: E = M.A. x 100% V.R.\n(c) An inclined plane of angle 15° is used raise a load of 4500N through a height of 2m. If the plane is 75% efficient calculate:\n(i) velocity ratio of the plane;\n(ii) work done on the load\n(d) Explain Charles\' law using the Kinetic theory of matter.\n\n', '', '(a) Explain (i) work: (ii) power\n(b) Show that the efficiency E, the force ratio M.A. and the velocity ratio V.R. of a machine are related by the equation: E = M.A. x 100% V.R.\n(c) An inclined plane of angle 15° is used raise a load of 4500N through a height of 2m. If the plane is 75% efficient calculate:\n(i) velocity ratio of the plane;\n(ii) work done on the load\n(d) Explain Charles\' law using the Kinetic theory of matter.', '(a) Explain (i) work: (ii) power~(b) Show that the efficiency E, the force ratio M.A. and the velocity ratio V.R. of a machine are related by the equation: E = M.A. x 100% V.R.\n(c) An inclined plane of angle 15° is used raise a load of 4500N through a height of 2m. If the plane is 75% efficient calculate:\n(i) velocity ratio of the plane;\n(ii) work done on the load\n(d) Explain Charles\' law using the Kinetic theory of matter.~(c) An inclined plane of angle 15° is used raise a load of 4500N through a height of 2m. If the plane is 75% efficient calculate:\n(i) velocity ratio of the plane;\n(ii) work done on the load\n(d) Explain Charles\' law using the Kinetic theory of matter.~(d) Explain Charles\' law using the Kinetic theory of matter.', NULL, '\n\nExplanation\n(a)(i) Work is the product of force and distance moved in the direction of the force\n(ii) Power is the time rate of doing work. Unit of power is watts or J / sec.\n\n(b) E = \\(\\frac{\\text{Work output}}{\\text{work input}}\\) x 100% work input\nE = \\(\\frac{\\text{load x distance moved by load}}{\\text{effort x distance moved by effort}}\\)\nBut Ma = \\(\\frac{Load}{effort}\\) and V.R = \\(\\frac{\\text{distance moved by effort}}{\\text{Effort distance moved by load}}\\)\nE = \\(\\frac{MA}{1} \\times \\frac{1}{VR}\\) x 100%\nE = \\(\\frac{MA}{V.R}\\) x 100%\n\n(c)(i) VR = \\(\\frac{1}{sin \\theta}\\) = \\(\\frac{1}{sin 45^o} = \\frac{1}{0.2588}\\) = 3.86\n\n75% = \\(\\frac{9000}{Win}\\) x 100%\nWin = \\(\\frac{9000 \\times 100}{75}\\) = 12,000J\nwork input = 12000 J.\n(d) At constant pressure, the number of collision per unit time is constant. An increase in temperature causes average Kinetic energy increases. The molecules then move faster. The volume therefore increases so as to maintain the number of collision per unit time.\n\n', '', 'Explanation\n(a)(i) Work is the product of force and distance moved in the direction of the force\n(ii) Power is the time rate of doing work. Unit of power is watts or J / sec.\n\n(b) E = \\(\\frac{\\text{Work output}}{\\text{work input}}\\) x 100% work input\nE = \\(\\frac{\\text{load x distance moved by load}}{\\text{effort x distance moved by effort}}\\)\nBut Ma = \\(\\frac{Load}{effort}\\) and V.R = \\(\\frac{\\text{distance moved by effort}}{\\text{Effort distance moved by load}}\\)\nE = \\(\\frac{MA}{1} \\times \\frac{1}{VR}\\) x 100%\nE = \\(\\frac{MA}{V.R}\\) x 100%\n\n(c)(i) VR = \\(\\frac{1}{sin \\theta}\\) = \\(\\frac{1}{sin 45^o} = \\frac{1}{0.2588}\\) = 3.86\n\n75% = \\(\\frac{9000}{Win}\\) x 100%\nWin = \\(\\frac{9000 \\times 100}{75}\\) = 12,000J\nwork input = 12000 J.\n(d) At constant pressure, the number of collision per unit time is constant. An increase in temperature causes average Kinetic energy increases. The molecules then move faster. The volume therefore increases so as to maintain the number of collision per unit time.\n\n', '(a)(i) Work is the product of force and distance moved in the direction of the force\n(ii) Power is the time rate of doing work. Unit of power is watts or J / sec.\n~(b) E = \\(\\frac{\\text{Work output}}{\\text{work input}}\\) x 100% work input\nE = \\(\\frac{\\text{load x distance moved by load}}{\\text{effort x distance moved by effort}}\\)\nBut Ma = \\(\\frac{Load}{effort}\\) and V.R = \\(\\frac{\\text{distance moved by effort}}{\\text{Effort distance moved by load}}\\)\nE = \\(\\frac{MA}{1} \\times \\frac{1}{VR}\\) x 100%\nE = \\(\\frac{MA}{V.R}\\) x 100%\n\n(c)(i) VR = \\(\\frac{1}{sin \\theta}\\) = \\(\\frac{1}{sin 45^o} = \\frac{1}{0.2588}\\) = 3.86\n\n75% = \\(\\frac{9000}{Win}\\) x 100%\nWin = \\(\\frac{9000 \\times 100}{75}\\) = 12,000J\nwork input = 12000 J.\n(d) At constant pressure, the number of collision per unit time is constant. An increase in temperature causes average Kinetic energy increases. The molecules then move faster. The volume therefore increases so as to maintain the number of collision per unit time.\n~(c)(i) VR = \\(\\frac{1}{sin \\theta}\\) = \\(\\frac{1}{sin 45^o} = \\frac{1}{0.2588}\\) = 3.86\n\n75% = \\(\\frac{9000}{Win}\\) x 100%\nWin = \\(\\frac{9000 \\times 100}{75}\\) = 12,000J\nwork input = 12000 J.\n(d) At constant pressure, the number of collision per unit time is constant. An increase in temperature causes average Kinetic energy increases. The molecules then move faster. The volume therefore increases so as to maintain the number of collision per unit time.\n~(d) At constant pressure, the number of collision per unit time is constant. An increase in temperature causes average Kinetic energy increases. The molecules then move faster. The volume therefore increases so as to maintain the number of collision per unit time.\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(266, '2019', '\n\nQuestion\n\n(a) State three properties of waves\n(b)(i) Describe, with the aid of a labelled diagram, an experiment to show how the frequency of the note emitted by a vibrating string depends on the tension in the string\n(ii) State two precautions necessary to ensure accurate results.\n(c) Draw a ray diagram showing how a virtual image of an object is formed by a concave mirror.\n\n', '', '(a) State three properties of waves\n(b)(i) Describe, with the aid of a labelled diagram, an experiment to show how the frequency of the note emitted by a vibrating string depends on the tension in the string\n(ii) State two precautions necessary to ensure accurate results.\n(c) Draw a ray diagram showing how a virtual image of an object is formed by a concave mirror.', '(a) State three properties of waves~(b)(i) Describe, with the aid of a labelled diagram, an experiment to show how the frequency of the note emitted by a vibrating string depends on the tension in the string\n(ii) State two precautions necessary to ensure accurate results.\n(c) Draw a ray diagram showing how a virtual image of an object is formed by a concave mirror.~(c) Draw a ray diagram showing how a virtual image of an object is formed by a concave mirror.', NULL, '\n\nExplanation\n(a) Reflection, refraction, polarisation\n= 3.86 9000 Win\n\n(b)(i)\n\n\nMethod: Keep the distance AB constant and tight. Sound fork of known frequency F and pluck the string in the middle. Then add load to scale pan until resonance occurs. Record the load in the scale pan and the corresponding frequency. Repeat the experiment for about four tuning forks of known frequencies. Graph: Plot a graph of frequency, F against the square root of the load. Observation: The straight line graph passing through the origin shows that frequency is proportional to the square root of tension (load). F \\(\\alpha\\) T\n\nPrecautions:\n- Sonometer wire is firmly fixed.\n- Strike the tuning fork on a pad\n\n(c)\n\n\n\n\n\n', '', 'Explanation\n(a) Reflection, refraction, polarisation\n= 3.86 9000 Win\n\n(b)(i)\n\n\nMethod: Keep the distance AB constant and tight. Sound fork of known frequency F and pluck the string in the middle. Then add load to scale pan until resonance occurs. Record the load in the scale pan and the corresponding frequency. Repeat the experiment for about four tuning forks of known frequencies. Graph: Plot a graph of frequency, F against the square root of the load. Observation: The straight line graph passing through the origin shows that frequency is proportional to the square root of tension (load). F \\(\\alpha\\) T\n\nPrecautions:\n- Sonometer wire is firmly fixed.\n- Strike the tuning fork on a pad\n\n(c)\n\n\n\n\n\n', '(a) Reflection, refraction, polarisation\n= 3.86 9000 Win\n~(b)(i)\n\n\nMethod: Keep the distance AB constant and tight. Sound fork of known frequency F and pluck the string in the middle. Then add load to scale pan until resonance occurs. Record the load in the scale pan and the corresponding frequency. Repeat the experiment for about four tuning forks of known frequencies. Graph: Plot a graph of frequency, F against the square root of the load. Observation: The straight line graph passing through the origin shows that frequency is proportional to the square root of tension (load). F \\(\\alpha\\) T\n\nPrecautions:\n- Sonometer wire is firmly fixed.\n- Strike the tuning fork on a pad\n\n(c)\n\n\n\n\n~(c)\n\n\n\n\n~d). F \\(\\alpha\\) T\n\nPrecautions:\n- Sonometer wire is firmly fixed.\n- Strike the tuning fork on a pad\n\n(c)\n\n\n\n\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(267, '2019', '\n\nQuestion\n\n(a) Sketch the magnetic flux pattern around a long; straight, current-carrying wire\n(b) State two methods by which the sensitivity of a moving-coil galvanometer can be increased.\n(c) A series RLC circuit comprises a 100-\\(\\Omega\\) resistor, a 3-H inductor and a 4-\\(\\mu\\)f capacitor. The a.c source of tile circuit has an e.m.f of 100V and a frequency of 160 Hz.\n(i) Draw the circuit diagram of the arrangement. Calculate the:\n(ii) capacitive reactance;\n(iii) inductive reactance:\n(iv) impedance of the circuit;\n(v) current in the circuit;\n(vi) average power dissipated in the circuit.\n\n', '', '(a) Sketch the magnetic flux pattern around a long; straight, current-carrying wire\n(b) State two methods by which the sensitivity of a moving-coil galvanometer can be increased.\n(c) A series RLC circuit comprises a 100-\\(\\Omega\\) resistor, a 3-H inductor and a 4-\\(\\mu\\)f capacitor. The a.c source of tile circuit has an e.m.f of 100V and a frequency of 160 Hz.\n(i) Draw the circuit diagram of the arrangement. Calculate the:\n(ii) capacitive reactance;\n(iii) inductive reactance:\n(iv) impedance of the circuit;\n(v) current in the circuit;\n(vi) average power dissipated in the circuit.', '(a) Sketch the magnetic flux pattern around a long; straight, current-carrying wire~(b) State two methods by which the sensitivity of a moving-coil galvanometer can be increased.\n(c) A series RLC circuit comprises a 100-\\(\\Omega\\) resistor, a 3-H inductor and a 4-\\(\\mu\\)f capacitor. The a.c source of tile circuit has an e.m.f of 100V and a frequency of 160 Hz.\n(i) Draw the circuit diagram of the arrangement. Calculate the:\n(ii) capacitive reactance;\n(iii) inductive reactance:\n(iv) impedance of the circuit;\n(v) current in the circuit;\n(vi) average power dissipated in the circuit.~(c) A series RLC circuit comprises a 100-\\(\\Omega\\) resistor, a 3-H inductor and a 4-\\(\\mu\\)f capacitor. The a.c source of tile circuit has an e.m.f of 100V and a frequency of 160 Hz.\n(i) Draw the circuit diagram of the arrangement. Calculate the:\n(ii) capacitive reactance;\n(iii) inductive reactance:\n(iv) impedance of the circuit;\n(v) current in the circuit;\n(vi) average power dissipated in the circuit.', NULL, '\n\nExplanation\n(a)\n\n\n(b) The sensitivity is increased by a weak spring and increasing the strength of the magnet C\n(c)(i)\n\n(ii) X = \\(\\frac{1}{2 \\pi FC} = \\frac{1}{2 \\pi \\frac{160}{\\pi} \\times 4}\\) = 781\\(\\Omega\\)\n(iii) X\\(_L\\) = 2\\(\\pi\\)fL = 2\\(pi\\) x \\(\\frac{160}{\\pi}\\) x 3 = 960\\(\\Omega\\)\n(iv) Z = \\(\\sqrt{R^2 + (X_L - X_c)^2} = \\sqrt{100^2 + (960 - 781)^2}\\) = 205\\(\\Omega\\)\n(v) I = \\(\\frac{v}{z} = \\frac{100}{205} = 0.49A\\)\n(vi) P = I\\(^2R\\) = (0.49)\\(^{2}\\) x 100\n= 23.795 W = 24W\n\n\n\n\n', '', 'Explanation\n(a)\n\n\n(b) The sensitivity is increased by a weak spring and increasing the strength of the magnet C\n(c)(i)\n\n(ii) X = \\(\\frac{1}{2 \\pi FC} = \\frac{1}{2 \\pi \\frac{160}{\\pi} \\times 4}\\) = 781\\(\\Omega\\)\n(iii) X\\(_L\\) = 2\\(\\pi\\)fL = 2\\(pi\\) x \\(\\frac{160}{\\pi}\\) x 3 = 960\\(\\Omega\\)\n(iv) Z = \\(\\sqrt{R^2 + (X_L - X_c)^2} = \\sqrt{100^2 + (960 - 781)^2}\\) = 205\\(\\Omega\\)\n(v) I = \\(\\frac{v}{z} = \\frac{100}{205} = 0.49A\\)\n(vi) P = I\\(^2R\\) = (0.49)\\(^{2}\\) x 100\n= 23.795 W = 24W\n\n\n\n\n', '(a)\n\n~(b) The sensitivity is increased by a weak spring and increasing the strength of the magnet C\n(c)(i)\n\n(ii) X = \\(\\frac{1}{2 \\pi FC} = \\frac{1}{2 \\pi \\frac{160}{\\pi} \\times 4}\\) = 781\\(\\Omega\\)\n(iii) X\\(_L\\) = 2\\(\\pi\\)fL = 2\\(pi\\) x \\(\\frac{160}{\\pi}\\) x 3 = 960\\(\\Omega\\)\n(iv) Z = \\(\\sqrt{R^2 + (X_L - X_c)^2} = \\sqrt{100^2 + (960 - 781)^2}\\) = 205\\(\\Omega\\)\n(v) I = \\(\\frac{v}{z} = \\frac{100}{205} = 0.49A\\)\n(vi) P = I\\(^2R\\) = (0.49)\\(^{2}\\) x 100\n= 23.795 W = 24W\n\n\n\n~(c)(i)\n\n(ii) X = \\(\\frac{1}{2 \\pi FC} = \\frac{1}{2 \\pi \\frac{160}{\\pi} \\times 4}\\) = 781\\(\\Omega\\)\n(iii) X\\(_L\\) = 2\\(\\pi\\)fL = 2\\(pi\\) x \\(\\frac{160}{\\pi}\\) x 3 = 960\\(\\Omega\\)\n(iv) Z = \\(\\sqrt{R^2 + (X_L - X_c)^2} = \\sqrt{100^2 + (960 - 781)^2}\\) = 205\\(\\Omega\\)\n(v) I = \\(\\frac{v}{z} = \\frac{100}{205} = 0.49A\\)\n(vi) P = I\\(^2R\\) = (0.49)\\(^{2}\\) x 100\n= 23.795 W = 24W\n\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
(268, '2019', '\n\nQuestion\n\n(a) Explain: (i) Fusion. (ii) Fission.\n(b) State three advantages of fusion over fission in the generation of power\n(c) Calculate in joules, the binding energy for\n\\(\\frac{59}{27}\\) Co\n(Atomic mass of \\(^{59}_{2} Co\\) = 58.9332 u)\n(Mass of proton = 1.00783u)\n(Mass of neutron = 1.00867 u)\n(Unified atomic mass unit, U = 931 MeV)\n(1 eV = 1.6 x 10\\(^{-19}\\) J)\n\n', '', '(a) Explain: (i) Fusion. (ii) Fission.\n(b) State three advantages of fusion over fission in the generation of power\n(c) Calculate in joules, the binding energy for\n\\(\\frac{59}{27}\\) Co\n(Atomic mass of \\(^{59}_{2} Co\\) = 58.9332 u)\n(Mass of proton = 1.00783u)\n(Mass of neutron = 1.00867 u)\n(Unified atomic mass unit, U = 931 MeV)\n(1 eV = 1.6 x 10\\(^{-19}\\) J)', '(a) Explain: (i) Fusion. (ii) Fission.~(b) State three advantages of fusion over fission in the generation of power\n(c) Calculate in joules, the binding energy for\n\\(\\frac{59}{27}\\) Co\n(Atomic mass of \\(^{59}_{2} Co\\) = 58.9332 u)\n(Mass of proton = 1.00783u)\n(Mass of neutron = 1.00867 u)\n(Unified atomic mass unit, U = 931 MeV)\n(1 eV = 1.6 x 10\\(^{-19}\\) J)~(c) Calculate in joules, the binding energy for\n\\(\\frac{59}{27}\\) Co\n(Atomic mass of \\(^{59}_{2} Co\\) = 58.9332 u)\n(Mass of proton = 1.00783u)\n(Mass of neutron = 1.00867 u)\n(Unified atomic mass unit, U = 931 MeV)\n(1 eV = 1.6 x 10\\(^{-19}\\) J)', NULL, '\n\nExplanation\n(a)(i) Fusion is the coming together of two light nuclei to form a heavier nucleus with the release of large amount of energy\n\\(^2_1H + ^3_1H \\to ^4_2He\\) + energy\n(ii) Fission is the breaking up of heavy nucleus into two nuclei of comparable masses when bombarded with neutrons releasing energy and neutrons in the process.\n\\(^{235}_{92}U + ^1_0n \\to ^{90}_{36}kr + ^{144}_{56}Ba + 2^1_0H + energy)\\)\n(b) - Binding materials are cheap and easy to come by\n- it produces less dangerous products.\n- There is no e mass of hydrogen that can be obtained\n(c) Bonding energy = Mass of (Protons + neutron) - atomis mass = 27(1.00783u) + 32(1.00867u) - 58.93332 u\n= 27.21121u + 32.27744u - 58.9332 u\n= 59.48865 u - 58.9332 u\n= 0.55545 u\n= 0.55545 x 931 MeV = 517.12 x 10\\(^6\\) eV\n= 517.12 x 10\\(^6\\) x 1.6 x 10\\(^{19}\\) J\n= 8.27 x 10\\(^{-11}\\)J.\n\n', '', 'Explanation\n(a)(i) Fusion is the coming together of two light nuclei to form a heavier nucleus with the release of large amount of energy\n\\(^2_1H + ^3_1H \\to ^4_2He\\) + energy\n(ii) Fission is the breaking up of heavy nucleus into two nuclei of comparable masses when bombarded with neutrons releasing energy and neutrons in the process.\n\\(^{235}_{92}U + ^1_0n \\to ^{90}_{36}kr + ^{144}_{56}Ba + 2^1_0H + energy)\\)\n(b) - Binding materials are cheap and easy to come by\n- it produces less dangerous products.\n- There is no e mass of hydrogen that can be obtained\n(c) Bonding energy = Mass of (Protons + neutron) - atomis mass = 27(1.00783u) + 32(1.00867u) - 58.93332 u\n= 27.21121u + 32.27744u - 58.9332 u\n= 59.48865 u - 58.9332 u\n= 0.55545 u\n= 0.55545 x 931 MeV = 517.12 x 10\\(^6\\) eV\n= 517.12 x 10\\(^6\\) x 1.6 x 10\\(^{19}\\) J\n= 8.27 x 10\\(^{-11}\\)J.\n\n', '(a)(i) Fusion is the coming together of two light nuclei to form a heavier nucleus with the release of large amount of energy\n\\(^2_1H + ^3_1H \\to ^4_2He\\) + energy\n(ii) Fission is the breaking up of heavy nucleus into two nuclei of comparable masses when bombarded with neutrons releasing energy and neutrons in the process.\n\\(^{235}_{92}U + ^1_0n \\to ^{90}_{36}kr + ^{144}_{56}Ba + 2^1_0H + energy)\\)~(b) - Binding materials are cheap and easy to come by\n- it produces less dangerous products.\n- There is no e mass of hydrogen that can be obtained\n(c) Bonding energy = Mass of (Protons + neutron) - atomis mass = 27(1.00783u) + 32(1.00867u) - 58.93332 u\n= 27.21121u + 32.27744u - 58.9332 u\n= 59.48865 u - 58.9332 u\n= 0.55545 u\n= 0.55545 x 931 MeV = 517.12 x 10\\(^6\\) eV\n= 517.12 x 10\\(^6\\) x 1.6 x 10\\(^{19}\\) J\n= 8.27 x 10\\(^{-11}\\)J.\n~(c) Bonding energy = Mass of (Protons + neutron) - atomis mass = 27(1.00783u) + 32(1.00867u) - 58.93332 u\n= 27.21121u + 32.27744u - 58.9332 u\n= 59.48865 u - 58.9332 u\n= 0.55545 u\n= 0.55545 x 931 MeV = 517.12 x 10\\(^6\\) eV\n= 517.12 x 10\\(^6\\) x 1.6 x 10\\(^{19}\\) J\n= 8.27 x 10\\(^{-11}\\)J.\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(269, '2019', '\n\nQuestion\n\n(a) Explain the term uniform acceleration\n(b)(i) Sketch and describe the velocity-time graph for the motion of a ball from the time it is projected vertically upwards until it returns to the point of projection.\n(ii) Neglecting air resistance and using ycur sketch, explain how the acceleration of free fall due to gravity g, and the maximum height attained when the ball is projected vertically upwards can be determined.\n(c) A stone is projected vertically upwards with a velocity of 20ms\\(^{-1}\\). Two seconds later, a second stone is similarly projected with the same velocity. When the two stones meet, the second one is rising at a velocity of 10ms\\(^{-1}\\). Neglecting air resistance, calculate the:\n(i) length of time the second stone is in motion before they meet,\n(ii) velocity of the first stone when they meet (Take g as 10ms\\(^{-2}\\))\n\n\n', '', '(a) Explain the term uniform acceleration\n(b)(i) Sketch and describe the velocity-time graph for the motion of a ball from the time it is projected vertically upwards until it returns to the point of projection.\n(ii) Neglecting air resistance and using ycur sketch, explain how the acceleration of free fall due to gravity g, and the maximum height attained when the ball is projected vertically upwards can be determined.\n(c) A stone is projected vertically upwards with a velocity of 20ms\\(^{-1}\\). Two seconds later, a second stone is similarly projected with the same velocity. When the two stones meet, the second one is rising at a velocity of 10ms\\(^{-1}\\). Neglecting air resistance, calculate the:\n(i) length of time the second stone is in motion before they meet,\n(ii) velocity of the first stone when they meet (Take g as 10ms\\(^{-2}\\))\n', '(a) Explain the term uniform acceleration~(b)(i) Sketch and describe the velocity-time graph for the motion of a ball from the time it is projected vertically upwards until it returns to the point of projection.\n(ii) Neglecting air resistance and using ycur sketch, explain how the acceleration of free fall due to gravity g, and the maximum height attained when the ball is projected vertically upwards can be determined.\n(c) A stone is projected vertically upwards with a velocity of 20ms\\(^{-1}\\). Two seconds later, a second stone is similarly projected with the same velocity. When the two stones meet, the second one is rising at a velocity of 10ms\\(^{-1}\\). Neglecting air resistance, calculate the:\n(i) length of time the second stone is in motion before they meet,\n(ii) velocity of the first stone when they meet (Take g as 10ms\\(^{-2}\\))\n~(c) A stone is projected vertically upwards with a velocity of 20ms\\(^{-1}\\). Two seconds later, a second stone is similarly projected with the same velocity. When the two stones meet, the second one is rising at a velocity of 10ms\\(^{-1}\\). Neglecting air resistance, calculate the:\n(i) length of time the second stone is in motion before they meet,\n(ii) velocity of the first stone when they meet (Take g as 10ms\\(^{-2}\\))\n', NULL, '\n\nExplanation\n(a) A body is said to be moving with uniform acceleration when its time rate of increase of velocity is constant. It is measured in m/s\\(^2\\). It is a vector quantity.\n\nThe above sketch shows the v-t graph of the ball. The velocity decreases uniformly from A to zero (p) at the top of K its flight in a time kp. P , Va It then increases in velocity uniformly as it returns to the point of projection A.\n(ii) The magnitude of the slope of the line AP = g or Ap\\(^-\\) = g, where g = acceleration of free fall due to gravity. The height attained when the ball is thrown up is equal to the area of the triangle AKP. The distance it falls from its maximum height is equal to the area of the triangle A\\(^-\\)PC which is equal to triangle AKP.\n(c)(i) The second stone moves upwards and its acceleration g = 10m/s\\(^2\\). From equation v = u - gt\n10 = 20 - 10t -10 = -10t\nt = 1\nThe second stone has been in air for 1s before they met.\n(ii) The first stone has been moving for (1 + 2)s = 3s\nFrom equation v = u + gt\n= 2 - (10 x 3) = 20 - 30 = -10m/s\nThe minus sign means the first stone is moving down and its velocity is 10m/s.\n\n\n', '', 'Explanation\n(a) A body is said to be moving with uniform acceleration when its time rate of increase of velocity is constant. It is measured in m/s\\(^2\\). It is a vector quantity.\n\nThe above sketch shows the v-t graph of the ball. The velocity decreases uniformly from A to zero (p) at the top of K its flight in a time kp. P , Va It then increases in velocity uniformly as it returns to the point of projection A.\n(ii) The magnitude of the slope of the line AP = g or Ap\\(^-\\) = g, where g = acceleration of free fall due to gravity. The height attained when the ball is thrown up is equal to the area of the triangle AKP. The distance it falls from its maximum height is equal to the area of the triangle A\\(^-\\)PC which is equal to triangle AKP.\n(c)(i) The second stone moves upwards and its acceleration g = 10m/s\\(^2\\). From equation v = u - gt\n10 = 20 - 10t -10 = -10t\nt = 1\nThe second stone has been in air for 1s before they met.\n(ii) The first stone has been moving for (1 + 2)s = 3s\nFrom equation v = u + gt\n= 2 - (10 x 3) = 20 - 30 = -10m/s\nThe minus sign means the first stone is moving down and its velocity is 10m/s.\n\n\n', '~(c)(i) The second stone moves upwards and its acceleration g = 10m/s\\(^2\\). From equation v = u - gt\n10 = 20 - 10t -10 = -10t\nt = 1\nThe second stone has been in air for 1s before they met.\n(ii) The first stone has been moving for (1 + 2)s = 3s\nFrom equation v = u + gt\n= 2 - (10 x 3) = 20 - 30 = -10m/s\nThe minus sign means the first stone is moving down and its velocity is 10m/s.\n\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(270, '2019', '\n\nQuestion\n\n(a) Distinguish between heat and temperature.\n(b) State two physical properties of substances which may be used to measure temperature.\n(c) State two reasons why mercury is preferred to alcohol as a thermometric liquid.\n(d)(i) Describe how a mercury-in-glass thermometer is calibrated.\n(ii) State two precautions necessary to ensure an accurate result.\n(e) Explain how land and sea breezes occur.\n\n', '', '(a) Distinguish between heat and temperature.\n(b) State two physical properties of substances which may be used to measure temperature.\n(c) State two reasons why mercury is preferred to alcohol as a thermometric liquid.\n(d)(i) Describe how a mercury-in-glass thermometer is calibrated.\n(ii) State two precautions necessary to ensure an accurate result.\n(e) Explain how land and sea breezes occur.', '(a) Distinguish between heat and temperature.~(b) State two physical properties of substances which may be used to measure temperature.\n(c) State two reasons why mercury is preferred to alcohol as a thermometric liquid.\n(d)(i) Describe how a mercury-in-glass thermometer is calibrated.\n(ii) State two precautions necessary to ensure an accurate result.\n(e) Explain how land and sea breezes occur.~(c) State two reasons why mercury is preferred to alcohol as a thermometric liquid.\n(d)(i) Describe how a mercury-in-glass thermometer is calibrated.\n(ii) State two precautions necessary to ensure an accurate result.\n(e) Explain how land and sea breezes occur.~(d)(i) Describe how a mercury-in-glass thermometer is calibrated.\n(ii) State two precautions necessary to ensure an accurate result.\n(e) Explain how land and sea breezes occur.~(e) Explain how land and sea breezes occur.', NULL, '\n\nExplanation\n(a) Heat is a form of energy due to temperature gradient while temperature is the degree of hot-ness or cold-ness of a body.\n(b)(i) The change in volume expansion of a fixed mass of a liquid.\n(ii) The change in volume of a gas at constant pressure.\n(iii) The change in electrical resistance of a substance.\n(iv) The change in e.m.f generated in a thermo-couple. (Note: Any correct two)\n\n(c)(i) It does not wet glass (unlike alcohol) (ii) It is opaque and easily seen (unlike alcohol) (iii) It is a\nbetter conductor of heat. (iv) It measures higher temperature than alcohol. (Note: Any correct two)\n(d) Support the thermometer vertically ensuring the bulb is covered with melting pure ice shavings. Allow the mercury thread to fall till its level remains steady. Mark this position on the thermometer as the lower fixed point. Suspend the thermometer in the steam of a hypsometer. Allow the mercury thread to rise until its level remains steady. Mark this position on the thermometer as the steam point. Sub-divide the fundamental interval into a number of equal parts. Precautions\n(i) The thermometer bulb should not be placed in the boiling water of the hypsometer.\n(ii) The thermometer should be inserted in the hypsometer such that the mercury thread is visible just above the top of the apparatus.\n(iii) Some few minutes should be allowed to pass when the mercury thread is steady before making the thermometer.\n(iv) One should ensure that the steam is at atmospheric pressure. (Note: Any correct two)\n(e) The breezes are due to the fact that the specific heat capacity of water is greater than that of land. During the day the land is hotter than the sea. A low pressure area is created on the land and cooler air from the sea blows into the land. This wind is called sea breeze. At night the sea is warmer than the land. A low pressure area is created over the sea, resulting in cooler air blowing from the land where the pressure is higher into the sea. This is called land breeze.\n\n', '', 'Explanation\n(a) Heat is a form of energy due to temperature gradient while temperature is the degree of hot-ness or cold-ness of a body.\n(b)(i) The change in volume expansion of a fixed mass of a liquid.\n(ii) The change in volume of a gas at constant pressure.\n(iii) The change in electrical resistance of a substance.\n(iv) The change in e.m.f generated in a thermo-couple. (Note: Any correct two)\n\n(c)(i) It does not wet glass (unlike alcohol) (ii) It is opaque and easily seen (unlike alcohol) (iii) It is a\nbetter conductor of heat. (iv) It measures higher temperature than alcohol. (Note: Any correct two)\n(d) Support the thermometer vertically ensuring the bulb is covered with melting pure ice shavings. Allow the mercury thread to fall till its level remains steady. Mark this position on the thermometer as the lower fixed point. Suspend the thermometer in the steam of a hypsometer. Allow the mercury thread to rise until its level remains steady. Mark this position on the thermometer as the steam point. Sub-divide the fundamental interval into a number of equal parts. Precautions\n(i) The thermometer bulb should not be placed in the boiling water of the hypsometer.\n(ii) The thermometer should be inserted in the hypsometer such that the mercury thread is visible just above the top of the apparatus.\n(iii) Some few minutes should be allowed to pass when the mercury thread is steady before making the thermometer.\n(iv) One should ensure that the steam is at atmospheric pressure. (Note: Any correct two)\n(e) The breezes are due to the fact that the specific heat capacity of water is greater than that of land. During the day the land is hotter than the sea. A low pressure area is created on the land and cooler air from the sea blows into the land. This wind is called sea breeze. At night the sea is warmer than the land. A low pressure area is created over the sea, resulting in cooler air blowing from the land where the pressure is higher into the sea. This is called land breeze.\n\n', '(a) Heat is a form of energy due to temperature gradient while temperature is the degree of hot-ness or cold-ness of a body.~(b)(i) The change in volume expansion of a fixed mass of a liquid.\n(ii) The change in volume of a gas at constant pressure.\n(iii) The change in electrical resistance of a substance.\n(iv) The change in e.m.f generated in a thermo-couple. (Note: Any correct two)\n\n(c)(i) It does not wet glass (unlike alcohol) (ii) It is opaque and easily seen (unlike alcohol) (iii) It is a\nbetter conductor of heat. (iv) It measures higher temperature than alcohol. (Note: Any correct two)\n(d) Support the thermometer vertically ensuring the bulb is covered with melting pure ice shavings. Allow the mercury thread to fall till its level remains steady. Mark this position on the thermometer as the lower fixed point. Suspend the thermometer in the steam of a hypsometer. Allow the mercury thread to rise until its level remains steady. Mark this position on the thermometer as the steam point. Sub-divide the fundamental interval into a number of equal parts. Precautions\n(i) The thermometer bulb should not be placed in the boiling water of the hypsometer.\n(ii) The thermometer should be inserted in the hypsometer such that the mercury thread is visible just above the top of the apparatus.\n(iii) Some few minutes should be allowed to pass when the mercury thread is steady before making the thermometer.\n(iv) One should ensure that the steam is at atmospheric pressure. (Note: Any correct two)\n(e) The breezes are due to the fact that the specific heat capacity of water is greater than that of land. During the day the land is hotter than the sea. A low pressure area is created on the land and cooler air from the sea blows into the land. This wind is called sea breeze. At night the sea is warmer than the land. A low pressure area is created over the sea, resulting in cooler air blowing from the land where the pressure is higher into the sea. This is called land breeze.\n~(c)(i) It does not wet glass (unlike alcohol) (ii) It is opaque and easily seen (unlike alcohol) (iii) It is a\nbetter conductor of heat. (iv) It measures higher temperature than alcohol. (Note: Any correct two)\n(d) Support the thermometer vertically ensuring the bulb is covered with melting pure ice shavings. Allow the mercury thread to fall till its level remains steady. Mark this position on the thermometer as the lower fixed point. Suspend the thermometer in the steam of a hypsometer. Allow the mercury thread to rise until its level remains steady. Mark this position on the thermometer as the steam point. Sub-divide the fundamental interval into a number of equal parts. Precautions\n(i) The thermometer bulb should not be placed in the boiling water of the hypsometer.\n(ii) The thermometer should be inserted in the hypsometer such that the mercury thread is visible just above the top of the apparatus.\n(iii) Some few minutes should be allowed to pass when the mercury thread is steady before making the thermometer.\n(iv) One should ensure that the steam is at atmospheric pressure. (Note: Any correct two)\n(e) The breezes are due to the fact that the specific heat capacity of water is greater than that of land. During the day the land is hotter than the sea. A low pressure area is created on the land and cooler air from the sea blows into the land. This wind is called sea breeze. At night the sea is warmer than the land. A low pressure area is created over the sea, resulting in cooler air blowing from the land where the pressure is higher into the sea. This is called land breeze.\n~(d) Support the thermometer vertically ensuring the bulb is covered with melting pure ice shavings. Allow the mercury thread to fall till its level remains steady. Mark this position on the thermometer as the lower fixed point. Suspend the thermometer in the steam of a hypsometer. Allow the mercury thread to rise until its level remains steady. Mark this position on the thermometer as the steam point. Sub-divide the fundamental interval into a number of equal parts. Precautions\n(i) The thermometer bulb should not be placed in the boiling water of the hypsometer.\n(ii) The thermometer should be inserted in the hypsometer such that the mercury thread is visible just above the top of the apparatus.\n(iii) Some few minutes should be allowed to pass when the mercury thread is steady before making the thermometer.\n(iv) One should ensure that the steam is at atmospheric pressure. (Note: Any correct two)\n(e) The breezes are due to the fact that the specific heat capacity of water is greater than that of land. During the day the land is hotter than the sea. A low pressure area is created on the land and cooler air from the sea blows into the land. This wind is called sea breeze. At night the sea is warmer than the land. A low pressure area is created over the sea, resulting in cooler air blowing from the land where the pressure is higher into the sea. This is called land breeze.\n~(e) The breezes are due to the fact that the specific heat capacity of water is greater than that of land. During the day the land is hotter than the sea. A low pressure area is created on the land and cooler air from the sea blows into the land. This wind is called sea breeze. At night the sea is warmer than the land. A low pressure area is created over the sea, resulting in cooler air blowing from the land where the pressure is higher into the sea. This is called land breeze.\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(271, '2019', '\n\nQuestion\n\n(a)(i) Describe, with the aid of a circuit diagram, an experiment to measure the resistance of a wire given an ammeter of low resistance, a battery, a key, a rheostat, va high-resistance voltmeter and some connecting wires.\n(ii) State two precautions necessary to obtain an accurate result.\n(b) Using the experimental result and any necessary measurements, explain how the resistivity of the wire may be determined.\n(c) Two cells, each of e.m.f. 2V and internal resistance 0.552, are connected in series. They are made to supply current to a combination of three resistors, one of resistance 20 connected in series to a parallel combination of two other resistors each of resistance 3Q. Draw the circuit diagram and calculate the:\n(i) current in the circuit\n(ii) potential difference across the parallel combination of the resistors\n(iii) lost volts of the battery.\n\n', '', '(a)(i) Describe, with the aid of a circuit diagram, an experiment to measure the resistance of a wire given an ammeter of low resistance, a battery, a key, a rheostat, va high-resistance voltmeter and some connecting wires.\n(ii) State two precautions necessary to obtain an accurate result.\n(b) Using the experimental result and any necessary measurements, explain how the resistivity of the wire may be determined.\n(c) Two cells, each of e.m.f. 2V and internal resistance 0.552, are connected in series. They are made to supply current to a combination of three resistors, one of resistance 20 connected in series to a parallel combination of two other resistors each of resistance 3Q. Draw the circuit diagram and calculate the:\n(i) current in the circuit\n(ii) potential difference across the parallel combination of the resistors\n(iii) lost volts of the battery.', '(a)(i) Describe, with the aid of a circuit diagram, an experiment to measure the resistance of a wire given an ammeter of low resistance, a battery, a key, a rheostat, va high-resistance voltmeter and some connecting wires.\n(ii) State two precautions necessary to obtain an accurate result.~(b) Using the experimental result and any necessary measurements, explain how the resistivity of the wire may be determined.\n(c) Two cells, each of e.m.f. 2V and internal resistance 0.552, are connected in series. They are made to supply current to a combination of three resistors, one of resistance 20 connected in series to a parallel combination of two other resistors each of resistance 3Q. Draw the circuit diagram and calculate the:\n(i) current in the circuit\n(ii) potential difference across the parallel combination of the resistors\n(iii) lost volts of the battery.~(c) Two cells, each of e.m.f. 2V and internal resistance 0.552, are connected in series. They are made to supply current to a combination of three resistors, one of resistance 20 connected in series to a parallel combination of two other resistors each of resistance 3Q. Draw the circuit diagram and calculate the:\n(i) current in the circuit\n(ii) potential difference across the parallel combination of the resistors\n(iii) lost volts of the battery.', NULL, '\n\nExplanation\n(a)\n\n\nDescription: The circuit is set up as shown above. The key (k) is closed. The rheostat (R) is adjusted so that lowest current is flowing. Ammeter and voltmeter readings (I and V) are recorded. Adjust rheostat to increase current. Ammeter and voltmeter readings are recorded for other positions of the rheostat. Values of (are plotted against values of I. Slope, which is the value of the resistance is determine.\nPrecautions;\ni) Kcy should be removed when readings are not taken\nii) Connections should be tight\n(iii) Parallax\' error when reading ammeter and voltmeter should be avoided\n(iv) Only low current should be used\nv) Zero error of the meters should be avoided. Note: Any correct two) Resistance, R = P\'/A where = length of wire, A = cross sectional area, p = resistivity )f wire. But slope, S of graph = R. Hence S = P\'/A\nMeasure I with a metre rule. Measure diameter with a micrometer screw gauge and determine A. Substitute p = \\(\\frac{SA}{I}\\)\n\nc) Diagram\n\ni) Total e.m.f. = 4V\nTotal resistance = 1 + 2 + \\(\\frac{3 \\times 3}{ 2 \\times 3}\\)\n= 3 + 1\\(\\frac{1}{2}\\)\n= \\(\\frac{1}{2} \\Omega\\)\nE = IR\\(_T\\),\n= E/R\n:.Current, I = 4/9 x 2/1\n= 8/9;\nA = 0.9A\n\n(ii) P.d = IR = 8/9 x 3/2 = 4/3 = 1.3V\n(iii) Lost volts = Ir = \\(\\frac{8}{9}\\) x 1 = \\(\\frac{8}{9} = 0.9V\\)\n\n\n', '', 'Explanation\n(a)\n\n\nDescription: The circuit is set up as shown above. The key (k) is closed. The rheostat (R) is adjusted so that lowest current is flowing. Ammeter and voltmeter readings (I and V) are recorded. Adjust rheostat to increase current. Ammeter and voltmeter readings are recorded for other positions of the rheostat. Values of (are plotted against values of I. Slope, which is the value of the resistance is determine.\nPrecautions;\ni) Kcy should be removed when readings are not taken\nii) Connections should be tight\n(iii) Parallax\' error when reading ammeter and voltmeter should be avoided\n(iv) Only low current should be used\nv) Zero error of the meters should be avoided. Note: Any correct two) Resistance, R = P\'/A where = length of wire, A = cross sectional area, p = resistivity )f wire. But slope, S of graph = R. Hence S = P\'/A\nMeasure I with a metre rule. Measure diameter with a micrometer screw gauge and determine A. Substitute p = \\(\\frac{SA}{I}\\)\n\nc) Diagram\n\ni) Total e.m.f. = 4V\nTotal resistance = 1 + 2 + \\(\\frac{3 \\times 3}{ 2 \\times 3}\\)\n= 3 + 1\\(\\frac{1}{2}\\)\n= \\(\\frac{1}{2} \\Omega\\)\nE = IR\\(_T\\),\n= E/R\n:.Current, I = 4/9 x 2/1\n= 8/9;\nA = 0.9A\n\n(ii) P.d = IR = 8/9 x 3/2 = 4/3 = 1.3V\n(iii) Lost volts = Ir = \\(\\frac{8}{9}\\) x 1 = \\(\\frac{8}{9} = 0.9V\\)\n\n\n', '~c) Diagram\n\ni) Total e.m.f. = 4V\nTotal resistance = 1 + 2 + \\(\\frac{3 \\times 3}{ 2 \\times 3}\\)\n= 3 + 1\\(\\frac{1}{2}\\)\n= \\(\\frac{1}{2} \\Omega\\)\nE = IR\\(_T\\),\n= E/R\n:.Current, I = 4/9 x 2/1\n= 8/9;\nA = 0.9A\n\n(ii) P.d = IR = 8/9 x 3/2 = 4/3 = 1.3V\n(iii) Lost volts = Ir = \\(\\frac{8}{9}\\) x 1 = \\(\\frac{8}{9} = 0.9V\\)\n\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(272, '2019', '\n\nQuestion\n\n(a) Explain the term photoelectric effect.\n(b)\n\n\nThe diagram above represents a photocell with its associated electric circuit. Identify each of the physical quantities\nrepresented by the letters A, B, C, D, E and F\n(c) What factor determines the: (i) current produced by the photocell\n(ii) maximum kinetic energy of the photoelectrons?\n(d) State one similarity and one difference between photoemission and evaporation\n(e) Name two methods by which a beam of free electrons may be produced other than photoemission\n(f) State two applications of photoelectric effect.\n(g) A light wavelength 5.0 x 10\\(^{-7}\\) m is incident on metal resulting in photoemission of electrons. If the work function of the metal is 3.04 x 10\\(^{-19}\\)J, calculate the:\n(i) frequency of the light\n(ii) energy of the incident photon,\n(iii) maximum kinetic energy of the photoelectrons (Speed of light = 3.00 x 108ms\\(^{-1}\\); Planck\'s constant = 6.6 x 10\\(^{-34}\\)Js).\n\n\n(a) Explain the term photoelectric effect.\n(b)\n\n', '', '(a) Explain the term photoelectric effect.\n(b)\n\n\nThe diagram above represents a photocell with its associated electric circuit. Identify each of the physical quantities\nrepresented by the letters A, B, C, D, E and F\n(c) What factor determines the: (i) current produced by the photocell\n(ii) maximum kinetic energy of the photoelectrons?\n(d) State one similarity and one difference between photoemission and evaporation\n(e) Name two methods by which a beam of free electrons may be produced other than photoemission\n(f) State two applications of photoelectric effect.\n(g) A light wavelength 5.0 x 10\\(^{-7}\\) m is incident on metal resulting in photoemission of electrons. If the work function of the metal is 3.04 x 10\\(^{-19}\\)J, calculate the:\n(i) frequency of the light\n(ii) energy of the incident photon,\n(iii) maximum kinetic energy of the photoelectrons (Speed of light = 3.00 x 108ms\\(^{-1}\\); Planck\'s constant = 6.6 x 10\\(^{-34}\\)Js).\n\n\n(a) Explain the term photoelectric effect.\n(b)', '(a) Explain the term photoelectric effect.\n(b)\n\n\nThe diagram above represents a photocell with its associated electric circuit. Identify each of the physical quantities\nrepresented by the letters A, B, C, D, E and F\n(c) What factor determines the: (i) current produced by the photocell\n(ii) maximum kinetic energy of the photoelectrons?\n(d) State one similarity and one difference between photoemission and evaporation\n(e) Name two methods by which a beam of free electrons may be produced other than photoemission\n(f) State two applications of photoelectric effect.\n(g) A light wavelength 5.0 x 10\\(^{-7}\\) m is incident on metal resulting in photoemission of electrons. If the work function of the metal is 3.04 x 10\\(^{-19}\\)J, calculate the:\n(i) frequency of the light\n(ii) energy of the incident photon,\n(iii) maximum kinetic energy of the photoelectrons (Speed of light = 3.00 x 108ms\\(^{-1}\\); Planck\'s constant = 6.6 x 10\\(^{-34}\\)Js).\n\n\n(a) Explain the term photoelectric effect.~(b)\n\n\nThe diagram above represents a photocell with its associated electric circuit. Identify each of the physical quantities\nrepresented by the letters A, B, C, D, E and F\n(c) What factor determines the: (i) current produced by the photocell\n(ii) maximum kinetic energy of the photoelectrons?\n(d) State one similarity and one difference between photoemission and evaporation\n(e) Name two methods by which a beam of free electrons may be produced other than photoemission\n(f) State two applications of photoelectric effect.\n(g) A light wavelength 5.0 x 10\\(^{-7}\\) m is incident on metal resulting in photoemission of electrons. If the work function of the metal is 3.04 x 10\\(^{-19}\\)J, calculate the:\n(i) frequency of the light\n(ii) energy of the incident photon,\n(iii) maximum kinetic energy of the photoelectrons (Speed of light = 3.00 x 108ms\\(^{-1}\\); Planck\'s constant = 6.6 x 10\\(^{-34}\\)Js).\n\n\n(a) Explain the term photoelectric effect.\n(b)~(c) What factor determines the: (i) current produced by the photocell\n(ii) maximum kinetic energy of the photoelectrons?\n(d) State one similarity and one difference between photoemission and evaporation\n(e) Name two methods by which a beam of free electrons may be produced other than photoemission\n(f) State two applications of photoelectric effect.\n(g) A light wavelength 5.0 x 10\\(^{-7}\\) m is incident on metal resulting in photoemission of electrons. If the work function of the metal is 3.04 x 10\\(^{-19}\\)J, calculate the:\n(i) frequency of the light\n(ii) energy of the incident photon,\n(iii) maximum kinetic energy of the photoelectrons (Speed of light = 3.00 x 108ms\\(^{-1}\\); Planck\'s constant = 6.6 x 10\\(^{-34}\\)Js).\n\n\n(a) Explain the term photoelectric effect.\n(b)~(d) State one similarity and one difference between photoemission and evaporation\n(e) Name two methods by which a beam of free electrons may be produced other than photoemission\n(f) State two applications of photoelectric effect.\n(g) A light wavelength 5.0 x 10\\(^{-7}\\) m is incident on metal resulting in photoemission of electrons. If the work function of the metal is 3.04 x 10\\(^{-19}\\)J, calculate the:\n(i) frequency of the light\n(ii) energy of the incident photon,\n(iii) maximum kinetic energy of the photoelectrons (Speed of light = 3.00 x 108ms\\(^{-1}\\); Planck\'s constant = 6.6 x 10\\(^{-34}\\)Js).\n\n\n(a) Explain the term photoelectric effect.\n(b)~(e) Name two methods by which a beam of free electrons may be produced other than photoemission\n(f) State two applications of photoelectric effect.\n(g) A light wavelength 5.0 x 10\\(^{-7}\\) m is incident on metal resulting in photoemission of electrons. If the work function of the metal is 3.04 x 10\\(^{-19}\\)J, calculate the:\n(i) frequency of the light\n(ii) energy of the incident photon,\n(iii) maximum kinetic energy of the photoelectrons (Speed of light = 3.00 x 108ms\\(^{-1}\\); Planck\'s constant = 6.6 x 10\\(^{-34}\\)Js).\n\n\n(a) Explain the term photoelectric effect.\n(b)~(f) State two applications of photoelectric effect.\n(g) A light wavelength 5.0 x 10\\(^{-7}\\) m is incident on metal resulting in photoemission of electrons. If the work function of the metal is 3.04 x 10\\(^{-19}\\)J, calculate the:\n(i) frequency of the light\n(ii) energy of the incident photon,\n(iii) maximum kinetic energy of the photoelectrons (Speed of light = 3.00 x 108ms\\(^{-1}\\); Planck\'s constant = 6.6 x 10\\(^{-34}\\)Js).\n\n\n(a) Explain the term photoelectric effect.\n(b)~(g) A light wavelength 5.0 x 10\\(^{-7}\\) m is incident on metal resulting in photoemission of electrons. If the work function of the metal is 3.04 x 10\\(^{-19}\\)J, calculate the:\n(i) frequency of the light\n(ii) energy of the incident photon,\n(iii) maximum kinetic energy of the photoelectrons (Speed of light = 3.00 x 108ms\\(^{-1}\\); Planck\'s constant = 6.6 x 10\\(^{-34}\\)Js).\n\n\n(a) Explain the term photoelectric effect.\n(b)', NULL, '\n\nExplanation\n(a) When EM radiation of appropriate frequency or wavelength is incident on a metal surface, it causes electrons to be emitted from the surface. This phenomenon is called photoelectric effect. It is a surface phenomenon. Emission is instantaneous and K.E. of emitted electrons depend on frequency or wavelength of incident photons.\n(b) A = Incident light or EM radiations\nB = Cathode\nC = Photoelectrons\nD = Anode\nE = Vacuum (Evacuated space)\nF = Glass tube\n\n(c)(i) Intensity of the incident radiation\n(ii) Energy or frequency of the incident radiation.\n\n(d) Similarities\n(i) In both, particles are emitted\n(ii) Both are surface phenomenon\n(iii) Both involves energy changes\nDifference\n(i) Evaporation involves molecules while photoemission involves electrons.\n(ii) Evaporation causes cooling while photoemission does not.\n\n(e)(i) Thermionic emission.\n(ii) Ionization of gases.\n(iii) Field emission\n\n(f) Photocells can be used.\n(i) To open or close doors automatically\n(ii) In automatic switching device.\n(iii) In television cameras\n(iv) As a burglar alarm\n(v) As safety devices in paper mills and related industries\n(vi) To produce sounds of varying intensities on film tracks.\n\n', '', 'Explanation\n(a) When EM radiation of appropriate frequency or wavelength is incident on a metal surface, it causes electrons to be emitted from the surface. This phenomenon is called photoelectric effect. It is a surface phenomenon. Emission is instantaneous and K.E. of emitted electrons depend on frequency or wavelength of incident photons.\n(b) A = Incident light or EM radiations\nB = Cathode\nC = Photoelectrons\nD = Anode\nE = Vacuum (Evacuated space)\nF = Glass tube\n\n(c)(i) Intensity of the incident radiation\n(ii) Energy or frequency of the incident radiation.\n\n(d) Similarities\n(i) In both, particles are emitted\n(ii) Both are surface phenomenon\n(iii) Both involves energy changes\nDifference\n(i) Evaporation involves molecules while photoemission involves electrons.\n(ii) Evaporation causes cooling while photoemission does not.\n\n(e)(i) Thermionic emission.\n(ii) Ionization of gases.\n(iii) Field emission\n\n(f) Photocells can be used.\n(i) To open or close doors automatically\n(ii) In automatic switching device.\n(iii) In television cameras\n(iv) As a burglar alarm\n(v) As safety devices in paper mills and related industries\n(vi) To produce sounds of varying intensities on film tracks.\n\n', '(a) When EM radiation of appropriate frequency or wavelength is incident on a metal surface, it causes electrons to be emitted from the surface. This phenomenon is called photoelectric effect. It is a surface phenomenon. Emission is instantaneous and K.E. of emitted electrons depend on frequency or wavelength of incident photons.~(b) A = Incident light or EM radiations\nB = Cathode\nC = Photoelectrons\nD = Anode\nE = Vacuum (Evacuated space)\nF = Glass tube\n\n(c)(i) Intensity of the incident radiation\n(ii) Energy or frequency of the incident radiation.\n\n(d) Similarities\n(i) In both, particles are emitted\n(ii) Both are surface phenomenon\n(iii) Both involves energy changes\nDifference\n(i) Evaporation involves molecules while photoemission involves electrons.\n(ii) Evaporation causes cooling while photoemission does not.\n\n(e)(i) Thermionic emission.\n(ii) Ionization of gases.\n(iii) Field emission\n\n(f) Photocells can be used.\n(i) To open or close doors automatically\n(ii) In automatic switching device.\n(iii) In television cameras\n(iv) As a burglar alarm\n(v) As safety devices in paper mills and related industries\n(vi) To produce sounds of varying intensities on film tracks.\n~(c)(i) Intensity of the incident radiation\n(ii) Energy or frequency of the incident radiation.\n\n(d) Similarities\n(i) In both, particles are emitted\n(ii) Both are surface phenomenon\n(iii) Both involves energy changes\nDifference\n(i) Evaporation involves molecules while photoemission involves electrons.\n(ii) Evaporation causes cooling while photoemission does not.\n\n(e)(i) Thermionic emission.\n(ii) Ionization of gases.\n(iii) Field emission\n\n(f) Photocells can be used.\n(i) To open or close doors automatically\n(ii) In automatic switching device.\n(iii) In television cameras\n(iv) As a burglar alarm\n(v) As safety devices in paper mills and related industries\n(vi) To produce sounds of varying intensities on film tracks.\n~(d) Similarities\n(i) In both, particles are emitted\n(ii) Both are surface phenomenon\n(iii) Both involves energy changes\nDifference\n(i) Evaporation involves molecules while photoemission involves electrons.\n(ii) Evaporation causes cooling while photoemission does not.\n\n(e)(i) Thermionic emission.\n(ii) Ionization of gases.\n(iii) Field emission\n\n(f) Photocells can be used.\n(i) To open or close doors automatically\n(ii) In automatic switching device.\n(iii) In television cameras\n(iv) As a burglar alarm\n(v) As safety devices in paper mills and related industries\n(vi) To produce sounds of varying intensities on film tracks.\n~(e)(i) Thermionic emission.\n(ii) Ionization of gases.\n(iii) Field emission\n\n(f) Photocells can be used.\n(i) To open or close doors automatically\n(ii) In automatic switching device.\n(iii) In television cameras\n(iv) As a burglar alarm\n(v) As safety devices in paper mills and related industries\n(vi) To produce sounds of varying intensities on film tracks.\n~(f) Photocells can be used.\n(i) To open or close doors automatically\n(ii) In automatic switching device.\n(iii) In television cameras\n(iv) As a burglar alarm\n(v) As safety devices in paper mills and related industries\n(vi) To produce sounds of varying intensities on film tracks.\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(273, '2019', '\n\nQuestion\n\n(a) Explain the term work\n(b) Draw a diagram of a pulley system with a velocity ratio of 5.\n(c) A man pulls up a box of mass 70kg using an inclined plane of effective length 5m onto a platform 2.5m high at uniform speed. If the frictional force between the box and the plane is 100N, draw the diagram of all the forces acting on the box when in motion and calculate the\n(i) minimum effort applied in pulling up the box\n(ii) velocity ratio of the plane\n(iii) mechanical advantage of the plane\n(iv) efficiency of the plane\n(v) energy lost in the system\n(vi) work output of the man\n(vii) total power developed by the man given that the time taken to raise the box onto the platform is 50s. (g = 10ms\\(^{-2}\\)).\n\n', '', '(a) Explain the term work\n(b) Draw a diagram of a pulley system with a velocity ratio of 5.\n(c) A man pulls up a box of mass 70kg using an inclined plane of effective length 5m onto a platform 2.5m high at uniform speed. If the frictional force between the box and the plane is 100N, draw the diagram of all the forces acting on the box when in motion and calculate the\n(i) minimum effort applied in pulling up the box\n(ii) velocity ratio of the plane\n(iii) mechanical advantage of the plane\n(iv) efficiency of the plane\n(v) energy lost in the system\n(vi) work output of the man\n(vii) total power developed by the man given that the time taken to raise the box onto the platform is 50s. (g = 10ms\\(^{-2}\\)).', '(a) Explain the term work~(b) Draw a diagram of a pulley system with a velocity ratio of 5.\n(c) A man pulls up a box of mass 70kg using an inclined plane of effective length 5m onto a platform 2.5m high at uniform speed. If the frictional force between the box and the plane is 100N, draw the diagram of all the forces acting on the box when in motion and calculate the\n(i) minimum effort applied in pulling up the box\n(ii) velocity ratio of the plane\n(iii) mechanical advantage of the plane\n(iv) efficiency of the plane\n(v) energy lost in the system\n(vi) work output of the man\n(vii) total power developed by the man given that the time taken to raise the box onto the platform is 50s. (g = 10ms\\(^{-2}\\)).~(c) A man pulls up a box of mass 70kg using an inclined plane of effective length 5m onto a platform 2.5m high at uniform speed. If the frictional force between the box and the plane is 100N, draw the diagram of all the forces acting on the box when in motion and calculate the\n(i) minimum effort applied in pulling up the box\n(ii) velocity ratio of the plane\n(iii) mechanical advantage of the plane\n(iv) efficiency of the plane\n(v) energy lost in the system\n(vi) work output of the man\n(vii) total power developed by the man given that the time taken to raise the box onto the platform is 50s. (g = 10ms\\(^{-2}\\)).', NULL, '\n\nExplanation\n(a) Work is done whenever a force moves a body through a certain distance in the direction of the force. It is measured in joule and it is a scalar quantity.\n\n(i) Minimum effort (E) applied in pulling up the box = F = mgsin\\(\\theta\\) = 100 + 700 sin30° = 100 + 700 x 0.5 = 100 + 350=450N.\n(ii) Velocity ratio (V.R) of the plane = 1/sin\\(\\theta\\) - 5/2.5 = 2\n(iii) Mechanical advantage of the plane (M.A) = L/E = 1.56 .\n(iv) Efficiency of the plane = \\(\\frac{M.A}{V.R}\\) x 100%\n= \\(\\frac{1.56}{2}\\) x 100% = 78%\n(v) Energy lost in the system = work input - work output = (450 x 5) - (700/2 x 5) = 2250 - 1750 = 550J.\n(vi) Work output of the man = mgh = 70 x 10 x 2.5 = 1750J\n(vii) Total power developed by the man\n= \\(\\frac{\\text{work done by effort}}{time}\\) = \\(\\frac{450 \\times 5}{50}\\) = 45 watts\n\n', '', 'Explanation\n(a) Work is done whenever a force moves a body through a certain distance in the direction of the force. It is measured in joule and it is a scalar quantity.\n\n(i) Minimum effort (E) applied in pulling up the box = F = mgsin\\(\\theta\\) = 100 + 700 sin30° = 100 + 700 x 0.5 = 100 + 350=450N.\n(ii) Velocity ratio (V.R) of the plane = 1/sin\\(\\theta\\) - 5/2.5 = 2\n(iii) Mechanical advantage of the plane (M.A) = L/E = 1.56 .\n(iv) Efficiency of the plane = \\(\\frac{M.A}{V.R}\\) x 100%\n= \\(\\frac{1.56}{2}\\) x 100% = 78%\n(v) Energy lost in the system = work input - work output = (450 x 5) - (700/2 x 5) = 2250 - 1750 = 550J.\n(vi) Work output of the man = mgh = 70 x 10 x 2.5 = 1750J\n(vii) Total power developed by the man\n= \\(\\frac{\\text{work done by effort}}{time}\\) = \\(\\frac{450 \\times 5}{50}\\) = 45 watts\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
(274, '2019', '\n\nQuestion\n\n(a) Explain the term resonance and give two examples\n(b)(i) Describe, with the aid of a labelled diagram, an experiment to show how the frequency of the note emitted by a vibrating string depends on the length of the string.\n(ii) State two precautions necessary to obtain an accurate result.\n(c) A sonometer wire is plucked and it vibrates emitting a fundamental note. State the effect on the frequency of the note if the\n(i) tension in the wire were made nine times as large with no change in the length of the wire;\n(ii) length of the wire were doubled with no change in the tension.\n\n', '', '(a) Explain the term resonance and give two examples\n(b)(i) Describe, with the aid of a labelled diagram, an experiment to show how the frequency of the note emitted by a vibrating string depends on the length of the string.\n(ii) State two precautions necessary to obtain an accurate result.\n(c) A sonometer wire is plucked and it vibrates emitting a fundamental note. State the effect on the frequency of the note if the\n(i) tension in the wire were made nine times as large with no change in the length of the wire;\n(ii) length of the wire were doubled with no change in the tension.', '(a) Explain the term resonance and give two examples~(b)(i) Describe, with the aid of a labelled diagram, an experiment to show how the frequency of the note emitted by a vibrating string depends on the length of the string.\n(ii) State two precautions necessary to obtain an accurate result.\n(c) A sonometer wire is plucked and it vibrates emitting a fundamental note. State the effect on the frequency of the note if the\n(i) tension in the wire were made nine times as large with no change in the length of the wire;\n(ii) length of the wire were doubled with no change in the tension.~(c) A sonometer wire is plucked and it vibrates emitting a fundamental note. State the effect on the frequency of the note if the\n(i) tension in the wire were made nine times as large with no change in the length of the wire;\n(ii) length of the wire were doubled with no change in the tension.', NULL, '\n\nExplanation\n(a) Resonance occurs whenever a particular body or system is set in oscillation at its own natural frequency as a result of impulses received from some other body or system which is vibrating with the same frequency. At resonance, the amplitude is maximum. Two examples of resonance are tuning a radio and divers board.\n\nProcedure: In the diagram above the string is kept taut by using a constant weight\n(w) between the two bridges A and B. The length, L between the bridges is adjusted until the wire resonates with the vibrating tuning fork. This happens when a paper rider flies off or when a loud sound is heard. Then, measure length, L of the resonating wire and frequency f. The experiment is repeated for at least four other tuning forks of known frequencies. In each case the values of the frequency, f and the corresponding length, L are recorded. A graph of f is plotted against 1/L or L against 1/f. A straight-line graph passing through origin is obtained showing that frequency is proportional to the reciprocal of the length.\n(iii) Precautions:\n(1) A smooth pulley should be used\n(2) the tuning fork should be struck on a rubber bung\n(c) For a fundamental note\nf\\(_o\\) = \\(\\frac{1}{20}\\) = \\(\\sqrt{\\frac{T}{m}}\\)\nwhere f\\(_o\\) = fundamental frequency\nL = Length of string\nT = Tension m = linear density or mass per unit length of the string\n(i) \\(f_1 = \\frac{1}{20} \\sqrt{\\frac{f}{m}}\\) = 3(\\(\\frac{1}{20} \\sqrt{\\frac{1}{20}} \\sqrt{\\frac{1}{m}})\\)\nf\\(_1 = 3f_o\\)\ni.e = the fundamental frequency, \\(f_o\\) is tripled\n(ii) \\(f_2 = \\frac{1}{20} (\\frac{1}{20} \\sqrt{\\frac{T}{m}}) = \\frac{1}{2} \\times \\frac{1}{20} \\sqrt{\\frac{1}{m}} = f_2 = \\frac{1}{2} f_o\\)\ni.e the fundamental frequency, \\(f_o\\) is halved\n\n', '', 'Explanation\n(a) Resonance occurs whenever a particular body or system is set in oscillation at its own natural frequency as a result of impulses received from some other body or system which is vibrating with the same frequency. At resonance, the amplitude is maximum. Two examples of resonance are tuning a radio and divers board.\n\nProcedure: In the diagram above the string is kept taut by using a constant weight\n(w) between the two bridges A and B. The length, L between the bridges is adjusted until the wire resonates with the vibrating tuning fork. This happens when a paper rider flies off or when a loud sound is heard. Then, measure length, L of the resonating wire and frequency f. The experiment is repeated for at least four other tuning forks of known frequencies. In each case the values of the frequency, f and the corresponding length, L are recorded. A graph of f is plotted against 1/L or L against 1/f. A straight-line graph passing through origin is obtained showing that frequency is proportional to the reciprocal of the length.\n(iii) Precautions:\n(1) A smooth pulley should be used\n(2) the tuning fork should be struck on a rubber bung\n(c) For a fundamental note\nf\\(_o\\) = \\(\\frac{1}{20}\\) = \\(\\sqrt{\\frac{T}{m}}\\)\nwhere f\\(_o\\) = fundamental frequency\nL = Length of string\nT = Tension m = linear density or mass per unit length of the string\n(i) \\(f_1 = \\frac{1}{20} \\sqrt{\\frac{f}{m}}\\) = 3(\\(\\frac{1}{20} \\sqrt{\\frac{1}{20}} \\sqrt{\\frac{1}{m}})\\)\nf\\(_1 = 3f_o\\)\ni.e = the fundamental frequency, \\(f_o\\) is tripled\n(ii) \\(f_2 = \\frac{1}{20} (\\frac{1}{20} \\sqrt{\\frac{T}{m}}) = \\frac{1}{2} \\times \\frac{1}{20} \\sqrt{\\frac{1}{m}} = f_2 = \\frac{1}{2} f_o\\)\ni.e the fundamental frequency, \\(f_o\\) is halved\n\n', '~(c) For a fundamental note\nf\\(_o\\) = \\(\\frac{1}{20}\\) = \\(\\sqrt{\\frac{T}{m}}\\)\nwhere f\\(_o\\) = fundamental frequency\nL = Length of string\nT = Tension m = linear density or mass per unit length of the string\n(i) \\(f_1 = \\frac{1}{20} \\sqrt{\\frac{f}{m}}\\) = 3(\\(\\frac{1}{20} \\sqrt{\\frac{1}{20}} \\sqrt{\\frac{1}{m}})\\)\nf\\(_1 = 3f_o\\)\ni.e = the fundamental frequency, \\(f_o\\) is tripled\n(ii) \\(f_2 = \\frac{1}{20} (\\frac{1}{20} \\sqrt{\\frac{T}{m}}) = \\frac{1}{2} \\times \\frac{1}{20} \\sqrt{\\frac{1}{m}} = f_2 = \\frac{1}{2} f_o\\)\ni.e the fundamental frequency, \\(f_o\\) is halved\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(275, '2019', '\n\nQuestion\n\n(a) Sketch the form of the magnetic flux pattern due to a current flowing\n(i) in a long solenoid\n(ii) through two long straight parallel wires when the directions of the current are opposite. (Neglect the earth\'s magnetic field).\n(b) Draw a labelled diagram of an electric bell and explain how it works\n(c) An electric bell takes a current of 0.2A groin a battery of two dry cells connected in series. Each cell has an e.m.f. of 1.5V and an internal resistance of 1.0\\(\\Omega\\).\n(i) Calculate the effective resistance of the bell\n(ii) What current would the bells take in the cells were arranged in parallel?\n\n', '', '(a) Sketch the form of the magnetic flux pattern due to a current flowing\n(i) in a long solenoid\n(ii) through two long straight parallel wires when the directions of the current are opposite. (Neglect the earth\'s magnetic field).\n(b) Draw a labelled diagram of an electric bell and explain how it works\n(c) An electric bell takes a current of 0.2A groin a battery of two dry cells connected in series. Each cell has an e.m.f. of 1.5V and an internal resistance of 1.0\\(\\Omega\\).\n(i) Calculate the effective resistance of the bell\n(ii) What current would the bells take in the cells were arranged in parallel?', '(a) Sketch the form of the magnetic flux pattern due to a current flowing\n(i) in a long solenoid\n(ii) through two long straight parallel wires when the directions of the current are opposite. (Neglect the earth\'s magnetic field).~(b) Draw a labelled diagram of an electric bell and explain how it works\n(c) An electric bell takes a current of 0.2A groin a battery of two dry cells connected in series. Each cell has an e.m.f. of 1.5V and an internal resistance of 1.0\\(\\Omega\\).\n(i) Calculate the effective resistance of the bell\n(ii) What current would the bells take in the cells were arranged in parallel?~(c) An electric bell takes a current of 0.2A groin a battery of two dry cells connected in series. Each cell has an e.m.f. of 1.5V and an internal resistance of 1.0\\(\\Omega\\).\n(i) Calculate the effective resistance of the bell\n(ii) What current would the bells take in the cells were arranged in parallel?', NULL, '\n\nExplanation\nExplanation: When the key is closed, current flows in the circuit thereby magnetising the electromagnet. The soft iron armature is thus attached to the electromagnet, causing the hammer to strike the gong.\nWhen the armature is attracted, a gap is created at the contact and the circuit breaks. The electromagnet loses its magnetism and the armature falls back to its original position by the pull of this spring. This process of make and break is repeated thereby making the bell to ring continuously\n\n(i) E = IR - 0.2(2 + R) R = Ressistence of bell 3 = 0.2(2 + R)\n15 = 2 + R\nR\\(_\\theta\\) = 15 - 2 = 13e\n(ii) E = 1(0.5 _ R) 1.5 = I(0.5 _ 13) = I(13.5)\nI = \\(\\frac{15}{135}\\) = 0.11A\n\n\n\n', '', 'Explanation\nExplanation: When the key is closed, current flows in the circuit thereby magnetising the electromagnet. The soft iron armature is thus attached to the electromagnet, causing the hammer to strike the gong.\nWhen the armature is attracted, a gap is created at the contact and the circuit breaks. The electromagnet loses its magnetism and the armature falls back to its original position by the pull of this spring. This process of make and break is repeated thereby making the bell to ring continuously\n\n(i) E = IR - 0.2(2 + R) R = Ressistence of bell 3 = 0.2(2 + R)\n15 = 2 + R\nR\\(_\\theta\\) = 15 - 2 = 13e\n(ii) E = 1(0.5 _ R) 1.5 = I(0.5 _ 13) = I(13.5)\nI = \\(\\frac{15}{135}\\) = 0.11A\n\n\n\n', '', 'WAEC-PK-1', 'WAEC-PHY', 1),
(276, '2019', '\n\nQuestion\n\n(a) Briefly explain what would happen to a stable element if it is bombarded by \\(\\alpha\\)-particles\n(b) Explain how the bombardment of Uranium with neutrons could lead to nuclear fission chain reaction and hence nuclear explosion.\n(c) State three characteristics of nuclear activity\n(d) State three applications of atomic energy\n(e) State two postulates of Bohr\'s model of the atom and two limitations of such a model\n\n', '', '(a) Briefly explain what would happen to a stable element if it is bombarded by \\(\\alpha\\)-particles\n(b) Explain how the bombardment of Uranium with neutrons could lead to nuclear fission chain reaction and hence nuclear explosion.\n(c) State three characteristics of nuclear activity\n(d) State three applications of atomic energy\n(e) State two postulates of Bohr\'s model of the atom and two limitations of such a model', '(a) Briefly explain what would happen to a stable element if it is bombarded by \\(\\alpha\\)-particles~(b) Explain how the bombardment of Uranium with neutrons could lead to nuclear fission chain reaction and hence nuclear explosion.\n(c) State three characteristics of nuclear activity\n(d) State three applications of atomic energy\n(e) State two postulates of Bohr\'s model of the atom and two limitations of such a model~(c) State three characteristics of nuclear activity\n(d) State three applications of atomic energy\n(e) State two postulates of Bohr\'s model of the atom and two limitations of such a model~(d) State three applications of atomic energy\n(e) State two postulates of Bohr\'s model of the atom and two limitations of such a model~(e) State two postulates of Bohr\'s model of the atom and two limitations of such a model', NULL, '\n\nExplanation\n(a) If a stable element is bombarded by \\(\\alpha\\) - particles the nucleus will become unstable and decay into other elements with the release of energy.\nExample; \\(^{14}_{7}N + ^4_2He \\to ^{17}_8 O + ^1_1H\\) + Energy\n(b) When the nucleus of Uranium is bombarded with a neutron, it splits into two nearly equal parts with a release of energy and neutrons. These excess neutron, if not controlled, further bombard the remaining nuclei, causing a chain reaction and consequently an explosion, provided the critical mass of the Uranium is, exceeded\n\\(^{235}_{92}U + ^1_0U \\to ^{236}_{92}U \\to ^{141}_{56}Ba + ^{92}_{36}Kr + 3^1_0n\\) + Energy\n(c)(i) Radioactive substances give off identifiable radiation i.e \\(\\alpha, \\beta\\) and 8 radiations\n(ii) They are chemically identical to their non-radioactive counterparts and later can be made radioactive by bombardment with nuclear particles\n(iii) Radioactive materials decay with specific half-life.\n\n(d) Three applications of atomic energy\n(i) It is used for the production of nuclear weapons for warfare\n(ii) It is used in the generation of electric power\n(iii) It is used to detect leakages in underground pipes carrying oil or gases.\n\n(e) Two Postulates of Bohr\'s Model of the Atom:\n(i) Electrons occupy certain fixed orbits around the nucleus of an atom. When an electron is in one of these orbits it does not radiate any energy\n(ii) The angular momentum is quantized in the allowed orbit.\n\nTwo Limitations To Bohr\'s Model:\n(i) It could not interpret the details of the spectra of complex atoms\n(ii) There is no theoretical basis for selecting the allowed orbits postulated.\n\n', '', 'Explanation\n(a) If a stable element is bombarded by \\(\\alpha\\) - particles the nucleus will become unstable and decay into other elements with the release of energy.\nExample; \\(^{14}_{7}N + ^4_2He \\to ^{17}_8 O + ^1_1H\\) + Energy\n(b) When the nucleus of Uranium is bombarded with a neutron, it splits into two nearly equal parts with a release of energy and neutrons. These excess neutron, if not controlled, further bombard the remaining nuclei, causing a chain reaction and consequently an explosion, provided the critical mass of the Uranium is, exceeded\n\\(^{235}_{92}U + ^1_0U \\to ^{236}_{92}U \\to ^{141}_{56}Ba + ^{92}_{36}Kr + 3^1_0n\\) + Energy\n(c)(i) Radioactive substances give off identifiable radiation i.e \\(\\alpha, \\beta\\) and 8 radiations\n(ii) They are chemically identical to their non-radioactive counterparts and later can be made radioactive by bombardment with nuclear particles\n(iii) Radioactive materials decay with specific half-life.\n\n(d) Three applications of atomic energy\n(i) It is used for the production of nuclear weapons for warfare\n(ii) It is used in the generation of electric power\n(iii) It is used to detect leakages in underground pipes carrying oil or gases.\n\n(e) Two Postulates of Bohr\'s Model of the Atom:\n(i) Electrons occupy certain fixed orbits around the nucleus of an atom. When an electron is in one of these orbits it does not radiate any energy\n(ii) The angular momentum is quantized in the allowed orbit.\n\nTwo Limitations To Bohr\'s Model:\n(i) It could not interpret the details of the spectra of complex atoms\n(ii) There is no theoretical basis for selecting the allowed orbits postulated.\n\n', '(a) If a stable element is bombarded by \\(\\alpha\\) - particles the nucleus will become unstable and decay into other elements with the release of energy.\nExample; \\(^{14}_{7}N + ^4_2He \\to ^{17}_8 O + ^1_1H\\) + Energy~(b) When the nucleus of Uranium is bombarded with a neutron, it splits into two nearly equal parts with a release of energy and neutrons. These excess neutron, if not controlled, further bombard the remaining nuclei, causing a chain reaction and consequently an explosion, provided the critical mass of the Uranium is, exceeded\n\\(^{235}_{92}U + ^1_0U \\to ^{236}_{92}U \\to ^{141}_{56}Ba + ^{92}_{36}Kr + 3^1_0n\\) + Energy\n(c)(i) Radioactive substances give off identifiable radiation i.e \\(\\alpha, \\beta\\) and 8 radiations\n(ii) They are chemically identical to their non-radioactive counterparts and later can be made radioactive by bombardment with nuclear particles\n(iii) Radioactive materials decay with specific half-life.\n\n(d) Three applications of atomic energy\n(i) It is used for the production of nuclear weapons for warfare\n(ii) It is used in the generation of electric power\n(iii) It is used to detect leakages in underground pipes carrying oil or gases.\n\n(e) Two Postulates of Bohr\'s Model of the Atom:\n(i) Electrons occupy certain fixed orbits around the nucleus of an atom. When an electron is in one of these orbits it does not radiate any energy\n(ii) The angular momentum is quantized in the allowed orbit.\n\nTwo Limitations To Bohr\'s Model:\n(i) It could not interpret the details of the spectra of complex atoms\n(ii) There is no theoretical basis for selecting the allowed orbits postulated.\n~(c)(i) Radioactive substances give off identifiable radiation i.e \\(\\alpha, \\beta\\) and 8 radiations\n(ii) They are chemically identical to their non-radioactive counterparts and later can be made radioactive by bombardment with nuclear particles\n(iii) Radioactive materials decay with specific half-life.\n\n(d) Three applications of atomic energy\n(i) It is used for the production of nuclear weapons for warfare\n(ii) It is used in the generation of electric power\n(iii) It is used to detect leakages in underground pipes carrying oil or gases.\n\n(e) Two Postulates of Bohr\'s Model of the Atom:\n(i) Electrons occupy certain fixed orbits around the nucleus of an atom. When an electron is in one of these orbits it does not radiate any energy\n(ii) The angular momentum is quantized in the allowed orbit.\n\nTwo Limitations To Bohr\'s Model:\n(i) It could not interpret the details of the spectra of complex atoms\n(ii) There is no theoretical basis for selecting the allowed orbits postulated.\n~(d) Three applications of atomic energy\n(i) It is used for the production of nuclear weapons for warfare\n(ii) It is used in the generation of electric power\n(iii) It is used to detect leakages in underground pipes carrying oil or gases.\n\n(e) Two Postulates of Bohr\'s Model of the Atom:\n(i) Electrons occupy certain fixed orbits around the nucleus of an atom. When an electron is in one of these orbits it does not radiate any energy\n(ii) The angular momentum is quantized in the allowed orbit.\n\nTwo Limitations To Bohr\'s Model:\n(i) It could not interpret the details of the spectra of complex atoms\n(ii) There is no theoretical basis for selecting the allowed orbits postulated.\n~(e) Two Postulates of Bohr\'s Model of the Atom:\n(i) Electrons occupy certain fixed orbits around the nucleus of an atom. When an electron is in one of these orbits it does not radiate any energy\n(ii) The angular momentum is quantized in the allowed orbit.\n\nTwo Limitations To Bohr\'s Model:\n(i) It could not interpret the details of the spectra of complex atoms\n(ii) There is no theoretical basis for selecting the allowed orbits postulated.\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(277, '2019', '\n\nQuestion\n\n(a) Explain with the aid of a diagram what is meant by the moment of a force about a point.\n(b) State the conditions of equilibrium for a number of coplanar parallel forces.\nA metre rule is found to balance at the 48cm mark. When a body of mass 60g is suspended at the 6cm mark, the balance point is found to be at the 30cm mark. Calculate:\n(i) the mass of the metre rule; (ii) the distance of the balance point from the zero end, if the body were moved to the 13cm mark.\n(c) Show that the efficiency E, the force ratio M.A and the velocity ratio V.R of a machine are related by the equation \\(E = \\frac{M.A}{V.R} \\times 100%\\)\nThe efficiency of a machine is 80%. Determine the work done by a person using this machine to raise a load of 200kg through a vertical distance of 3.0m.\n[Take g = 10ms\\(^{-2}\\)]\n\n', '', '(a) Explain with the aid of a diagram what is meant by the moment of a force about a point.\n(b) State the conditions of equilibrium for a number of coplanar parallel forces.\nA metre rule is found to balance at the 48cm mark. When a body of mass 60g is suspended at the 6cm mark, the balance point is found to be at the 30cm mark. Calculate:\n(i) the mass of the metre rule; (ii) the distance of the balance point from the zero end, if the body were moved to the 13cm mark.\n(c) Show that the efficiency E, the force ratio M.A and the velocity ratio V.R of a machine are related by the equation \\(E = \\frac{M.A}{V.R} \\times 100%\\)\nThe efficiency of a machine is 80%. Determine the work done by a person using this machine to raise a load of 200kg through a vertical distance of 3.0m.\n[Take g = 10ms\\(^{-2}\\)]', '(a) Explain with the aid of a diagram what is meant by the moment of a force about a point.~(b) State the conditions of equilibrium for a number of coplanar parallel forces.\nA metre rule is found to balance at the 48cm mark. When a body of mass 60g is suspended at the 6cm mark, the balance point is found to be at the 30cm mark. Calculate:\n(i) the mass of the metre rule; (ii) the distance of the balance point from the zero end, if the body were moved to the 13cm mark.\n(c) Show that the efficiency E, the force ratio M.A and the velocity ratio V.R of a machine are related by the equation \\(E = \\frac{M.A}{V.R} \\times 100%\\)\nThe efficiency of a machine is 80%. Determine the work done by a person using this machine to raise a load of 200kg through a vertical distance of 3.0m.\n[Take g = 10ms\\(^{-2}\\)]~(c) Show that the efficiency E, the force ratio M.A and the velocity ratio V.R of a machine are related by the equation \\(E = \\frac{M.A}{V.R} \\times 100%\\)\nThe efficiency of a machine is 80%. Determine the work done by a person using this machine to raise a load of 200kg through a vertical distance of 3.0m.\n[Take g = 10ms\\(^{-2}\\)]', NULL, '\n\nExplanation\n(a) Moment of a force is the product of force and perpendicular distance from the turning point.\n\nMoment = F x D\nThe S.I unit is Nm.\n(b) Conditions: (i) The sum of forces in one direction must be equal to sum of forces in the opposite direction.\n            (ii) Sum of clockwise forces must be equal to sum of anti-clockwise forces about the same point.\n\nSum of clockwise moment = Sum of anti- clockwise moment.\n\\(M \\times 18 = 60 \\times 24\\)\nMass of metre rule \\(M = \\frac{60 \\times 24}{18} = 80g\\)\n\nClockwise moment = anti- clockwise moment\n\\(80 \\times 48 - (13 + x) = 60 \\times x\\)\n\\(80 \\times 35 - x = 60x\\)\n\\(2800 - 80x = 60x \\implies 2800 = 60x + 80x\\)\n\\(2800 = 140x \\implies x = 20cm\\)\nDistance of balance point from the zero end = 20cm + 13cm = 33cm.\n(c) \\(Efficiency = \\frac{\\text{Work output}}{\\text{Work input}} \\times 100%\\)\n= \\(\\frac{\\text{Load x Distance moved by load}}{\\text{Effort x Distance moved by effort}}\\)\n= \\(M.A \\times \\frac{1}{V.R} \\times 100%\\)\n(ii) \\(80% = \\frac{200 \\times 3.0}{\\text{Work done by effort}} \\times100% \\)\n\\(0.8 = \\frac{600}{\\text{Work done by effort}} \\implies \\text{Work done by effort} = \\frac{600}{0.8} = 750J\\)\n\n', '', 'Explanation\n(a) Moment of a force is the product of force and perpendicular distance from the turning point.\n\nMoment = F x D\nThe S.I unit is Nm.\n(b) Conditions: (i) The sum of forces in one direction must be equal to sum of forces in the opposite direction.\n            (ii) Sum of clockwise forces must be equal to sum of anti-clockwise forces about the same point.\n\nSum of clockwise moment = Sum of anti- clockwise moment.\n\\(M \\times 18 = 60 \\times 24\\)\nMass of metre rule \\(M = \\frac{60 \\times 24}{18} = 80g\\)\n\nClockwise moment = anti- clockwise moment\n\\(80 \\times 48 - (13 + x) = 60 \\times x\\)\n\\(80 \\times 35 - x = 60x\\)\n\\(2800 - 80x = 60x \\implies 2800 = 60x + 80x\\)\n\\(2800 = 140x \\implies x = 20cm\\)\nDistance of balance point from the zero end = 20cm + 13cm = 33cm.\n(c) \\(Efficiency = \\frac{\\text{Work output}}{\\text{Work input}} \\times 100%\\)\n= \\(\\frac{\\text{Load x Distance moved by load}}{\\text{Effort x Distance moved by effort}}\\)\n= \\(M.A \\times \\frac{1}{V.R} \\times 100%\\)\n(ii) \\(80% = \\frac{200 \\times 3.0}{\\text{Work done by effort}} \\times100% \\)\n\\(0.8 = \\frac{600}{\\text{Work done by effort}} \\implies \\text{Work done by effort} = \\frac{600}{0.8} = 750J\\)\n\n', '(a) Moment of a force is the product of force and perpendicular distance from the turning point.\n\nMoment = F x D\nThe S.I unit is Nm.~(b) Conditions: (i) The sum of forces in one direction must be equal to sum of forces in the opposite direction.\n            (ii) Sum of clockwise forces must be equal to sum of anti-clockwise forces about the same point.\n\nSum of clockwise moment = Sum of anti- clockwise moment.\n\\(M \\times 18 = 60 \\times 24\\)\nMass of metre rule \\(M = \\frac{60 \\times 24}{18} = 80g\\)\n\nClockwise moment = anti- clockwise moment\n\\(80 \\times 48 - (13 + x) = 60 \\times x\\)\n\\(80 \\times 35 - x = 60x\\)\n\\(2800 - 80x = 60x \\implies 2800 = 60x + 80x\\)\n\\(2800 = 140x \\implies x = 20cm\\)\nDistance of balance point from the zero end = 20cm + 13cm = 33cm.\n(c) \\(Efficiency = \\frac{\\text{Work output}}{\\text{Work input}} \\times 100%\\)\n= \\(\\frac{\\text{Load x Distance moved by load}}{\\text{Effort x Distance moved by effort}}\\)\n= \\(M.A \\times \\frac{1}{V.R} \\times 100%\\)\n(ii) \\(80% = \\frac{200 \\times 3.0}{\\text{Work done by effort}} \\times100% \\)\n\\(0.8 = \\frac{600}{\\text{Work done by effort}} \\implies \\text{Work done by effort} = \\frac{600}{0.8} = 750J\\)\n~(c) \\(Efficiency = \\frac{\\text{Work output}}{\\text{Work input}} \\times 100%\\)\n= \\(\\frac{\\text{Load x Distance moved by load}}{\\text{Effort x Distance moved by effort}}\\)\n= \\(M.A \\times \\frac{1}{V.R} \\times 100%\\)\n(ii) \\(80% = \\frac{200 \\times 3.0}{\\text{Work done by effort}} \\times100% \\)\n\\(0.8 = \\frac{600}{\\text{Work done by effort}} \\implies \\text{Work done by effort} = \\frac{600}{0.8} = 750J\\)\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(278, '2019', '\n\nQuestion\n\n(a) Draw a labelled diagram of a vacuum flask. Explain how its construction minimizes heat exchange with the surroundings.\n(b) State Boyle\'s law. A thread of mercury of length 15cm 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 20cm. Calculate the length of the air column when the tube is held:\n(i) Horizontally ; (ii) vertically with the open end underneath. [Atmospheric pressure = 76cm Hg]\n(c) Explain why it is not advisable to sterilize a clinical thermometer in boiling water at normal atmospheric pressure.\n\n', '', '(a) Draw a labelled diagram of a vacuum flask. Explain how its construction minimizes heat exchange with the surroundings.\n(b) State Boyle\'s law. A thread of mercury of length 15cm 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 20cm. Calculate the length of the air column when the tube is held:\n(i) Horizontally ; (ii) vertically with the open end underneath. [Atmospheric pressure = 76cm Hg]\n(c) Explain why it is not advisable to sterilize a clinical thermometer in boiling water at normal atmospheric pressure.', '(a) Draw a labelled diagram of a vacuum flask. Explain how its construction minimizes heat exchange with the surroundings.~(b) State Boyle\'s law. A thread of mercury of length 15cm 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 20cm. Calculate the length of the air column when the tube is held:\n(i) Horizontally ; (ii) vertically with the open end underneath. [Atmospheric pressure = 76cm Hg]\n(c) Explain why it is not advisable to sterilize a clinical thermometer in boiling water at normal atmospheric pressure.~(c) Explain why it is not advisable to sterilize a clinical thermometer in boiling water at normal atmospheric pressure.', NULL, '\n\nExplanation\n(a)\nVacuum between the double walls prevents heat loss by conduction and convection. The walls are silvered to prevent heat loss by radiation. The cover is made of poor conductor (cork) and it prevents evaporation.\n(b) Boyle\'s law states that the volume of a given mass of gas is inversely proportional to its pressure provided its temperature is constant. Mathematically,\n\\(V = \\frac{K}{P}\\) where K is a constant of proportionality; P = Pressure ; V = Volume.\n\\(\\therefore PV = K \\implies P_{1} V_{1} = P_{2} V_{2}\\)\n(i) Atmospheric pressure (76cm Hg)\n \n\\(V_{1} = 20cm, V_{2} = ? , P_{1} = (15cm + 76cm) = 91 cmHg\\)\n\\(P_{2} = 76 cmHg\\)\nUsing Boyle\'s law,\n\\(P_{1} V_{1} = P_{2} V_{2}\\)\n\\(V_{2} = \\frac{P_{1} V_{1}}{P_{2}}\\)\n\\(V_{2} = \\frac{91 \\times 20}{76} = \\frac{1820}{76}\\)\n= \\(23.95 cm \\approxeq 24cm\\)\nNote: We have assumed that volume is proportional to length. Also, \\(P_{2} = 76 cmHg\\) because only the atmospheric pressure is acting in this case.\n(ii)\nIn this case, \\(V_{1} = 20cm\\), \\(P_{1} = 91 cmHg \\) as above ; In this case, \\(V_{2} = ?\\) but \\(P_{2} = (76 - 15) cmHg = 61 cmHg\\)\n\\(V_{2} = \\frac{91 \\times 20}{61} = \\frac{1820}{61}\\)\n= \\(29.84 cmHg \\approxeq 30 cmHg\\)\n\nClinical thermometer is used to determine the temperature of human body which is about 37°C for a healthy person. It is not advisable to sterilize a clinical thermometer in boiling water at normal atmospheric pressure because\n(1) The range of the clinical thermometer is usually between 40°C and 50°C. Hence when put in boiling water it over expands since the temperature of boiling water is 100°C.\n(2) The boiling water temperature will be too high for the glass material thus cracking it.\n\n', '', 'Explanation\n(a)\nVacuum between the double walls prevents heat loss by conduction and convection. The walls are silvered to prevent heat loss by radiation. The cover is made of poor conductor (cork) and it prevents evaporation.\n(b) Boyle\'s law states that the volume of a given mass of gas is inversely proportional to its pressure provided its temperature is constant. Mathematically,\n\\(V = \\frac{K}{P}\\) where K is a constant of proportionality; P = Pressure ; V = Volume.\n\\(\\therefore PV = K \\implies P_{1} V_{1} = P_{2} V_{2}\\)\n(i) Atmospheric pressure (76cm Hg)\n \n\\(V_{1} = 20cm, V_{2} = ? , P_{1} = (15cm + 76cm) = 91 cmHg\\)\n\\(P_{2} = 76 cmHg\\)\nUsing Boyle\'s law,\n\\(P_{1} V_{1} = P_{2} V_{2}\\)\n\\(V_{2} = \\frac{P_{1} V_{1}}{P_{2}}\\)\n\\(V_{2} = \\frac{91 \\times 20}{76} = \\frac{1820}{76}\\)\n= \\(23.95 cm \\approxeq 24cm\\)\nNote: We have assumed that volume is proportional to length. Also, \\(P_{2} = 76 cmHg\\) because only the atmospheric pressure is acting in this case.\n(ii)\nIn this case, \\(V_{1} = 20cm\\), \\(P_{1} = 91 cmHg \\) as above ; In this case, \\(V_{2} = ?\\) but \\(P_{2} = (76 - 15) cmHg = 61 cmHg\\)\n\\(V_{2} = \\frac{91 \\times 20}{61} = \\frac{1820}{61}\\)\n= \\(29.84 cmHg \\approxeq 30 cmHg\\)\n\nClinical thermometer is used to determine the temperature of human body which is about 37°C for a healthy person. It is not advisable to sterilize a clinical thermometer in boiling water at normal atmospheric pressure because\n(1) The range of the clinical thermometer is usually between 40°C and 50°C. Hence when put in boiling water it over expands since the temperature of boiling water is 100°C.\n(2) The boiling water temperature will be too high for the glass material thus cracking it.\n\n', '(a)\nVacuum between the double walls prevents heat loss by conduction and convection. The walls are silvered to prevent heat loss by radiation. The cover is made of poor conductor (cork) and it prevents evaporation.~(b) Boyle\'s law states that the volume of a given mass of gas is inversely proportional to its pressure provided its temperature is constant. Mathematically,\n\\(V = \\frac{K}{P}\\) where K is a constant of proportionality; P = Pressure ; V = Volume.\n\\(\\therefore PV = K \\implies P_{1} V_{1} = P_{2} V_{2}\\)\n(i) Atmospheric pressure (76cm Hg)\n \n\\(V_{1} = 20cm, V_{2} = ? , P_{1} = (15cm + 76cm) = 91 cmHg\\)\n\\(P_{2} = 76 cmHg\\)\nUsing Boyle\'s law,\n\\(P_{1} V_{1} = P_{2} V_{2}\\)\n\\(V_{2} = \\frac{P_{1} V_{1}}{P_{2}}\\)\n\\(V_{2} = \\frac{91 \\times 20}{76} = \\frac{1820}{76}\\)\n= \\(23.95 cm \\approxeq 24cm\\)\nNote: We have assumed that volume is proportional to length. Also, \\(P_{2} = 76 cmHg\\) because only the atmospheric pressure is acting in this case.\n(ii)\nIn this case, \\(V_{1} = 20cm\\), \\(P_{1} = 91 cmHg \\) as above ; In this case, \\(V_{2} = ?\\) but \\(P_{2} = (76 - 15) cmHg = 61 cmHg\\)\n\\(V_{2} = \\frac{91 \\times 20}{61} = \\frac{1820}{61}\\)\n= \\(29.84 cmHg \\approxeq 30 cmHg\\)\n\nClinical thermometer is used to determine the temperature of human body which is about 37°C for a healthy person. It is not advisable to sterilize a clinical thermometer in boiling water at normal atmospheric pressure because\n(1) The range of the clinical thermometer is usually between 40°C and 50°C. Hence when put in boiling water it over expands since the temperature of boiling water is 100°C.\n(2) The boiling water temperature will be too high for the glass material thus cracking it.\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(279, '2019', '\n\nQuestion\n\n(a) Explain with the aid of a diagram how a converging lens could be used to : (i) ignite a piece of carbon paper ; (ii) produce an enlarged picture on a screen ; (iii) correct an eye defect.\n(b) What is a mechanical wave? Describe with the aid of a diagram, an experiment to show that sound needs a material medium for transmission.\nState 3 characteristics of sound and mention the factor on which each depends.\n\n', '', '(a) Explain with the aid of a diagram how a converging lens could be used to : (i) ignite a piece of carbon paper ; (ii) produce an enlarged picture on a screen ; (iii) correct an eye defect.\n(b) What is a mechanical wave? Describe with the aid of a diagram, an experiment to show that sound needs a material medium for transmission.\nState 3 characteristics of sound and mention the factor on which each depends.', '(a) Explain with the aid of a diagram how a converging lens could be used to : (i) ignite a piece of carbon paper ; (ii) produce an enlarged picture on a screen ; (iii) correct an eye defect.~(b) What is a mechanical wave? Describe with the aid of a diagram, an experiment to show that sound needs a material medium for transmission.\nState 3 characteristics of sound and mention the factor on which each depends.', NULL, '\n\nExplanation\n(a)(i)\nWhen the parallel rays of light from the sun pass through the converging lens, they converge at the principal focus. This converging rays cause the paper to ignite after a while.\n(ii)\nProduction of an enlarged picture\nas in the magnifying glass. I - enlarged image; O - object. Alternatively, as in the slide projector below\n\nThe lens causes the image to be magnified but inverted in this case.\n(iii) Correction of eye defect:\nConverging lens can be used for correcting long sightedness. Long sightedness occurs when the eyeball is too short. Rays from the normal near point (25cm from the eye) are brought to focus beyond the eyeball instead of on the retina. The correction is done as shown below.\n\n(b) A mechanical wave is one that requires material medium for their propagation e.g. sound waves, water waves. Waves which do not require material medium for their propagation i.e. they can pass through vacuum are electromagnetic waves.\nExperiment to show that sound requires material medium for transmission.\n\nWhen the key is closed, current will flow through the electric bell and the hammer will be seen, striking the gong. Sound will thus be produced. However as the air inside the bell jar is gradually pumped out using a vacuum pump via the opening, the intensity of the sound produced by the electric bell gradually dies down even though the hammer is still seen striking the gong. When the air has been pumped out and the bell jar becomes empty (vacuum), the intensity of the sound dies almost completely. This shows that sound requires a material medium (air in this case) for transmission.\nThe three characteristics of sound are:\n1. Intensity or Loudness: depends on amplitude.\n2. Pitch : depends on frequency\n3. Quality or Timbre : depends on overtones.\n\n\n', '', 'Explanation\n(a)(i)\nWhen the parallel rays of light from the sun pass through the converging lens, they converge at the principal focus. This converging rays cause the paper to ignite after a while.\n(ii)\nProduction of an enlarged picture\nas in the magnifying glass. I - enlarged image; O - object. Alternatively, as in the slide projector below\n\nThe lens causes the image to be magnified but inverted in this case.\n(iii) Correction of eye defect:\nConverging lens can be used for correcting long sightedness. Long sightedness occurs when the eyeball is too short. Rays from the normal near point (25cm from the eye) are brought to focus beyond the eyeball instead of on the retina. The correction is done as shown below.\n\n(b) A mechanical wave is one that requires material medium for their propagation e.g. sound waves, water waves. Waves which do not require material medium for their propagation i.e. they can pass through vacuum are electromagnetic waves.\nExperiment to show that sound requires material medium for transmission.\n\nWhen the key is closed, current will flow through the electric bell and the hammer will be seen, striking the gong. Sound will thus be produced. However as the air inside the bell jar is gradually pumped out using a vacuum pump via the opening, the intensity of the sound produced by the electric bell gradually dies down even though the hammer is still seen striking the gong. When the air has been pumped out and the bell jar becomes empty (vacuum), the intensity of the sound dies almost completely. This shows that sound requires a material medium (air in this case) for transmission.\nThe three characteristics of sound are:\n1. Intensity or Loudness: depends on amplitude.\n2. Pitch : depends on frequency\n3. Quality or Timbre : depends on overtones.\n\n\n', '(a)(i)\nWhen the parallel rays of light from the sun pass through the converging lens, they converge at the principal focus. This converging rays cause the paper to ignite after a while.\n(ii)\nProduction of an enlarged picture\nas in the magnifying glass. I - enlarged image; O - object. Alternatively, as in the slide projector below\n\nThe lens causes the image to be magnified but inverted in this case.\n(iii) Correction of eye defect:\nConverging lens can be used for correcting long sightedness. Long sightedness occurs when the eyeball is too short. Rays from the normal near point (25cm from the eye) are brought to focus beyond the eyeball instead of on the retina. The correction is done as shown below.\n~(b) A mechanical wave is one that requires material medium for their propagation e.g. sound waves, water waves. Waves which do not require material medium for their propagation i.e. they can pass through vacuum are electromagnetic waves.\nExperiment to show that sound requires material medium for transmission.\n\nWhen the key is closed, current will flow through the electric bell and the hammer will be seen, striking the gong. Sound will thus be produced. However as the air inside the bell jar is gradually pumped out using a vacuum pump via the opening, the intensity of the sound produced by the electric bell gradually dies down even though the hammer is still seen striking the gong. When the air has been pumped out and the bell jar becomes empty (vacuum), the intensity of the sound dies almost completely. This shows that sound requires a material medium (air in this case) for transmission.\nThe three characteristics of sound are:\n1. Intensity or Loudness: depends on amplitude.\n2. Pitch : depends on frequency\n3. Quality or Timbre : depends on overtones.\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
(280, '2019', '\n\nQuestion\n\n(a) Explain what is meant by: (i) electric field intensity ; (ii) electric lines of force.\n(b) Two similar but opposite point charges -q and +q each of magnitude \\(5 \\times 10^{-8} C\\) are seperated by a distance of 8.0cm in vacuum as shown in the diagram below.\n\nCalculate the magnitude and direction of the resultant electric field intensity E at the point P. Draw the lines of force due to this system of charges. [Take \\(\\frac{1}{4 \\pi \\varepsilon _{0}}\\)]\n(c)\nCalculate the following in the series circuit shown above: (i) reactance of the capacitor ; (ii) impedance of the circuit ; (iii) current through the circuit ; (iv) voltage across the capacitor ; (v) average power used in the circuit.\n\n', '', '(a) Explain what is meant by: (i) electric field intensity ; (ii) electric lines of force.\n(b) Two similar but opposite point charges -q and +q each of magnitude \\(5 \\times 10^{-8} C\\) are seperated by a distance of 8.0cm in vacuum as shown in the diagram below.\n\nCalculate the magnitude and direction of the resultant electric field intensity E at the point P. Draw the lines of force due to this system of charges. [Take \\(\\frac{1}{4 \\pi \\varepsilon _{0}}\\)]\n(c)\nCalculate the following in the series circuit shown above: (i) reactance of the capacitor ; (ii) impedance of the circuit ; (iii) current through the circuit ; (iv) voltage across the capacitor ; (v) average power used in the circuit.', '(a) Explain what is meant by: (i) electric field intensity ; (ii) electric lines of force.~(b) Two similar but opposite point charges -q and +q each of magnitude \\(5 \\times 10^{-8} C\\) are seperated by a distance of 8.0cm in vacuum as shown in the diagram below.\n\nCalculate the magnitude and direction of the resultant electric field intensity E at the point P. Draw the lines of force due to this system of charges. [Take \\(\\frac{1}{4 \\pi \\varepsilon _{0}}\\)]\n(c)\nCalculate the following in the series circuit shown above: (i) reactance of the capacitor ; (ii) impedance of the circuit ; (iii) current through the circuit ; (iv) voltage across the capacitor ; (v) average power used in the circuit.~(c)\nCalculate the following in the series circuit shown above: (i) reactance of the capacitor ; (ii) impedance of the circuit ; (iii) current through the circuit ; (iv) voltage across the capacitor ; (v) average power used in the circuit.', NULL, '\n\nExplanation\n(a)(i) Electric Field Intensity: An electric field is a region in which a force of electrical origin is experienced. Electric Field Intensity (E) at any point is the force per unit charge which the field exerts at that point. The S.I unit is NC\\(^{-1}\\). It is a vector quantity and its direction is the same as that of the force or a small positive test charge placed in the field. Mathematically,\n\\(E = \\frac{F}{q}\\), F - force and q - charge.\n(ii) Electric Lines of Force: Electric lines of force are imaginary lines used to represent electric field patterns. They have the following characteristics:\n(1) At any point, the tangent to a line of force gives the direction of E at that point.\n(2) The number of lines per unit cross-sectional area is proportional to the magnitude of electric field. The closer the lines, the larger the electric field.\n(b)\nField due to -q at p:\n\\(E_{1} = \\frac{q}{4 \\pi \\varepsilon _{0} r^{2}}\\)\n= \\(\\frac{-5.0 \\times 10^{-8}}{0.05^{2}} \\times 9 \\times 10^{9}\\)\n= \\(- 180000 NC^{-1}\\)\n\\(E_{2} = \\frac{+ q}{4 \\pi \\varepsilon _{0} r^{2}}\\)\n= \\(\\frac{5.0 \\times 10^{-8}}{0.03^{2}} \\times 9 \\times 10^{9}\\)\n= \\(+ 500000 NC^{-1}\\)\n\\(\\therefore\\) Resultant of the electric field intensity E at the point P is\n\\(E = E_{1} + E_{2} = 500000 - 180000 = 320000 NC^{-1}\\)\nIn the direction of the positive charge.\n(c)\n(i) Reactance of the capacitor \\(X_{c} = \\frac{1}{\\omega c} = \\frac{1}{2 \\pi f c}\\)\n= \\(\\frac{1}{2 \\times \\pi \\times 25 \\times 10 \\times 10^{-6}} = \\frac{2000}{\\pi} \\Omega\\)\n(ii) Impedance of the circuit, \\(Z= \\sqrt{X_{c} ^{2} + R^{2}}\\)\n= \\(\\sqrt{(\\frac{2000}{\\pi})^{2} + 1000^{2}}\\)\n= \\(\\sqrt{(636.62)^{2} + (1000)^{2}}\\)\n= \\(\\sqrt{1405285.0244}\\)\n= \\(1185.45 \\Omega\\)\n(iii) Current through the circuit, \\(I = \\frac{E}{Z}\\)\n= \\(\\frac{90}{1185.45} = 0.0759 A_{rms}\\)\n(iv) Voltage across the capacitor \\(V = IX_{c}\\)\n= \\(0.0759 \\times 636.62\\)\n= \\(48.319 V_{rms}\\)\n(v) Average power used in the circuit \\(P = IV\\)\n= \\(I^{2} Z\\)\n= \\((0.0759)^{2} \\times 1185.45\\)\n= \\(6.829 W\\)\n\n', '', 'Explanation\n(a)(i) Electric Field Intensity: An electric field is a region in which a force of electrical origin is experienced. Electric Field Intensity (E) at any point is the force per unit charge which the field exerts at that point. The S.I unit is NC\\(^{-1}\\). It is a vector quantity and its direction is the same as that of the force or a small positive test charge placed in the field. Mathematically,\n\\(E = \\frac{F}{q}\\), F - force and q - charge.\n(ii) Electric Lines of Force: Electric lines of force are imaginary lines used to represent electric field patterns. They have the following characteristics:\n(1) At any point, the tangent to a line of force gives the direction of E at that point.\n(2) The number of lines per unit cross-sectional area is proportional to the magnitude of electric field. The closer the lines, the larger the electric field.\n(b)\nField due to -q at p:\n\\(E_{1} = \\frac{q}{4 \\pi \\varepsilon _{0} r^{2}}\\)\n= \\(\\frac{-5.0 \\times 10^{-8}}{0.05^{2}} \\times 9 \\times 10^{9}\\)\n= \\(- 180000 NC^{-1}\\)\n\\(E_{2} = \\frac{+ q}{4 \\pi \\varepsilon _{0} r^{2}}\\)\n= \\(\\frac{5.0 \\times 10^{-8}}{0.03^{2}} \\times 9 \\times 10^{9}\\)\n= \\(+ 500000 NC^{-1}\\)\n\\(\\therefore\\) Resultant of the electric field intensity E at the point P is\n\\(E = E_{1} + E_{2} = 500000 - 180000 = 320000 NC^{-1}\\)\nIn the direction of the positive charge.\n(c)\n(i) Reactance of the capacitor \\(X_{c} = \\frac{1}{\\omega c} = \\frac{1}{2 \\pi f c}\\)\n= \\(\\frac{1}{2 \\times \\pi \\times 25 \\times 10 \\times 10^{-6}} = \\frac{2000}{\\pi} \\Omega\\)\n(ii) Impedance of the circuit, \\(Z= \\sqrt{X_{c} ^{2} + R^{2}}\\)\n= \\(\\sqrt{(\\frac{2000}{\\pi})^{2} + 1000^{2}}\\)\n= \\(\\sqrt{(636.62)^{2} + (1000)^{2}}\\)\n= \\(\\sqrt{1405285.0244}\\)\n= \\(1185.45 \\Omega\\)\n(iii) Current through the circuit, \\(I = \\frac{E}{Z}\\)\n= \\(\\frac{90}{1185.45} = 0.0759 A_{rms}\\)\n(iv) Voltage across the capacitor \\(V = IX_{c}\\)\n= \\(0.0759 \\times 636.62\\)\n= \\(48.319 V_{rms}\\)\n(v) Average power used in the circuit \\(P = IV\\)\n= \\(I^{2} Z\\)\n= \\((0.0759)^{2} \\times 1185.45\\)\n= \\(6.829 W\\)\n\n', '(a)(i) Electric Field Intensity: An electric field is a region in which a force of electrical origin is experienced. Electric Field Intensity (E) at any point is the force per unit charge which the field exerts at that point. The S.I unit is NC\\(^{-1}\\). It is a vector quantity and its direction is the same as that of the force or a small positive test charge placed in the field. Mathematically,\n\\(E = \\frac{F}{q}\\), F - force and q - charge.\n(ii) Electric Lines of Force: Electric lines of force are imaginary lines used to represent electric field patterns. They have the following characteristics:\n(1) At any point, the tangent to a line of force gives the direction of E at that point.\n(2) The number of lines per unit cross-sectional area is proportional to the magnitude of electric field. The closer the lines, the larger the electric field.~(b)\nField due to -q at p:\n\\(E_{1} = \\frac{q}{4 \\pi \\varepsilon _{0} r^{2}}\\)\n= \\(\\frac{-5.0 \\times 10^{-8}}{0.05^{2}} \\times 9 \\times 10^{9}\\)\n= \\(- 180000 NC^{-1}\\)\n\\(E_{2} = \\frac{+ q}{4 \\pi \\varepsilon _{0} r^{2}}\\)\n= \\(\\frac{5.0 \\times 10^{-8}}{0.03^{2}} \\times 9 \\times 10^{9}\\)\n= \\(+ 500000 NC^{-1}\\)\n\\(\\therefore\\) Resultant of the electric field intensity E at the point P is\n\\(E = E_{1} + E_{2} = 500000 - 180000 = 320000 NC^{-1}\\)\nIn the direction of the positive charge.\n(c)\n(i) Reactance of the capacitor \\(X_{c} = \\frac{1}{\\omega c} = \\frac{1}{2 \\pi f c}\\)\n= \\(\\frac{1}{2 \\times \\pi \\times 25 \\times 10 \\times 10^{-6}} = \\frac{2000}{\\pi} \\Omega\\)\n(ii) Impedance of the circuit, \\(Z= \\sqrt{X_{c} ^{2} + R^{2}}\\)\n= \\(\\sqrt{(\\frac{2000}{\\pi})^{2} + 1000^{2}}\\)\n= \\(\\sqrt{(636.62)^{2} + (1000)^{2}}\\)\n= \\(\\sqrt{1405285.0244}\\)\n= \\(1185.45 \\Omega\\)\n(iii) Current through the circuit, \\(I = \\frac{E}{Z}\\)\n= \\(\\frac{90}{1185.45} = 0.0759 A_{rms}\\)\n(iv) Voltage across the capacitor \\(V = IX_{c}\\)\n= \\(0.0759 \\times 636.62\\)\n= \\(48.319 V_{rms}\\)\n(v) Average power used in the circuit \\(P = IV\\)\n= \\(I^{2} Z\\)\n= \\((0.0759)^{2} \\times 1185.45\\)\n= \\(6.829 W\\)\n~(c)\n(i) Reactance of the capacitor \\(X_{c} = \\frac{1}{\\omega c} = \\frac{1}{2 \\pi f c}\\)\n= \\(\\frac{1}{2 \\times \\pi \\times 25 \\times 10 \\times 10^{-6}} = \\frac{2000}{\\pi} \\Omega\\)\n(ii) Impedance of the circuit, \\(Z= \\sqrt{X_{c} ^{2} + R^{2}}\\)\n= \\(\\sqrt{(\\frac{2000}{\\pi})^{2} + 1000^{2}}\\)\n= \\(\\sqrt{(636.62)^{2} + (1000)^{2}}\\)\n= \\(\\sqrt{1405285.0244}\\)\n= \\(1185.45 \\Omega\\)\n(iii) Current through the circuit, \\(I = \\frac{E}{Z}\\)\n= \\(\\frac{90}{1185.45} = 0.0759 A_{rms}\\)\n(iv) Voltage across the capacitor \\(V = IX_{c}\\)\n= \\(0.0759 \\times 636.62\\)\n= \\(48.319 V_{rms}\\)\n(v) Average power used in the circuit \\(P = IV\\)\n= \\(I^{2} Z\\)\n= \\((0.0759)^{2} \\times 1185.45\\)\n= \\(6.829 W\\)\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(281, '2019', '\n\nQuestion\n\n(a) Define the apparent cubic expansivity of a liquid\n(b)(i) Describe with the aid of a labelled diagram, an experiment to determine the apparent cubic expansivity of a liquid.\n(ii) State two precuations that should be taken to ensure accurate results.\n(c) A density glass bottle contains 44.25g of a liquid at 0°C and 42.02g at 50°C. Calculate the real cubic expansivity of the liquid. (Linear expansivity of glass = 1.0 x 10-5K\\(^{-1}\\))\n\n', '', '(a) Define the apparent cubic expansivity of a liquid\n(b)(i) Describe with the aid of a labelled diagram, an experiment to determine the apparent cubic expansivity of a liquid.\n(ii) State two precuations that should be taken to ensure accurate results.\n(c) A density glass bottle contains 44.25g of a liquid at 0°C and 42.02g at 50°C. Calculate the real cubic expansivity of the liquid. (Linear expansivity of glass = 1.0 x 10-5K\\(^{-1}\\))', '(a) Define the apparent cubic expansivity of a liquid~(b)(i) Describe with the aid of a labelled diagram, an experiment to determine the apparent cubic expansivity of a liquid.\n(ii) State two precuations that should be taken to ensure accurate results.\n(c) A density glass bottle contains 44.25g of a liquid at 0°C and 42.02g at 50°C. Calculate the real cubic expansivity of the liquid. (Linear expansivity of glass = 1.0 x 10-5K\\(^{-1}\\))~(c) A density glass bottle contains 44.25g of a liquid at 0°C and 42.02g at 50°C. Calculate the real cubic expansivity of the liquid. (Linear expansivity of glass = 1.0 x 10-5K\\(^{-1}\\))', NULL, '\n\nExplanation\n(a) The apparent cubic expansivity of a liquid is defined as the apparent increase in volume per unit volume per unit rise in temperature.\n(b)(i)\n\nProcedure: Density bottle dried and weighed, M,, fill bottle with liquid, replace the stopper and wipe the bottle dry. Weigh bottle and content M\\(_2\\), suspend the bottle with content in a water bath. Record the initial temperature of the water in the bath T\\(_1\\). Then heat water gradually up to boiling point. Notice that some liquid is expelled through the orifice of the stopper and the water in the bath is allowed to continue boiling until no more liquid is expelled through the orifice of the stopper.Record the final temperature T\\(_2\\) of the water bath. Remove the bottle from the bath, allow it to cool, wipe, dry it and reweigh M\\(_3\\).\nCalculation: Mass of the liquid expelled from the bottle = (M\\(_2\\) - M\\(_3\\)) Mass of the liquid left in the bottle = (M\\(_3\\) - M\\(_1\\) Temperature change, (T\\(_2\\) - T\\(_1\\)) Apparent cubic expansivity\n\\(\\alpha\\) a = \\(\\frac{\\text{Mass of the liquid expelled}}{\\text{ Mass of remaining liquid \\times change in temperature}}\\)\n\\(\\alpha\\) a = \\(\\frac{M_2 - M_3}{(M_3 - M_1) - (T_2 - T_1)}\\)\n(ii) Precautions;\n(1) Avoid parallax in reading thermometer\n(2) Bottle should not be submerged\n\n(c) Apparent cubic expansivity\n\\(\\alpha\\) a = \\(\\frac{M_2 - M_3}{(M_3 - M_1) - (T_2 - T_1)} = \\frac{44.25 - 42.02}{42.02 \\times 50}\\)\n\\(\\alpha\\) a = 1.06 x 10\\(^{-3} k^{-1}\\)\nBut real cubic expansivity = \\(\\alpha\\) a =+ \\(\\alpha\\) g\nWhere \\(\\alpha\\) g is the cubic expansivity of the containing vessel i.e. glass\n\\(\\alpha\\) a = 1.09 x 10\\(^3 K^{-1}\\)\n\n', '', 'Explanation\n(a) The apparent cubic expansivity of a liquid is defined as the apparent increase in volume per unit volume per unit rise in temperature.\n(b)(i)\n\nProcedure: Density bottle dried and weighed, M,, fill bottle with liquid, replace the stopper and wipe the bottle dry. Weigh bottle and content M\\(_2\\), suspend the bottle with content in a water bath. Record the initial temperature of the water in the bath T\\(_1\\). Then heat water gradually up to boiling point. Notice that some liquid is expelled through the orifice of the stopper and the water in the bath is allowed to continue boiling until no more liquid is expelled through the orifice of the stopper.Record the final temperature T\\(_2\\) of the water bath. Remove the bottle from the bath, allow it to cool, wipe, dry it and reweigh M\\(_3\\).\nCalculation: Mass of the liquid expelled from the bottle = (M\\(_2\\) - M\\(_3\\)) Mass of the liquid left in the bottle = (M\\(_3\\) - M\\(_1\\) Temperature change, (T\\(_2\\) - T\\(_1\\)) Apparent cubic expansivity\n\\(\\alpha\\) a = \\(\\frac{\\text{Mass of the liquid expelled}}{\\text{ Mass of remaining liquid \\times change in temperature}}\\)\n\\(\\alpha\\) a = \\(\\frac{M_2 - M_3}{(M_3 - M_1) - (T_2 - T_1)}\\)\n(ii) Precautions;\n(1) Avoid parallax in reading thermometer\n(2) Bottle should not be submerged\n\n(c) Apparent cubic expansivity\n\\(\\alpha\\) a = \\(\\frac{M_2 - M_3}{(M_3 - M_1) - (T_2 - T_1)} = \\frac{44.25 - 42.02}{42.02 \\times 50}\\)\n\\(\\alpha\\) a = 1.06 x 10\\(^{-3} k^{-1}\\)\nBut real cubic expansivity = \\(\\alpha\\) a =+ \\(\\alpha\\) g\nWhere \\(\\alpha\\) g is the cubic expansivity of the containing vessel i.e. glass\n\\(\\alpha\\) a = 1.09 x 10\\(^3 K^{-1}\\)\n\n', '(a) The apparent cubic expansivity of a liquid is defined as the apparent increase in volume per unit volume per unit rise in temperature.~(b)(i)\n\nProcedure: Density bottle dried and weighed, M,, fill bottle with liquid, replace the stopper and wipe the bottle dry. Weigh bottle and content M\\(_2\\), suspend the bottle with content in a water bath. Record the initial temperature of the water in the bath T\\(_1\\). Then heat water gradually up to boiling point. Notice that some liquid is expelled through the orifice of the stopper and the water in the bath is allowed to continue boiling until no more liquid is expelled through the orifice of the stopper.Record the final temperature T\\(_2\\) of the water bath. Remove the bottle from the bath, allow it to cool, wipe, dry it and reweigh M\\(_3\\).\nCalculation: Mass of the liquid expelled from the bottle = (M\\(_2\\) - M\\(_3\\)) Mass of the liquid left in the bottle = (M\\(_3\\) - M\\(_1\\) Temperature change, (T\\(_2\\) - T\\(_1\\)) Apparent cubic expansivity\n\\(\\alpha\\) a = \\(\\frac{\\text{Mass of the liquid expelled}}{\\text{ Mass of remaining liquid \\times change in temperature}}\\)\n\\(\\alpha\\) a = \\(\\frac{M_2 - M_3}{(M_3 - M_1) - (T_2 - T_1)}\\)\n(ii) Precautions;\n(1) Avoid parallax in reading thermometer\n(2) Bottle should not be submerged\n\n(c) Apparent cubic expansivity\n\\(\\alpha\\) a = \\(\\frac{M_2 - M_3}{(M_3 - M_1) - (T_2 - T_1)} = \\frac{44.25 - 42.02}{42.02 \\times 50}\\)\n\\(\\alpha\\) a = 1.06 x 10\\(^{-3} k^{-1}\\)\nBut real cubic expansivity = \\(\\alpha\\) a =+ \\(\\alpha\\) g\nWhere \\(\\alpha\\) g is the cubic expansivity of the containing vessel i.e. glass\n\\(\\alpha\\) a = 1.09 x 10\\(^3 K^{-1}\\)\n~(c) Apparent cubic expansivity\n\\(\\alpha\\) a = \\(\\frac{M_2 - M_3}{(M_3 - M_1) - (T_2 - T_1)} = \\frac{44.25 - 42.02}{42.02 \\times 50}\\)\n\\(\\alpha\\) a = 1.06 x 10\\(^{-3} k^{-1}\\)\nBut real cubic expansivity = \\(\\alpha\\) a =+ \\(\\alpha\\) g\nWhere \\(\\alpha\\) g is the cubic expansivity of the containing vessel i.e. glass\n\\(\\alpha\\) a = 1.09 x 10\\(^3 K^{-1}\\)\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(282, '2019', '\n\nQuestion\n\n(a) What is meant by dispersion of white light?\n(b) State the colours in the spectrum of white light in ascending order of their wavelengths\n(c) Which colour is deviated.\n(i) least\n(ii) most?\n(d) Explain why white light is dispersed when it passes through a glass prism.\n(e) Describe, with the aid of a labelled diagram, how a pure spectrum of white light can be produced on a screen.\n\n', '', '(a) What is meant by dispersion of white light?\n(b) State the colours in the spectrum of white light in ascending order of their wavelengths\n(c) Which colour is deviated.\n(i) least\n(ii) most?\n(d) Explain why white light is dispersed when it passes through a glass prism.\n(e) Describe, with the aid of a labelled diagram, how a pure spectrum of white light can be produced on a screen.', '(a) What is meant by dispersion of white light?~(b) State the colours in the spectrum of white light in ascending order of their wavelengths\n(c) Which colour is deviated.\n(i) least\n(ii) most?\n(d) Explain why white light is dispersed when it passes through a glass prism.\n(e) Describe, with the aid of a labelled diagram, how a pure spectrum of white light can be produced on a screen.~(c) Which colour is deviated.\n(i) least\n(ii) most?\n(d) Explain why white light is dispersed when it passes through a glass prism.\n(e) Describe, with the aid of a labelled diagram, how a pure spectrum of white light can be produced on a screen.~(d) Explain why white light is dispersed when it passes through a glass prism.\n(e) Describe, with the aid of a labelled diagram, how a pure spectrum of white light can be produced on a screen.~(e) Describe, with the aid of a labelled diagram, how a pure spectrum of white light can be produced on a screen.', NULL, '\n\nExplanation\n(a) This is the separation of white light into its component or constituent colours.\n(b) VIBGYOR (Violet, Indigo, Blue, Green, Yellow, Orange, Red).\n(c)(i) Red (ii) Violet.\n(d) White light consists of a mixture of different colours. The refractive index of glass is different for each colour so that when white light falls on the prism each colour in it is refracted at different angles with the result that the colours are spread out to form a spectrum\n\nDescription: An illuminated slit is placed at the principal focus of a converging lens, such that parallel beam of light emerging therefrom is incident on a prism which disperses the beam into separate parallel beams of different colours each of which is brought into its own focus in the focal plane of a second lens. Here the spectrum is formed on a white screen.\n\n\n', '', 'Explanation\n(a) This is the separation of white light into its component or constituent colours.\n(b) VIBGYOR (Violet, Indigo, Blue, Green, Yellow, Orange, Red).\n(c)(i) Red (ii) Violet.\n(d) White light consists of a mixture of different colours. The refractive index of glass is different for each colour so that when white light falls on the prism each colour in it is refracted at different angles with the result that the colours are spread out to form a spectrum\n\nDescription: An illuminated slit is placed at the principal focus of a converging lens, such that parallel beam of light emerging therefrom is incident on a prism which disperses the beam into separate parallel beams of different colours each of which is brought into its own focus in the focal plane of a second lens. Here the spectrum is formed on a white screen.\n\n\n', '(a) This is the separation of white light into its component or constituent colours.~(b) VIBGYOR (Violet, Indigo, Blue, Green, Yellow, Orange, Red).\n(c)(i) Red (ii) Violet.\n(d) White light consists of a mixture of different colours. The refractive index of glass is different for each colour so that when white light falls on the prism each colour in it is refracted at different angles with the result that the colours are spread out to form a spectrum\n\nDescription: An illuminated slit is placed at the principal focus of a converging lens, such that parallel beam of light emerging therefrom is incident on a prism which disperses the beam into separate parallel beams of different colours each of which is brought into its own focus in the focal plane of a second lens. Here the spectrum is formed on a white screen.\n\n~(c)(i) Red (ii) Violet.\n(d) White light consists of a mixture of different colours. The refractive index of glass is different for each colour so that when white light falls on the prism each colour in it is refracted at different angles with the result that the colours are spread out to form a spectrum\n\nDescription: An illuminated slit is placed at the principal focus of a converging lens, such that parallel beam of light emerging therefrom is incident on a prism which disperses the beam into separate parallel beams of different colours each of which is brought into its own focus in the focal plane of a second lens. Here the spectrum is formed on a white screen.\n\n~d).\n(c)(i) Red (ii) Violet.\n(d) White light consists of a mixture of different colours. The refractive index of glass is different for each colour so that when white light falls on the prism each colour in it is refracted at different angles with the result that the colours are spread out to form a spectrum\n\nDescription: An illuminated slit is placed at the principal focus of a converging lens, such that parallel beam of light emerging therefrom is incident on a prism which disperses the beam into separate parallel beams of different colours each of which is brought into its own focus in the focal plane of a second lens. Here the spectrum is formed on a white screen.\n\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(283, '2019', '\n\nQuestion\n\n(a) Explain what is meant by the statement: The capacitance of a parallel-plate capacitor is 2\\(\\mu\\)F\n(b) State: (i) three factors on which its capacitance depends\n(ii) three uses of capacitors.\n(c) Derive a formula for the energy W stored in a charged capacitor of capacitance C carrying a charge Q on either plate.,\n(d) Two parallel-plate capacitors of capacitances 2\\(\\mu\\)F and 3\\(\\mu\\)F are connected in parallel and the combination is connected to a 50V d.c. source. Draw the circuit diagram of the arrangement and determine the:\n(i) charge on either plate of each capacitor\n(ii) potential difference across each capacitor\n(iii) energy of the combinad capacitors.\n\n', '', '(a) Explain what is meant by the statement: The capacitance of a parallel-plate capacitor is 2\\(\\mu\\)F\n(b) State: (i) three factors on which its capacitance depends\n(ii) three uses of capacitors.\n(c) Derive a formula for the energy W stored in a charged capacitor of capacitance C carrying a charge Q on either plate.,\n(d) Two parallel-plate capacitors of capacitances 2\\(\\mu\\)F and 3\\(\\mu\\)F are connected in parallel and the combination is connected to a 50V d.c. source. Draw the circuit diagram of the arrangement and determine the:\n(i) charge on either plate of each capacitor\n(ii) potential difference across each capacitor\n(iii) energy of the combinad capacitors.', '(a) Explain what is meant by the statement: The capacitance of a parallel-plate capacitor is 2\\(\\mu\\)F~(b) State: (i) three factors on which its capacitance depends\n(ii) three uses of capacitors.\n(c) Derive a formula for the energy W stored in a charged capacitor of capacitance C carrying a charge Q on either plate.,\n(d) Two parallel-plate capacitors of capacitances 2\\(\\mu\\)F and 3\\(\\mu\\)F are connected in parallel and the combination is connected to a 50V d.c. source. Draw the circuit diagram of the arrangement and determine the:\n(i) charge on either plate of each capacitor\n(ii) potential difference across each capacitor\n(iii) energy of the combinad capacitors.~(c) Derive a formula for the energy W stored in a charged capacitor of capacitance C carrying a charge Q on either plate.,\n(d) Two parallel-plate capacitors of capacitances 2\\(\\mu\\)F and 3\\(\\mu\\)F are connected in parallel and the combination is connected to a 50V d.c. source. Draw the circuit diagram of the arrangement and determine the:\n(i) charge on either plate of each capacitor\n(ii) potential difference across each capacitor\n(iii) energy of the combinad capacitors.~(d) Two parallel-plate capacitors of capacitances 2\\(\\mu\\)F and 3\\(\\mu\\)F are connected in parallel and the combination is connected to a 50V d.c. source. Draw the circuit diagram of the arrangement and determine the:\n(i) charge on either plate of each capacitor\n(ii) potential difference across each capacitor\n(iii) energy of the combinad capacitors.', NULL, '\n\nExplanation\n3(a) This means that the ratio of the charge on the capacitor to the potential difference between the plates is 4LF.\n(b)(i) (1) Area of plates\n(2) Dielectric between plates\n(3) Distance between plates\n\n(b)(ii) (1) To separate a.c from d.c.\n(2) To control current in a.c circuit\n(3) To store charges or energy\n(c) Workdone, W = QV\nAverage work done is given by \\(\\frac{1}{2}\\) x QV .........(i)\nBut V = \\(\\frac{Q}{V}\\) ...............(ii)\nSubstituting (ii) in (i)\nW = \\(\\frac{1}{2}\\) x Q x \\(\\frac{Q}{C}\\)\nEnergy stored W = \\(\\frac{1}{2} \\times \\frac{Q^2}{2} \\frac{1}{2} \\frac{Q^2}{C}\\)\n\n(i) Q\\(_2 = C_2V\\)\n= 100 x 10\\(^{-6}\\) C = 100\\(mu\\)C\n\\(Q_1 = C_1 V\\)\n= 3 10\\(^{-6}\\) x 50 = 150\\(\\mu\\)C\n(iii) P.d across each capacitor is 50V\n(iii) E = \\(\\frac{1}{2} CV^2\\) Since capacitors are in parallel\nC = \\(\\frac{1}{2}\\) x 5 \\(\\mu\\)F x 50 x 50\n= \\(\\frac{1}{2}\\) x 5 x 10\\(^{-6}\\) x 50 x 50 = 6.25 x 10\\(^{-3}\\)J\n\n\n\n', '', 'Explanation\n3(a) This means that the ratio of the charge on the capacitor to the potential difference between the plates is 4LF.\n(b)(i) (1) Area of plates\n(2) Dielectric between plates\n(3) Distance between plates\n\n(b)(ii) (1) To separate a.c from d.c.\n(2) To control current in a.c circuit\n(3) To store charges or energy\n(c) Workdone, W = QV\nAverage work done is given by \\(\\frac{1}{2}\\) x QV .........(i)\nBut V = \\(\\frac{Q}{V}\\) ...............(ii)\nSubstituting (ii) in (i)\nW = \\(\\frac{1}{2}\\) x Q x \\(\\frac{Q}{C}\\)\nEnergy stored W = \\(\\frac{1}{2} \\times \\frac{Q^2}{2} \\frac{1}{2} \\frac{Q^2}{C}\\)\n\n(i) Q\\(_2 = C_2V\\)\n= 100 x 10\\(^{-6}\\) C = 100\\(mu\\)C\n\\(Q_1 = C_1 V\\)\n= 3 10\\(^{-6}\\) x 50 = 150\\(\\mu\\)C\n(iii) P.d across each capacitor is 50V\n(iii) E = \\(\\frac{1}{2} CV^2\\) Since capacitors are in parallel\nC = \\(\\frac{1}{2}\\) x 5 \\(\\mu\\)F x 50 x 50\n= \\(\\frac{1}{2}\\) x 5 x 10\\(^{-6}\\) x 50 x 50 = 6.25 x 10\\(^{-3}\\)J\n\n\n\n', '(a) This means that the ratio of the charge on the capacitor to the potential difference between the plates is 4LF.\n(b)(i) (1) Area of plates\n(2) Dielectric between plates\n(3) Distance between plates\n~(b)(i) (1) Area of plates\n(2) Dielectric between plates\n(3) Distance between plates\n\n(b)(ii) (1) To separate a.c from d.c.\n(2) To control current in a.c circuit\n(3) To store charges or energy\n(c) Workdone, W = QV\nAverage work done is given by \\(\\frac{1}{2}\\) x QV .........(i)\nBut V = \\(\\frac{Q}{V}\\) ...............(ii)\nSubstituting (ii) in (i)\nW = \\(\\frac{1}{2}\\) x Q x \\(\\frac{Q}{C}\\)\nEnergy stored W = \\(\\frac{1}{2} \\times \\frac{Q^2}{2} \\frac{1}{2} \\frac{Q^2}{C}\\)\n\n(i) Q\\(_2 = C_2V\\)\n= 100 x 10\\(^{-6}\\) C = 100\\(mu\\)C\n\\(Q_1 = C_1 V\\)\n= 3 10\\(^{-6}\\) x 50 = 150\\(\\mu\\)C\n(iii) P.d across each capacitor is 50V\n(iii) E = \\(\\frac{1}{2} CV^2\\) Since capacitors are in parallel\nC = \\(\\frac{1}{2}\\) x 5 \\(\\mu\\)F x 50 x 50\n= \\(\\frac{1}{2}\\) x 5 x 10\\(^{-6}\\) x 50 x 50 = 6.25 x 10\\(^{-3}\\)J\n\n\n~(c) Workdone, W = QV\nAverage work done is given by \\(\\frac{1}{2}\\) x QV .........(i)\nBut V = \\(\\frac{Q}{V}\\) ...............(ii)\nSubstituting (ii) in (i)\nW = \\(\\frac{1}{2}\\) x Q x \\(\\frac{Q}{C}\\)\nEnergy stored W = \\(\\frac{1}{2} \\times \\frac{Q^2}{2} \\frac{1}{2} \\frac{Q^2}{C}\\)\n\n(i) Q\\(_2 = C_2V\\)\n= 100 x 10\\(^{-6}\\) C = 100\\(mu\\)C\n\\(Q_1 = C_1 V\\)\n= 3 10\\(^{-6}\\) x 50 = 150\\(\\mu\\)C\n(iii) P.d across each capacitor is 50V\n(iii) E = \\(\\frac{1}{2} CV^2\\) Since capacitors are in parallel\nC = \\(\\frac{1}{2}\\) x 5 \\(\\mu\\)F x 50 x 50\n= \\(\\frac{1}{2}\\) x 5 x 10\\(^{-6}\\) x 50 x 50 = 6.25 x 10\\(^{-3}\\)J\n\n\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(284, '2019', '\n\nQuestion\n\n(a) Explain the following terms:\n(i) viscosity;\n(ii) terminal velocity.\n(b)(i) Describe an experiment to determine the terminal velocity of a steel ball falling through a jar of glycerine..\n(ii) State two precautions that should be taken to ensure accurate result.\n(c) State two\n(i) effects of viscosity\n(ii) applications of viscosity.\n\n', '', '(a) Explain the following terms:\n(i) viscosity;\n(ii) terminal velocity.\n(b)(i) Describe an experiment to determine the terminal velocity of a steel ball falling through a jar of glycerine..\n(ii) State two precautions that should be taken to ensure accurate result.\n(c) State two\n(i) effects of viscosity\n(ii) applications of viscosity.', '(a) Explain the following terms:\n(i) viscosity;\n(ii) terminal velocity.~(b)(i) Describe an experiment to determine the terminal velocity of a steel ball falling through a jar of glycerine..\n(ii) State two precautions that should be taken to ensure accurate result.\n(c) State two\n(i) effects of viscosity\n(ii) applications of viscosity.~(c) State two\n(i) effects of viscosity\n(ii) applications of viscosity.', NULL, '\n\nExplanation\n(a)(i) Viscosity is the property of a fluid which tends to prevent or oppose motion of one layer or part over another. It varies with the temperature and nature of the fluid. (ii) A body falling through a fluid eventually attains a steady velocity called terminal velocity. At this point the weight of the body equals the upthrust and viscous force. (b)(i)\n\n\nDescription: A tall glass jar is filled with glycerine anJ supported vertically. Bands (acting as markers) are attached to points A.B.C.D.E and F at equal intervals. The ball is allowed to fall from the top of the jar while the time is noted as the ball passed each marker in turn. It will be observed that the time intervals between A and B, B and C are different while the time intervals for C and D, D and E, E and F are equal. Showing that a steady terminal velocity is reached from C. This ;Weans that\n\\(\\frac{CD}{t} = \\frac{DE}{t} = \\frac{EF}{t}\\) is the terminal velocity.\n(ii) Precautions:\n(1) Drop the ball gently.\n(2) Ball must fall centrally down the glycerine\n\n(c)(i) (1) Viscosity is responsible for the different rates of flow of fluid\n(2) Viscosity affects motion of bodies in fluids.\n\n(ii) (1) Viscous liquids are used as lubricants\n(2) The knowledge of viscous drag is applied in the design of ships and aircraft.\n\n', '', 'Explanation\n(a)(i) Viscosity is the property of a fluid which tends to prevent or oppose motion of one layer or part over another. It varies with the temperature and nature of the fluid. (ii) A body falling through a fluid eventually attains a steady velocity called terminal velocity. At this point the weight of the body equals the upthrust and viscous force. (b)(i)\n\n\nDescription: A tall glass jar is filled with glycerine anJ supported vertically. Bands (acting as markers) are attached to points A.B.C.D.E and F at equal intervals. The ball is allowed to fall from the top of the jar while the time is noted as the ball passed each marker in turn. It will be observed that the time intervals between A and B, B and C are different while the time intervals for C and D, D and E, E and F are equal. Showing that a steady terminal velocity is reached from C. This ;Weans that\n\\(\\frac{CD}{t} = \\frac{DE}{t} = \\frac{EF}{t}\\) is the terminal velocity.\n(ii) Precautions:\n(1) Drop the ball gently.\n(2) Ball must fall centrally down the glycerine\n\n(c)(i) (1) Viscosity is responsible for the different rates of flow of fluid\n(2) Viscosity affects motion of bodies in fluids.\n\n(ii) (1) Viscous liquids are used as lubricants\n(2) The knowledge of viscous drag is applied in the design of ships and aircraft.\n\n', '~(b)(i)\n\n\nDescription: A tall glass jar is filled with glycerine anJ supported vertically. Bands (acting as markers) are attached to points A.B.C.D.E and F at equal intervals. The ball is allowed to fall from the top of the jar while the time is noted as the ball passed each marker in turn. It will be observed that the time intervals between A and B, B and C are different while the time intervals for C and D, D and E, E and F are equal. Showing that a steady terminal velocity is reached from C. This ;Weans that\n\\(\\frac{CD}{t} = \\frac{DE}{t} = \\frac{EF}{t}\\) is the terminal velocity.\n(ii) Precautions:\n(1) Drop the ball gently.\n(2) Ball must fall centrally down the glycerine\n\n(c)(i) (1) Viscosity is responsible for the different rates of flow of fluid\n(2) Viscosity affects motion of bodies in fluids.\n\n(ii) (1) Viscous liquids are used as lubricants\n(2) The knowledge of viscous drag is applied in the design of ships and aircraft.\n~(c)(i) (1) Viscosity is responsible for the different rates of flow of fluid\n(2) Viscosity affects motion of bodies in fluids.\n\n(ii) (1) Viscous liquids are used as lubricants\n(2) The knowledge of viscous drag is applied in the design of ships and aircraft.\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(285, '2019', '\n\nQuestion\n\n(a) What is meant by the statement: The linear expansivity of a solid is \\(1.0 \\times10^{-5}K^{-1}\\)?\n(b)(i) Describe an experiment to determine the linear expansivity of a steel rod. (ii) Steel bars, each of length 3m at 29°C are to be used for constructing a rail line. If the linear expansivity of steel is \\(1.0 \\times10^{-5}K^{-1}\\), calculate the safety gap that must be left between successive bars if the highest temperature expected is 41°C.\n(c) State three advantages and two disadvantages of thermal expansion of solids.\n\n', '', '(a) What is meant by the statement: The linear expansivity of a solid is \\(1.0 \\times10^{-5}K^{-1}\\)?\n(b)(i) Describe an experiment to determine the linear expansivity of a steel rod. (ii) Steel bars, each of length 3m at 29°C are to be used for constructing a rail line. If the linear expansivity of steel is \\(1.0 \\times10^{-5}K^{-1}\\), calculate the safety gap that must be left between successive bars if the highest temperature expected is 41°C.\n(c) State three advantages and two disadvantages of thermal expansion of solids.', '(a) What is meant by the statement: The linear expansivity of a solid is \\(1.0 \\times10^{-5}K^{-1}\\)?~(b)(i) Describe an experiment to determine the linear expansivity of a steel rod. (ii) Steel bars, each of length 3m at 29°C are to be used for constructing a rail line. If the linear expansivity of steel is \\(1.0 \\times10^{-5}K^{-1}\\), calculate the safety gap that must be left between successive bars if the highest temperature expected is 41°C.\n(c) State three advantages and two disadvantages of thermal expansion of solids.~(c) State three advantages and two disadvantages of thermal expansion of solids.', NULL, '\n\nExplanation\n(a) The statement: The linear expansivity of a solid is \\(1.0 \\times10^{-5}K^{-1}\\)means that, A unit length of the solid will expand in length by a fraction \\(1.0 \\times10^{-5}\\) of the original per kelvin rise in temperature.\n(Note: \\(1.0 \\times10-5 = \\frac{1}{100,000}\\), a fraction).\n(b) Experiment to determine the linear expansivity of a steel rod.\n\nThe apparatus is set up as shown above. The length of the steel rod used is first measured with a meter rule and it is then placed in the jacket with one end fixed. The micrometer screw gauge is turned until the end of the screw just touches the end of the rod, and the gauge reading is noted. It is then turned back enough to make sure that the end of the rod can expand freely. The reading on the thermometer is noted and steam is now passed through the jacket for several minutes. The micrometer is used to note the new reading. More steam is passed again and the reading is taken until a constant reading is obtained. The reading is noted and the temperature is noted from the thermometer.\nThe formula \\(\\alpha = \\frac{l_{f} - l_{i}}{l_{i} (\\theta_{f} - \\theta_{i})}\\) is used to determine the linear expansivity.\n\\(l_{f} - \\text{final length} ; l_{i} - \\text{initial length} ; \\theta_{f} - \\text{final temperature} ; \\theta_{i} - \\text{initial temperature}\\)\n(b)(ii) \\(l = 3m ; \\theta_{i} = 29°C ; \\theta_{f} = 41°C\\)\n\\(\\alpha = 1.0 \\times 10^{-5} K^{-1}\\)\nSafety gap = change in length.\n\\(\\Delta l = \\alpha l \\Delta \\theta\\)\n= \\(1 \\times 10^{-5} \\times 3.0 \\times (41 - 29)\\)\n= \\(36 \\times 10^{-5} = 0.00036m\\)\n(c) Advantages:\n(1) Used for marking bimetallic strip which is used in thermostat. (2) Removal of tight glass stopper (3) Red-hot rivets in ship building. (4) Expansion of metals is used in bimetallic thermometer.\nDisadvantages:\n(1) Expansion of metals or concrete bridges. (2) Cracking of glass cup when hot water is poured into the glass cup. (3) Expansion of balance wheel or wrist-watch. (4) Sagging of overhead wire. (5) Expansion of railway lines.\n\n', '', 'Explanation\n(a) The statement: The linear expansivity of a solid is \\(1.0 \\times10^{-5}K^{-1}\\)means that, A unit length of the solid will expand in length by a fraction \\(1.0 \\times10^{-5}\\) of the original per kelvin rise in temperature.\n(Note: \\(1.0 \\times10-5 = \\frac{1}{100,000}\\), a fraction).\n(b) Experiment to determine the linear expansivity of a steel rod.\n\nThe apparatus is set up as shown above. The length of the steel rod used is first measured with a meter rule and it is then placed in the jacket with one end fixed. The micrometer screw gauge is turned until the end of the screw just touches the end of the rod, and the gauge reading is noted. It is then turned back enough to make sure that the end of the rod can expand freely. The reading on the thermometer is noted and steam is now passed through the jacket for several minutes. The micrometer is used to note the new reading. More steam is passed again and the reading is taken until a constant reading is obtained. The reading is noted and the temperature is noted from the thermometer.\nThe formula \\(\\alpha = \\frac{l_{f} - l_{i}}{l_{i} (\\theta_{f} - \\theta_{i})}\\) is used to determine the linear expansivity.\n\\(l_{f} - \\text{final length} ; l_{i} - \\text{initial length} ; \\theta_{f} - \\text{final temperature} ; \\theta_{i} - \\text{initial temperature}\\)\n(b)(ii) \\(l = 3m ; \\theta_{i} = 29°C ; \\theta_{f} = 41°C\\)\n\\(\\alpha = 1.0 \\times 10^{-5} K^{-1}\\)\nSafety gap = change in length.\n\\(\\Delta l = \\alpha l \\Delta \\theta\\)\n= \\(1 \\times 10^{-5} \\times 3.0 \\times (41 - 29)\\)\n= \\(36 \\times 10^{-5} = 0.00036m\\)\n(c) Advantages:\n(1) Used for marking bimetallic strip which is used in thermostat. (2) Removal of tight glass stopper (3) Red-hot rivets in ship building. (4) Expansion of metals is used in bimetallic thermometer.\nDisadvantages:\n(1) Expansion of metals or concrete bridges. (2) Cracking of glass cup when hot water is poured into the glass cup. (3) Expansion of balance wheel or wrist-watch. (4) Sagging of overhead wire. (5) Expansion of railway lines.\n\n', '(a) The statement: The linear expansivity of a solid is \\(1.0 \\times10^{-5}K^{-1}\\)means that, A unit length of the solid will expand in length by a fraction \\(1.0 \\times10^{-5}\\) of the original per kelvin rise in temperature.\n(Note: \\(1.0 \\times10-5 = \\frac{1}{100,000}\\), a fraction).\n(b) Experiment to determine the linear expansivity of a steel rod.\n\nThe apparatus is set up as shown above. The length of the steel rod used is first measured with a meter rule and it is then placed in the jacket with one end fixed. The micrometer screw gauge is turned until the end of the screw just touches the end of the rod, and the gauge reading is noted. It is then turned back enough to make sure that the end of the rod can expand freely. The reading on the thermometer is noted and steam is now passed through the jacket for several minutes. The micrometer is used to note the new reading. More steam is passed again and the reading is taken until a constant reading is obtained. The reading is noted and the temperature is noted from the thermometer.\nThe formula \\(\\alpha = \\frac{l_{f} - l_{i}}{l_{i} (\\theta_{f} - \\theta_{i})}\\) is used to determine the linear expansivity.\n\\(l_{f} - \\text{final length} ; l_{i} - \\text{initial length} ; \\theta_{f} - \\text{final temperature} ; \\theta_{i} - \\text{initial temperature}\\)~(b) Experiment to determine the linear expansivity of a steel rod.\n\nThe apparatus is set up as shown above. The length of the steel rod used is first measured with a meter rule and it is then placed in the jacket with one end fixed. The micrometer screw gauge is turned until the end of the screw just touches the end of the rod, and the gauge reading is noted. It is then turned back enough to make sure that the end of the rod can expand freely. The reading on the thermometer is noted and steam is now passed through the jacket for several minutes. The micrometer is used to note the new reading. More steam is passed again and the reading is taken until a constant reading is obtained. The reading is noted and the temperature is noted from the thermometer.\nThe formula \\(\\alpha = \\frac{l_{f} - l_{i}}{l_{i} (\\theta_{f} - \\theta_{i})}\\) is used to determine the linear expansivity.\n\\(l_{f} - \\text{final length} ; l_{i} - \\text{initial length} ; \\theta_{f} - \\text{final temperature} ; \\theta_{i} - \\text{initial temperature}\\)\n(b)(ii) \\(l = 3m ; \\theta_{i} = 29°C ; \\theta_{f} = 41°C\\)\n\\(\\alpha = 1.0 \\times 10^{-5} K^{-1}\\)\nSafety gap = change in length.\n\\(\\Delta l = \\alpha l \\Delta \\theta\\)\n= \\(1 \\times 10^{-5} \\times 3.0 \\times (41 - 29)\\)\n= \\(36 \\times 10^{-5} = 0.00036m\\)\n(c) Advantages:\n(1) Used for marking bimetallic strip which is used in thermostat. (2) Removal of tight glass stopper (3) Red-hot rivets in ship building. (4) Expansion of metals is used in bimetallic thermometer.\nDisadvantages:\n(1) Expansion of metals or concrete bridges. (2) Cracking of glass cup when hot water is poured into the glass cup. (3) Expansion of balance wheel or wrist-watch. (4) Sagging of overhead wire. (5) Expansion of railway lines.\n~(c) Advantages:\n(1) Used for marking bimetallic strip which is used in thermostat. (2) Removal of tight glass stopper (3) Red-hot rivets in ship building. (4) Expansion of metals is used in bimetallic thermometer.\nDisadvantages:\n(1) Expansion of metals or concrete bridges. (2) Cracking of glass cup when hot water is poured into the glass cup. (3) Expansion of balance wheel or wrist-watch. (4) Sagging of overhead wire. (5) Expansion of railway lines.\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
(286, '2019', '\n\nQuestion\n\n(a) What is a wave motion?\nThe equation \\(y = A \\sin \\frac{2\\pi}{\\lambda}(Vt-X)\\) represents a wavetrain in which y is the vertical displacement of a particle at distance X from the origin in the medium through which the wave is travelling. Explain, with the aid of a diagram, what A and \\(\\lambda\\)represent.\n(b) (i) Describe an experiment to determine the frequency of a note emitted by a source of sound\n(ii) A pipe closed at one end is 1 m long. The air in the pipe is set into vibration and a fundamental note is produced. If the velocity of sound in air is 340ms\\(^{-1}\\), calculate the frequency of the note\n(c) State two differences between a sound wave and a radio wave.\n\n', '', '(a) What is a wave motion?\nThe equation \\(y = A \\sin \\frac{2\\pi}{\\lambda}(Vt-X)\\) represents a wavetrain in which y is the vertical displacement of a particle at distance X from the origin in the medium through which the wave is travelling. Explain, with the aid of a diagram, what A and \\(\\lambda\\)represent.\n(b) (i) Describe an experiment to determine the frequency of a note emitted by a source of sound\n(ii) A pipe closed at one end is 1 m long. The air in the pipe is set into vibration and a fundamental note is produced. If the velocity of sound in air is 340ms\\(^{-1}\\), calculate the frequency of the note\n(c) State two differences between a sound wave and a radio wave.', '(a) What is a wave motion?\nThe equation \\(y = A \\sin \\frac{2\\pi}{\\lambda}(Vt-X)\\) represents a wavetrain in which y is the vertical displacement of a particle at distance X from the origin in the medium through which the wave is travelling. Explain, with the aid of a diagram, what A and \\(\\lambda\\)represent.~(b) (i) Describe an experiment to determine the frequency of a note emitted by a source of sound\n(ii) A pipe closed at one end is 1 m long. The air in the pipe is set into vibration and a fundamental note is produced. If the velocity of sound in air is 340ms\\(^{-1}\\), calculate the frequency of the note\n(c) State two differences between a sound wave and a radio wave.~(c) State two differences between a sound wave and a radio wave.', NULL, '\n\nExplanation\n(a) A wave motion is succession of pulses propagated with time from one region of space to another. Energy is transmitted in the process.\n\nA is the amplitude. It is the maximum displacement from equilibrium position of the wave particle. \\(\\lambda\\)- the wavelength, is the distance between two successivepoints on the wave that is in phase.\n(b) Experiment to determine the frequency of a note emitted by a source of sound\n\nAssuming that the source of sound isa turning fork. The vibrating tube turning fork is held close to the top of a moveable tube as shown. The tube is deeped in a cylinder that is 3/4 filled with water. Starting with the air column of length, I,with the turning fork sounding gently close to the top, the length I is gradually increased; The, sound coming from the air column is heard to increase. When the sound is loudest the air has been set into resonance by the fork. The length of I of the air column ismeasured and recorded. The length of the air column can be shown to be \\(\\frac{\\lambda}{4}\\)neglecting end correction. Thus using the equation:\n\\(f = \\frac{V}{\\lambda} = \\frac{V}{4l}\\)\nwhere V is the velocity of sound in air, f the frequency of the turning-fork the source of sound, the frequency is then determined.\n(ii) Using \\(f = \\frac{V}{\\lambda} = \\frac{V}{4l}\\)\n= \\(\\frac{340}{4 \\times 1} = 85 Hz\\)\n(c) A sound wave :\n(i) requires material medium for propagation. (ii) is a longitudinal wave motion. While radio wave (i) does not require material medium for propagation. (ii) is a transverse wave motion.\n\n', '', 'Explanation\n(a) A wave motion is succession of pulses propagated with time from one region of space to another. Energy is transmitted in the process.\n\nA is the amplitude. It is the maximum displacement from equilibrium position of the wave particle. \\(\\lambda\\)- the wavelength, is the distance between two successivepoints on the wave that is in phase.\n(b) Experiment to determine the frequency of a note emitted by a source of sound\n\nAssuming that the source of sound isa turning fork. The vibrating tube turning fork is held close to the top of a moveable tube as shown. The tube is deeped in a cylinder that is 3/4 filled with water. Starting with the air column of length, I,with the turning fork sounding gently close to the top, the length I is gradually increased; The, sound coming from the air column is heard to increase. When the sound is loudest the air has been set into resonance by the fork. The length of I of the air column ismeasured and recorded. The length of the air column can be shown to be \\(\\frac{\\lambda}{4}\\)neglecting end correction. Thus using the equation:\n\\(f = \\frac{V}{\\lambda} = \\frac{V}{4l}\\)\nwhere V is the velocity of sound in air, f the frequency of the turning-fork the source of sound, the frequency is then determined.\n(ii) Using \\(f = \\frac{V}{\\lambda} = \\frac{V}{4l}\\)\n= \\(\\frac{340}{4 \\times 1} = 85 Hz\\)\n(c) A sound wave :\n(i) requires material medium for propagation. (ii) is a longitudinal wave motion. While radio wave (i) does not require material medium for propagation. (ii) is a transverse wave motion.\n\n', '(a) A wave motion is succession of pulses propagated with time from one region of space to another. Energy is transmitted in the process.\n\nA is the amplitude. It is the maximum displacement from equilibrium position of the wave particle. \\(\\lambda\\)- the wavelength, is the distance between two successivepoints on the wave that is in phase.~(b) Experiment to determine the frequency of a note emitted by a source of sound\n\nAssuming that the source of sound isa turning fork. The vibrating tube turning fork is held close to the top of a moveable tube as shown. The tube is deeped in a cylinder that is 3/4 filled with water. Starting with the air column of length, I,with the turning fork sounding gently close to the top, the length I is gradually increased; The, sound coming from the air column is heard to increase. When the sound is loudest the air has been set into resonance by the fork. The length of I of the air column ismeasured and recorded. The length of the air column can be shown to be \\(\\frac{\\lambda}{4}\\)neglecting end correction. Thus using the equation:\n\\(f = \\frac{V}{\\lambda} = \\frac{V}{4l}\\)\nwhere V is the velocity of sound in air, f the frequency of the turning-fork the source of sound, the frequency is then determined.\n(ii) Using \\(f = \\frac{V}{\\lambda} = \\frac{V}{4l}\\)\n= \\(\\frac{340}{4 \\times 1} = 85 Hz\\)\n(c) A sound wave :\n(i) requires material medium for propagation. (ii) is a longitudinal wave motion. While radio wave (i) does not require material medium for propagation. (ii) is a transverse wave motion.\n~(c) A sound wave :\n(i) requires material medium for propagation. (ii) is a longitudinal wave motion. While radio wave (i) does not require material medium for propagation. (ii) is a transverse wave motion.\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(287, '2019', '\n\nQuestion\n\n(a) State the laws of electromagnetic induction.\n(b) (i) Describe a simple experiment to show how an induced e.m.f, can be produced;(ii) State two factorson which the magnitude of the induced e.m.f. depends\n(c) Explain what is meant by the r.m.s. value of an alternating current\n(d) (i) If the alternating current is represented by \\(I = l_{o}\\sin \\omega t\\), state what the symbol \\(I, I_{o}, \\omega\\) and \\(\\omega t\\)represent.\n(ii) Calculate the instantaneous value of such a current, if in a circuit it has r.m.s value of 15.0A when its phase angle is 30°.\n\n', '', '(a) State the laws of electromagnetic induction.\n(b) (i) Describe a simple experiment to show how an induced e.m.f, can be produced;(ii) State two factorson which the magnitude of the induced e.m.f. depends\n(c) Explain what is meant by the r.m.s. value of an alternating current\n(d) (i) If the alternating current is represented by \\(I = l_{o}\\sin \\omega t\\), state what the symbol \\(I, I_{o}, \\omega\\) and \\(\\omega t\\)represent.\n(ii) Calculate the instantaneous value of such a current, if in a circuit it has r.m.s value of 15.0A when its phase angle is 30°.', '(a) State the laws of electromagnetic induction.~(b) (i) Describe a simple experiment to show how an induced e.m.f, can be produced;(ii) State two factorson which the magnitude of the induced e.m.f. depends\n(c) Explain what is meant by the r.m.s. value of an alternating current\n(d) (i) If the alternating current is represented by \\(I = l_{o}\\sin \\omega t\\), state what the symbol \\(I, I_{o}, \\omega\\) and \\(\\omega t\\)represent.\n(ii) Calculate the instantaneous value of such a current, if in a circuit it has r.m.s value of 15.0A when its phase angle is 30°.~(c) Explain what is meant by the r.m.s. value of an alternating current\n(d) (i) If the alternating current is represented by \\(I = l_{o}\\sin \\omega t\\), state what the symbol \\(I, I_{o}, \\omega\\) and \\(\\omega t\\)represent.\n(ii) Calculate the instantaneous value of such a current, if in a circuit it has r.m.s value of 15.0A when its phase angle is 30°.~(d) (i) If the alternating current is represented by \\(I = l_{o}\\sin \\omega t\\), state what the symbol \\(I, I_{o}, \\omega\\) and \\(\\omega t\\)represent.\n(ii) Calculate the instantaneous value of such a current, if in a circuit it has r.m.s value of 15.0A when its phase angle is 30°.', NULL, '\n\nExplanation\n(a) The laws of electromagnetic induction states that\n(1) whenever there is a change in the magnetic flux linked with a circuit, an e.m.f. is induced, the strength of which is proportional to the rate of change of the flux linked with the circuit. The above law is sometimes called faraday\'s law of electromagnetic induction.\n(2) The direction of the induced current is always such as to oppose the change producing it. This is called Lenz\'s law.\nb(i) Experiment to show how an induced e.m.f. can be produced.\n\nA coil with many turns is connected to centre-scale zero galvanometer G. As the bar magnet is moved rapidly close and through the turns, an induced e.m.f. is produced. The galvanometer pointer is deflected showing a flow of current. The same effects could be observed if the coil is moving towards the magnet instead. Or as long as there is relative motion between the two.\n(ii) The magnitude of the induced e.m.f. depends on:\n(1) The magnitude of flux linking the coil. (2) The relative speed between the magnet and the coil.\n(c) The root-mean-square (r.m.s.) value of an alternating current is the steady or direct current which produces the same heating effect per secoridin a given resistor. Its relationship with the peakvalue of alternating current \\(I_{o}\\)is given by\n\\(I_{r.m.s} = \\frac{1}{\\sqrt{2}} I_{o}\\)\nd(i) \\(I\\) is the instantaneous current\n\\(I_{o}\\)is the peak value of the current\n\\(\\alpha\\)is the angular frequency\n\\(\\alpha t\\) is the phase or the current (since it is wave-like).\n(ii) \\(I = I_{o} \\sin \\alpha t\\)\n= \\(I_{r.m.s} \\sqrt{2} \\sin 30°\\)\n= \\(\\sqrt{2} \\times 15 \\times \\frac{1}{2} = 10.6A\\)\n\n', '', 'Explanation\n(a) The laws of electromagnetic induction states that\n(1) whenever there is a change in the magnetic flux linked with a circuit, an e.m.f. is induced, the strength of which is proportional to the rate of change of the flux linked with the circuit. The above law is sometimes called faraday\'s law of electromagnetic induction.\n(2) The direction of the induced current is always such as to oppose the change producing it. This is called Lenz\'s law.\nb(i) Experiment to show how an induced e.m.f. can be produced.\n\nA coil with many turns is connected to centre-scale zero galvanometer G. As the bar magnet is moved rapidly close and through the turns, an induced e.m.f. is produced. The galvanometer pointer is deflected showing a flow of current. The same effects could be observed if the coil is moving towards the magnet instead. Or as long as there is relative motion between the two.\n(ii) The magnitude of the induced e.m.f. depends on:\n(1) The magnitude of flux linking the coil. (2) The relative speed between the magnet and the coil.\n(c) The root-mean-square (r.m.s.) value of an alternating current is the steady or direct current which produces the same heating effect per secoridin a given resistor. Its relationship with the peakvalue of alternating current \\(I_{o}\\)is given by\n\\(I_{r.m.s} = \\frac{1}{\\sqrt{2}} I_{o}\\)\nd(i) \\(I\\) is the instantaneous current\n\\(I_{o}\\)is the peak value of the current\n\\(\\alpha\\)is the angular frequency\n\\(\\alpha t\\) is the phase or the current (since it is wave-like).\n(ii) \\(I = I_{o} \\sin \\alpha t\\)\n= \\(I_{r.m.s} \\sqrt{2} \\sin 30°\\)\n= \\(\\sqrt{2} \\times 15 \\times \\frac{1}{2} = 10.6A\\)\n\n', '~(c) The root-mean-square (r.m.s.) value of an alternating current is the steady or direct current which produces the same heating effect per secoridin a given resistor. Its relationship with the peakvalue of alternating current \\(I_{o}\\)is given by\n\\(I_{r.m.s} = \\frac{1}{\\sqrt{2}} I_{o}\\)\nd(i) \\(I\\) is the instantaneous current\n\\(I_{o}\\)is the peak value of the current\n\\(\\alpha\\)is the angular frequency\n\\(\\alpha t\\) is the phase or the current (since it is wave-like).\n(ii) \\(I = I_{o} \\sin \\alpha t\\)\n= \\(I_{r.m.s} \\sqrt{2} \\sin 30°\\)\n= \\(\\sqrt{2} \\times 15 \\times \\frac{1}{2} = 10.6A\\)\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(288, '2019', '\n\nQuestion\n\n(a)(i) Explain the terms: photoelectric emission and threshold frequency; (ii) Einstein\'s photoelectric equation can be written as \\(E = hf - hf_{o}\\)What does each of the symbols used in the equation above represent?\n(b) Calculate the frequency of the proton whose energy is required to eject a surface electron with a kinetic energy of \\(1.97 \\times10^{-16} eV\\) if the work function of the metal is \\(1.33 \\times10^{-16}eV\\). \\((1 eV = 1.6 \\times 10^{-18}J; h = 6.60 \\times 10^{-34}JS)\\).\n(c) In a photoelectric cell, no electrons are emitted until the threshold frequency of light is reached. Explain what happens to the energy of the light before emission of electrons begins. State one factor that may affect the numbers of emitted electrons.\n(d) Explain what is meant by the duality of matter, illustrating your answer with observation phenomena.\n\n', '', '(a)(i) Explain the terms: photoelectric emission and threshold frequency; (ii) Einstein\'s photoelectric equation can be written as \\(E = hf - hf_{o}\\)What does each of the symbols used in the equation above represent?\n(b) Calculate the frequency of the proton whose energy is required to eject a surface electron with a kinetic energy of \\(1.97 \\times10^{-16} eV\\) if the work function of the metal is \\(1.33 \\times10^{-16}eV\\). \\((1 eV = 1.6 \\times 10^{-18}J; h = 6.60 \\times 10^{-34}JS)\\).\n(c) In a photoelectric cell, no electrons are emitted until the threshold frequency of light is reached. Explain what happens to the energy of the light before emission of electrons begins. State one factor that may affect the numbers of emitted electrons.\n(d) Explain what is meant by the duality of matter, illustrating your answer with observation phenomena.', '(a)(i) Explain the terms: photoelectric emission and threshold frequency; (ii) Einstein\'s photoelectric equation can be written as \\(E = hf - hf_{o}\\)What does each of the symbols used in the equation above represent?~(b) Calculate the frequency of the proton whose energy is required to eject a surface electron with a kinetic energy of \\(1.97 \\times10^{-16} eV\\) if the work function of the metal is \\(1.33 \\times10^{-16}eV\\). \\((1 eV = 1.6 \\times 10^{-18}J; h = 6.60 \\times 10^{-34}JS)\\).\n(c) In a photoelectric cell, no electrons are emitted until the threshold frequency of light is reached. Explain what happens to the energy of the light before emission of electrons begins. State one factor that may affect the numbers of emitted electrons.\n(d) Explain what is meant by the duality of matter, illustrating your answer with observation phenomena.~(c) In a photoelectric cell, no electrons are emitted until the threshold frequency of light is reached. Explain what happens to the energy of the light before emission of electrons begins. State one factor that may affect the numbers of emitted electrons.\n(d) Explain what is meant by the duality of matter, illustrating your answer with observation phenomena.~(d) Explain what is meant by the duality of matter, illustrating your answer with observation phenomena.', NULL, '\n\nExplanation\na(i) Photoelectric emission occurs when electrons are emitted from the surface of metal plate when it is illuminated by light of sufficient high frequency. Threshold frequency is the minimum frequency of the illuminating light which will just be sufficient to cause photoelectric emission. Below this frequency, electrons will not be emitted.\n(ii) In the Einstein\'s photoelectric equation \\(E = hf - hf_{0}\\),\nE - is the maximum kinetic energy of the emitted electron ;\nhf - is the energy of the illuminating light;\n\\(hf_{0}\\) - is the work function. It represents the minimum energy required to liberate electrons from the surface of the metal\n\\(f_{0}\\) - is the threshold frequency.\n(b) Using \\(E = hf - hf_{0}\\)\n\\(f = \\frac{E + hf_{0}}{h}\\)\n(d) The duality of matter means that matter can exhibit wave properties as well as have particle properties. This double nature of matter is general in nature e.g. particles like electrons can have wave properties. Waves such as X-rays can have particle properties. This is equally illustrated by photoelectric effect.\nDe Broglie equation for the waves carried with particle is\n\\(\\lambda = \\frac{h}{mv} = \\frac{h}{p}\\)\nmv = p is momentum of particle of mass m;h - planks constant.\n\n', '', 'Explanation\na(i) Photoelectric emission occurs when electrons are emitted from the surface of metal plate when it is illuminated by light of sufficient high frequency. Threshold frequency is the minimum frequency of the illuminating light which will just be sufficient to cause photoelectric emission. Below this frequency, electrons will not be emitted.\n(ii) In the Einstein\'s photoelectric equation \\(E = hf - hf_{0}\\),\nE - is the maximum kinetic energy of the emitted electron ;\nhf - is the energy of the illuminating light;\n\\(hf_{0}\\) - is the work function. It represents the minimum energy required to liberate electrons from the surface of the metal\n\\(f_{0}\\) - is the threshold frequency.\n(b) Using \\(E = hf - hf_{0}\\)\n\\(f = \\frac{E + hf_{0}}{h}\\)\n(d) The duality of matter means that matter can exhibit wave properties as well as have particle properties. This double nature of matter is general in nature e.g. particles like electrons can have wave properties. Waves such as X-rays can have particle properties. This is equally illustrated by photoelectric effect.\nDe Broglie equation for the waves carried with particle is\n\\(\\lambda = \\frac{h}{mv} = \\frac{h}{p}\\)\nmv = p is momentum of particle of mass m;h - planks constant.\n\n', '(i) Photoelectric emission occurs when electrons are emitted from the surface of metal plate when it is illuminated by light of sufficient high frequency. Threshold frequency is the minimum frequency of the illuminating light which will just be sufficient to cause photoelectric emission. Below this frequency, electrons will not be emitted.\n(ii) In the Einstein\'s photoelectric equation \\(E = hf - hf_{0}\\),\nE - is the maximum kinetic energy of the emitted electron ;\nhf - is the energy of the illuminating light;\n\\(hf_{0}\\) - is the work function. It represents the minimum energy required to liberate electrons from the surface of the metal\n\\(f_{0}\\) - is the threshold frequency.~(b) Using \\(E = hf - hf_{0}\\)\n\\(f = \\frac{E + hf_{0}}{h}\\)\n(d) The duality of matter means that matter can exhibit wave properties as well as have particle properties. This double nature of matter is general in nature e.g. particles like electrons can have wave properties. Waves such as X-rays can have particle properties. This is equally illustrated by photoelectric effect.\nDe Broglie equation for the waves carried with particle is\n\\(\\lambda = \\frac{h}{mv} = \\frac{h}{p}\\)\nmv = p is momentum of particle of mass m;h - planks constant.\n~(d) The duality of matter means that matter can exhibit wave properties as well as have particle properties. This double nature of matter is general in nature e.g. particles like electrons can have wave properties. Waves such as X-rays can have particle properties. This is equally illustrated by photoelectric effect.\nDe Broglie equation for the waves carried with particle is\n\\(\\lambda = \\frac{h}{mv} = \\frac{h}{p}\\)\nmv = p is momentum of particle of mass m;h - planks constant.\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(289, '2019', '\n\nQuestion\n\n(a) Using a suitable diagram, explain how the following can be obtained from a velocity-time graph (i) acceleration; (ii) retardation; (iii) total distance covered.\n(b) Show that the displacement of a body moving with uniform acceleration a is given by \\(s = ut + \\frac{1}{2} at^{2}\\)where u is the velocity of the body at time t = 0.\n(c) A particle moving in a straight line with uniform deceleration has a velocity of 40ms\\(^{-1}\\) at a point P, 20ms\\(^{-1}\\) at a point Q and comes to rest at a point R where QR = 50m. Calculate the: (i) distance PQ; (ii) time taken to cover PQ; (iii) time taken to cover PR.\n\n', '', '(a) Using a suitable diagram, explain how the following can be obtained from a velocity-time graph (i) acceleration; (ii) retardation; (iii) total distance covered.\n(b) Show that the displacement of a body moving with uniform acceleration a is given by \\(s = ut + \\frac{1}{2} at^{2}\\)where u is the velocity of the body at time t = 0.\n(c) A particle moving in a straight line with uniform deceleration has a velocity of 40ms\\(^{-1}\\) at a point P, 20ms\\(^{-1}\\) at a point Q and comes to rest at a point R where QR = 50m. Calculate the: (i) distance PQ; (ii) time taken to cover PQ; (iii) time taken to cover PR.', '(a) Using a suitable diagram, explain how the following can be obtained from a velocity-time graph (i) acceleration; (ii) retardation; (iii) total distance covered.~(b) Show that the displacement of a body moving with uniform acceleration a is given by \\(s = ut + \\frac{1}{2} at^{2}\\)where u is the velocity of the body at time t = 0.\n(c) A particle moving in a straight line with uniform deceleration has a velocity of 40ms\\(^{-1}\\) at a point P, 20ms\\(^{-1}\\) at a point Q and comes to rest at a point R where QR = 50m. Calculate the: (i) distance PQ; (ii) time taken to cover PQ; (iii) time taken to cover PR.~(c) A particle moving in a straight line with uniform deceleration has a velocity of 40ms\\(^{-1}\\) at a point P, 20ms\\(^{-1}\\) at a point Q and comes to rest at a point R where QR = 50m. Calculate the: (i) distance PQ; (ii) time taken to cover PQ; (iii) time taken to cover PR.', NULL, '\n\nExplanation\n(a)\nVelocity- time graph:\n(i) Acceleration : Gradient of AB in the graph above;(ii) retardation: Gradient of CD in the graph above; (iii) total distance covered is the area under the graph.\n(b) S = average velocity x time\n\\(S = (\\frac{u +v}{2}) \\times t ... (1)\\)\nbut \\(v = u + at ... (2)\\), substitute (2) in (1)\n\\(S = (\\frac{u + u + at}{2}) \\times t\\)\n\\(S = (\\frac{2u + at}{2}) \\times t \\)\n\\(S = ut + \\frac{1}{2} at^{2}\\)\n(c) P ----------- Q ----50m------- R\n  40 ms\\(^{-1}\\)   20 ms\\(^{-1}\\)\nSince the acceleration is uniform and using equation\n\\(v^{2} = u^{2} + 2as\\) for the motion along QR.\n\\(0 = 20^{2} -2a(50)\\) (retardation)\n\\(0 = 400 -100a \\implies 100a = 400\\)\n\\(a = 4ms^{-2}\\) (retardation, hence it is negative)\n\\(20^{2} = 40^{2} + 2(-4) s\\)\n\\(400 = 1600 - 8s \\implies 8s = 1200\\)\n\\(s = 150m\\)\n\n', '', 'Explanation\n(a)\nVelocity- time graph:\n(i) Acceleration : Gradient of AB in the graph above;(ii) retardation: Gradient of CD in the graph above; (iii) total distance covered is the area under the graph.\n(b) S = average velocity x time\n\\(S = (\\frac{u +v}{2}) \\times t ... (1)\\)\nbut \\(v = u + at ... (2)\\), substitute (2) in (1)\n\\(S = (\\frac{u + u + at}{2}) \\times t\\)\n\\(S = (\\frac{2u + at}{2}) \\times t \\)\n\\(S = ut + \\frac{1}{2} at^{2}\\)\n(c) P ----------- Q ----50m------- R\n  40 ms\\(^{-1}\\)   20 ms\\(^{-1}\\)\nSince the acceleration is uniform and using equation\n\\(v^{2} = u^{2} + 2as\\) for the motion along QR.\n\\(0 = 20^{2} -2a(50)\\) (retardation)\n\\(0 = 400 -100a \\implies 100a = 400\\)\n\\(a = 4ms^{-2}\\) (retardation, hence it is negative)\n\\(20^{2} = 40^{2} + 2(-4) s\\)\n\\(400 = 1600 - 8s \\implies 8s = 1200\\)\n\\(s = 150m\\)\n\n', '(a)\nVelocity- time graph:\n(i) Acceleration : Gradient of AB in the graph above;(ii) retardation: Gradient of CD in the graph above; (iii) total distance covered is the area under the graph.~(b) S = average velocity x time\n\\(S = (\\frac{u +v}{2}) \\times t ... (1)\\)\nbut \\(v = u + at ... (2)\\), substitute (2) in (1)\n\\(S = (\\frac{u + u + at}{2}) \\times t\\)\n\\(S = (\\frac{2u + at}{2}) \\times t \\)\n\\(S = ut + \\frac{1}{2} at^{2}\\)\n(c) P ----------- Q ----50m------- R\n  40 ms\\(^{-1}\\)   20 ms\\(^{-1}\\)\nSince the acceleration is uniform and using equation\n\\(v^{2} = u^{2} + 2as\\) for the motion along QR.\n\\(0 = 20^{2} -2a(50)\\) (retardation)\n\\(0 = 400 -100a \\implies 100a = 400\\)\n\\(a = 4ms^{-2}\\) (retardation, hence it is negative)\n\\(20^{2} = 40^{2} + 2(-4) s\\)\n\\(400 = 1600 - 8s \\implies 8s = 1200\\)\n\\(s = 150m\\)\n~(c) P ----------- Q ----50m------- R\n  40 ms\\(^{-1}\\)   20 ms\\(^{-1}\\)\nSince the acceleration is uniform and using equation\n\\(v^{2} = u^{2} + 2as\\) for the motion along QR.\n\\(0 = 20^{2} -2a(50)\\) (retardation)\n\\(0 = 400 -100a \\implies 100a = 400\\)\n\\(a = 4ms^{-2}\\) (retardation, hence it is negative)\n\\(20^{2} = 40^{2} + 2(-4) s\\)\n\\(400 = 1600 - 8s \\implies 8s = 1200\\)\n\\(s = 150m\\)\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(290, '2019', '\n\nQuestion\n\n(a) Define the boiling point of a liquid.\n(b) Describe an experiment to determine the boiling point of small quantity of a liquid.\n(c) A piece of copper of mass 300 g at a temperature of 950°C is quickly transferred to 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 the water that will boil away.\n[Specific heat capacity of copper = \\(4.0 \\times10^{2} Jkg^{-1}K^{-1}\\); Specific heat capacity of water = \\(4.2 \\times10^{3} Jkg^{-1} K^{-1}\\);Specific latent heat of vaporization of steam = \\(2.26 \\times10^{6} Jkg^{-1}\\)\n(d) State four other effects of heat on a substance other than expansion.\n\n', '', '(a) Define the boiling point of a liquid.\n(b) Describe an experiment to determine the boiling point of small quantity of a liquid.\n(c) A piece of copper of mass 300 g at a temperature of 950°C is quickly transferred to 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 the water that will boil away.\n[Specific heat capacity of copper = \\(4.0 \\times10^{2} Jkg^{-1}K^{-1}\\); Specific heat capacity of water = \\(4.2 \\times10^{3} Jkg^{-1} K^{-1}\\);Specific latent heat of vaporization of steam = \\(2.26 \\times10^{6} Jkg^{-1}\\)\n(d) State four other effects of heat on a substance other than expansion.', '(a) Define the boiling point of a liquid.~(b) Describe an experiment to determine the boiling point of small quantity of a liquid.\n(c) A piece of copper of mass 300 g at a temperature of 950°C is quickly transferred to 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 the water that will boil away.\n[Specific heat capacity of copper = \\(4.0 \\times10^{2} Jkg^{-1}K^{-1}\\); Specific heat capacity of water = \\(4.2 \\times10^{3} Jkg^{-1} K^{-1}\\);Specific latent heat of vaporization of steam = \\(2.26 \\times10^{6} Jkg^{-1}\\)\n(d) State four other effects of heat on a substance other than expansion.~(c) A piece of copper of mass 300 g at a temperature of 950°C is quickly transferred to 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 the water that will boil away.\n[Specific heat capacity of copper = \\(4.0 \\times10^{2} Jkg^{-1}K^{-1}\\); Specific heat capacity of water = \\(4.2 \\times10^{3} Jkg^{-1} K^{-1}\\);Specific latent heat of vaporization of steam = \\(2.26 \\times10^{6} Jkg^{-1}\\)\n(d) State four other effects of heat on a substance other than expansion.~(d) State four other effects of heat on a substance other than expansion.', NULL, '\n\nExplanation\n(a) The boiling point of a liquid is the temperature at which its s.v.p is equal to the external atmospheric pressure.\n(b) Experiment to determine the boiling point of a small quantity of a liquid.\n\nSome liquid is introduced into the shorter length of the J - tube above the mercury. The beaker and its contents are heated and stirred continuously. At a stage when the levels of mercury in the two limbs of the J - tube are the same the temperature is read. At this temperature the s.v.p. of the liquid is equal to the atmospheric pressure. This temperature is the boiling point of the liquid.\n(c) Heat lost by copper = Heat gained by water\n\\(mc \\theta = m_{1} c_{1} \\theta_{1} + ml\\)\n\\(0.3 \\times 400 \\times (950° - 100°) = 0.25 \\times 4200 \\times (100° - 25°) + 2.26 \\times 10^{6} M\\)\n\\(M = \\frac{102000 - 78750}{2.26 \\times 10^{6}}\\)\n\\(M = 10.3g\\)\nTherefore, the mass of water that will boil away is 10.3g.\n(d) Four effects of heat other than expansion are:\n(1) Change of state (2) Change in temperature (3) Chemical change (4) Change in electrical resistance of conductivity. (5) Melting sublimation.\n\n', '', 'Explanation\n(a) The boiling point of a liquid is the temperature at which its s.v.p is equal to the external atmospheric pressure.\n(b) Experiment to determine the boiling point of a small quantity of a liquid.\n\nSome liquid is introduced into the shorter length of the J - tube above the mercury. The beaker and its contents are heated and stirred continuously. At a stage when the levels of mercury in the two limbs of the J - tube are the same the temperature is read. At this temperature the s.v.p. of the liquid is equal to the atmospheric pressure. This temperature is the boiling point of the liquid.\n(c) Heat lost by copper = Heat gained by water\n\\(mc \\theta = m_{1} c_{1} \\theta_{1} + ml\\)\n\\(0.3 \\times 400 \\times (950° - 100°) = 0.25 \\times 4200 \\times (100° - 25°) + 2.26 \\times 10^{6} M\\)\n\\(M = \\frac{102000 - 78750}{2.26 \\times 10^{6}}\\)\n\\(M = 10.3g\\)\nTherefore, the mass of water that will boil away is 10.3g.\n(d) Four effects of heat other than expansion are:\n(1) Change of state (2) Change in temperature (3) Chemical change (4) Change in electrical resistance of conductivity. (5) Melting sublimation.\n\n', '(a) The boiling point of a liquid is the temperature at which its s.v.p is equal to the external atmospheric pressure.~(b) Experiment to determine the boiling point of a small quantity of a liquid.\n\nSome liquid is introduced into the shorter length of the J - tube above the mercury. The beaker and its contents are heated and stirred continuously. At a stage when the levels of mercury in the two limbs of the J - tube are the same the temperature is read. At this temperature the s.v.p. of the liquid is equal to the atmospheric pressure. This temperature is the boiling point of the liquid.\n(c) Heat lost by copper = Heat gained by water\n\\(mc \\theta = m_{1} c_{1} \\theta_{1} + ml\\)\n\\(0.3 \\times 400 \\times (950° - 100°) = 0.25 \\times 4200 \\times (100° - 25°) + 2.26 \\times 10^{6} M\\)\n\\(M = \\frac{102000 - 78750}{2.26 \\times 10^{6}}\\)\n\\(M = 10.3g\\)\nTherefore, the mass of water that will boil away is 10.3g.\n(d) Four effects of heat other than expansion are:\n(1) Change of state (2) Change in temperature (3) Chemical change (4) Change in electrical resistance of conductivity. (5) Melting sublimation.\n~(c) Heat lost by copper = Heat gained by water\n\\(mc \\theta = m_{1} c_{1} \\theta_{1} + ml\\)\n\\(0.3 \\times 400 \\times (950° - 100°) = 0.25 \\times 4200 \\times (100° - 25°) + 2.26 \\times 10^{6} M\\)\n\\(M = \\frac{102000 - 78750}{2.26 \\times 10^{6}}\\)\n\\(M = 10.3g\\)\nTherefore, the mass of water that will boil away is 10.3g.\n(d) Four effects of heat other than expansion are:\n(1) Change of state (2) Change in temperature (3) Chemical change (4) Change in electrical resistance of conductivity. (5) Melting sublimation.\n~(d) Four effects of heat other than expansion are:\n(1) Change of state (2) Change in temperature (3) Chemical change (4) Change in electrical resistance of conductivity. (5) Melting sublimation.\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(291, '2019', '\n\nQuestion\n\n(a) Define the capacitance of a capacitor.\n(b) State three factors on which the capacitance of a parallel plate capacitor depends.\n(c) Derive a formula for the energy W stored in a charged capacitor of capacitance C carrying a charge Q on either plate.\n(d) Two capacitors of capacitance 4\\(\\mu F\\) and 6\\(\\mu F\\)are connected in series to a 100V d.c supply. Draw the circuit diagram and calculate the (i) charge on either plate of each capacitor (ii) p.d. across each capacitor; (iii) energy of the combined capacitors.\n\n', '', '(a) Define the capacitance of a capacitor.\n(b) State three factors on which the capacitance of a parallel plate capacitor depends.\n(c) Derive a formula for the energy W stored in a charged capacitor of capacitance C carrying a charge Q on either plate.\n(d) Two capacitors of capacitance 4\\(\\mu F\\) and 6\\(\\mu F\\)are connected in series to a 100V d.c supply. Draw the circuit diagram and calculate the (i) charge on either plate of each capacitor (ii) p.d. across each capacitor; (iii) energy of the combined capacitors.', '(a) Define the capacitance of a capacitor.~(b) State three factors on which the capacitance of a parallel plate capacitor depends.\n(c) Derive a formula for the energy W stored in a charged capacitor of capacitance C carrying a charge Q on either plate.\n(d) Two capacitors of capacitance 4\\(\\mu F\\) and 6\\(\\mu F\\)are connected in series to a 100V d.c supply. Draw the circuit diagram and calculate the (i) charge on either plate of each capacitor (ii) p.d. across each capacitor; (iii) energy of the combined capacitors.~(c) Derive a formula for the energy W stored in a charged capacitor of capacitance C carrying a charge Q on either plate.\n(d) Two capacitors of capacitance 4\\(\\mu F\\) and 6\\(\\mu F\\)are connected in series to a 100V d.c supply. Draw the circuit diagram and calculate the (i) charge on either plate of each capacitor (ii) p.d. across each capacitor; (iii) energy of the combined capacitors.~(d) Two capacitors of capacitance 4\\(\\mu F\\) and 6\\(\\mu F\\)are connected in series to a 100V d.c supply. Draw the circuit diagram and calculate the (i) charge on either plate of each capacitor (ii) p.d. across each capacitor; (iii) energy of the combined capacitors.', NULL, '\n\nExplanation\n(a) Capacitance of a capacitor is the ratio of the charge Q on either plate of the capacitor to the p.d in volts between the plates. Mathematically,\n\\(Capacitance = \\frac{Q}{V}\\)\nQ - Charge ; V - voltage.\n(b) Three factors on which the capacitance of a parallel - plate capacitor depends on are: (1) Common area of the plates (2) Dielectric between the plates (3) Separation between the plates.\n(c) Work done W = QV\nAverage work done = \\(\\frac{1}{2} QV\\)\nBut \\(V = \\frac{Q}{C}\\)\n\\(\\therefore W = \\frac{Q}{2} \\times \\frac{Q}{C} = \\frac{Q^{2}}{2C}\\)\n\\(\\therefore \\text{Energy stored W = } \\frac{Q^{2}}{2C}\\)\n(d)\n\\(\\frac{1}{C} = \\frac{1}{C_{1}} + \\frac{1}{C_{2}}\\)\n= \\(\\frac{1}{6} + \\frac{1}{4} = \\frac{10}{24}\\)\n\\(C = 2.4 \\mu F\\)\n\\(Q = \\frac{24}{10} \\times 100 \\times 10^{-6} = 240 \\mu C\\)\n= \\(240 \\times 10^{-6} C\\)\n(ii) p.d across 4\\(\\mu F, V_{4} = \\frac{Q}{C}\\)\n\\(V_{4} = \\frac{240 \\times 10^{-6}}{4 \\times 10^{-6}} = 60V\\)\np.d across 6\\(\\mu F, V_{6} = \\frac{Q}{C}\\)\n= \\(\\frac{240 \\times 10^{-6}}{6 \\times 10^{-6}} = 40V\\)\n(iii) Energy stored = \\(\\frac{Q^{2}}{2C}\\)\n= \\(\\frac{(240 \\times 10^{-6})^{2}}{2 \\times 2.4 \\times 10^{-6}}\\)\n= \\(1.2 \\times 10^{-2} J\\)\n\n', '', 'Explanation\n(a) Capacitance of a capacitor is the ratio of the charge Q on either plate of the capacitor to the p.d in volts between the plates. Mathematically,\n\\(Capacitance = \\frac{Q}{V}\\)\nQ - Charge ; V - voltage.\n(b) Three factors on which the capacitance of a parallel - plate capacitor depends on are: (1) Common area of the plates (2) Dielectric between the plates (3) Separation between the plates.\n(c) Work done W = QV\nAverage work done = \\(\\frac{1}{2} QV\\)\nBut \\(V = \\frac{Q}{C}\\)\n\\(\\therefore W = \\frac{Q}{2} \\times \\frac{Q}{C} = \\frac{Q^{2}}{2C}\\)\n\\(\\therefore \\text{Energy stored W = } \\frac{Q^{2}}{2C}\\)\n(d)\n\\(\\frac{1}{C} = \\frac{1}{C_{1}} + \\frac{1}{C_{2}}\\)\n= \\(\\frac{1}{6} + \\frac{1}{4} = \\frac{10}{24}\\)\n\\(C = 2.4 \\mu F\\)\n\\(Q = \\frac{24}{10} \\times 100 \\times 10^{-6} = 240 \\mu C\\)\n= \\(240 \\times 10^{-6} C\\)\n(ii) p.d across 4\\(\\mu F, V_{4} = \\frac{Q}{C}\\)\n\\(V_{4} = \\frac{240 \\times 10^{-6}}{4 \\times 10^{-6}} = 60V\\)\np.d across 6\\(\\mu F, V_{6} = \\frac{Q}{C}\\)\n= \\(\\frac{240 \\times 10^{-6}}{6 \\times 10^{-6}} = 40V\\)\n(iii) Energy stored = \\(\\frac{Q^{2}}{2C}\\)\n= \\(\\frac{(240 \\times 10^{-6})^{2}}{2 \\times 2.4 \\times 10^{-6}}\\)\n= \\(1.2 \\times 10^{-2} J\\)\n\n', '(a) Capacitance of a capacitor is the ratio of the charge Q on either plate of the capacitor to the p.d in volts between the plates. Mathematically,\n\\(Capacitance = \\frac{Q}{V}\\)\nQ - Charge ; V - voltage.~(b) Three factors on which the capacitance of a parallel - plate capacitor depends on are: (1) Common area of the plates (2) Dielectric between the plates (3) Separation between the plates.\n(c) Work done W = QV\nAverage work done = \\(\\frac{1}{2} QV\\)\nBut \\(V = \\frac{Q}{C}\\)\n\\(\\therefore W = \\frac{Q}{2} \\times \\frac{Q}{C} = \\frac{Q^{2}}{2C}\\)\n\\(\\therefore \\text{Energy stored W = } \\frac{Q^{2}}{2C}\\)\n(d)\n\\(\\frac{1}{C} = \\frac{1}{C_{1}} + \\frac{1}{C_{2}}\\)\n= \\(\\frac{1}{6} + \\frac{1}{4} = \\frac{10}{24}\\)\n\\(C = 2.4 \\mu F\\)\n\\(Q = \\frac{24}{10} \\times 100 \\times 10^{-6} = 240 \\mu C\\)\n= \\(240 \\times 10^{-6} C\\)\n(ii) p.d across 4\\(\\mu F, V_{4} = \\frac{Q}{C}\\)\n\\(V_{4} = \\frac{240 \\times 10^{-6}}{4 \\times 10^{-6}} = 60V\\)\np.d across 6\\(\\mu F, V_{6} = \\frac{Q}{C}\\)\n= \\(\\frac{240 \\times 10^{-6}}{6 \\times 10^{-6}} = 40V\\)\n(iii) Energy stored = \\(\\frac{Q^{2}}{2C}\\)\n= \\(\\frac{(240 \\times 10^{-6})^{2}}{2 \\times 2.4 \\times 10^{-6}}\\)\n= \\(1.2 \\times 10^{-2} J\\)\n~(c) Work done W = QV\nAverage work done = \\(\\frac{1}{2} QV\\)\nBut \\(V = \\frac{Q}{C}\\)\n\\(\\therefore W = \\frac{Q}{2} \\times \\frac{Q}{C} = \\frac{Q^{2}}{2C}\\)\n\\(\\therefore \\text{Energy stored W = } \\frac{Q^{2}}{2C}\\)\n(d)\n\\(\\frac{1}{C} = \\frac{1}{C_{1}} + \\frac{1}{C_{2}}\\)\n= \\(\\frac{1}{6} + \\frac{1}{4} = \\frac{10}{24}\\)\n\\(C = 2.4 \\mu F\\)\n\\(Q = \\frac{24}{10} \\times 100 \\times 10^{-6} = 240 \\mu C\\)\n= \\(240 \\times 10^{-6} C\\)\n(ii) p.d across 4\\(\\mu F, V_{4} = \\frac{Q}{C}\\)\n\\(V_{4} = \\frac{240 \\times 10^{-6}}{4 \\times 10^{-6}} = 60V\\)\np.d across 6\\(\\mu F, V_{6} = \\frac{Q}{C}\\)\n= \\(\\frac{240 \\times 10^{-6}}{6 \\times 10^{-6}} = 40V\\)\n(iii) Energy stored = \\(\\frac{Q^{2}}{2C}\\)\n= \\(\\frac{(240 \\times 10^{-6})^{2}}{2 \\times 2.4 \\times 10^{-6}}\\)\n= \\(1.2 \\times 10^{-2} J\\)\n~(d)\n\\(\\frac{1}{C} = \\frac{1}{C_{1}} + \\frac{1}{C_{2}}\\)\n= \\(\\frac{1}{6} + \\frac{1}{4} = \\frac{10}{24}\\)\n\\(C = 2.4 \\mu F\\)\n\\(Q = \\frac{24}{10} \\times 100 \\times 10^{-6} = 240 \\mu C\\)\n= \\(240 \\times 10^{-6} C\\)\n(ii) p.d across 4\\(\\mu F, V_{4} = \\frac{Q}{C}\\)\n\\(V_{4} = \\frac{240 \\times 10^{-6}}{4 \\times 10^{-6}} = 60V\\)\np.d across 6\\(\\mu F, V_{6} = \\frac{Q}{C}\\)\n= \\(\\frac{240 \\times 10^{-6}}{6 \\times 10^{-6}} = 40V\\)\n(iii) Energy stored = \\(\\frac{Q^{2}}{2C}\\)\n= \\(\\frac{(240 \\times 10^{-6})^{2}}{2 \\times 2.4 \\times 10^{-6}}\\)\n= \\(1.2 \\times 10^{-2} J\\)\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(292, '2019', '\n\nQuestion\n\n(a) Explain the terms nuclear fission and nuclear fusion.\n(b) State two advantages of fusion over fission and explain briefly why, in spite of these advantages,fusion is not normally usedfor the generation of power.\n(c) \\(^{238} _{92} U\\) is a long half- life alpha emitter and decays to thorium Th which, in turn decays by beta emission with a small decay constantto an isotope of protactinium Pa. The protactinium decay scheme of \\(^{238} _{92} U \\) as stated above.\n(d) State three uses of radioisotopes.\n\n', '', '(a) Explain the terms nuclear fission and nuclear fusion.\n(b) State two advantages of fusion over fission and explain briefly why, in spite of these advantages,fusion is not normally usedfor the generation of power.\n(c) \\(^{238} _{92} U\\) is a long half- life alpha emitter and decays to thorium Th which, in turn decays by beta emission with a small decay constantto an isotope of protactinium Pa. The protactinium decay scheme of \\(^{238} _{92} U \\) as stated above.\n(d) State three uses of radioisotopes.', '(a) Explain the terms nuclear fission and nuclear fusion.~(b) State two advantages of fusion over fission and explain briefly why, in spite of these advantages,fusion is not normally usedfor the generation of power.\n(c) \\(^{238} _{92} U\\) is a long half- life alpha emitter and decays to thorium Th which, in turn decays by beta emission with a small decay constantto an isotope of protactinium Pa. The protactinium decay scheme of \\(^{238} _{92} U \\) as stated above.\n(d) State three uses of radioisotopes.~(c) \\(^{238} _{92} U\\) is a long half- life alpha emitter and decays to thorium Th which, in turn decays by beta emission with a small decay constantto an isotope of protactinium Pa. The protactinium decay scheme of \\(^{238} _{92} U \\) as stated above.\n(d) State three uses of radioisotopes.~(d) State three uses of radioisotopes.', NULL, '\n\nExplanation\n(a) Fission: Nuclear reaction in which a heavy nucleus is bombarded with neutrons and split into approximate equal parts emitting neutrons and releasing large amounts of energy.\nFusion: Nuclear reaction between two light nuclei forming a heavier nucleus with the release of a large amount of energy.\n(b) Advantages of Fusion over Fission:\n(1)By - products are non-radioactive. (2) Easily achieved with lightest elements so that nuclear repulsion is easily overcome as nuclei approach each other. (3) Raw materials are cheaply available.\nWhy fusion is not normally used : Very high temperature is required to overcome coulomb repulsive forces between the two light nuclei. It is difficult to get materials that will withstand such high temperature. Mathematically,\n\\(T^{\\frac{1}{2}} = \\frac{0.693}{\\lambda}\\), where \\(\\lambda\\) is the decay constant.\nDecay constant = \\(\\frac{\\text{no of atoms disintegrating per sec}}{\\text{no of atoms in the source at that instant}}\\)\nIsotopes: Isotopes of an element are atoms which have the same proton numbers/atoms number but different nucleon numbers/mass number.\nDecay scheme: \\(^{238} _{92} U \\to ^{234} _{90} Th + ^{4} _{2} He + Energy\\)\n\\(^{234} _{90} Th \\to ^{234} _{91} Pa + ^{0} _{-1} e + Energy\\)\n\\(^{234} _{91} \\to ^{0} _{-1} e + ^{234} _{92} + Energy\\)\n(d) Uses of radioisotopes:\n(1) Radio therapy (2) Radio dating (3) Used as tracers and preservatives in agriculture.\n\n', '', 'Explanation\n(a) Fission: Nuclear reaction in which a heavy nucleus is bombarded with neutrons and split into approximate equal parts emitting neutrons and releasing large amounts of energy.\nFusion: Nuclear reaction between two light nuclei forming a heavier nucleus with the release of a large amount of energy.\n(b) Advantages of Fusion over Fission:\n(1)By - products are non-radioactive. (2) Easily achieved with lightest elements so that nuclear repulsion is easily overcome as nuclei approach each other. (3) Raw materials are cheaply available.\nWhy fusion is not normally used : Very high temperature is required to overcome coulomb repulsive forces between the two light nuclei. It is difficult to get materials that will withstand such high temperature. Mathematically,\n\\(T^{\\frac{1}{2}} = \\frac{0.693}{\\lambda}\\), where \\(\\lambda\\) is the decay constant.\nDecay constant = \\(\\frac{\\text{no of atoms disintegrating per sec}}{\\text{no of atoms in the source at that instant}}\\)\nIsotopes: Isotopes of an element are atoms which have the same proton numbers/atoms number but different nucleon numbers/mass number.\nDecay scheme: \\(^{238} _{92} U \\to ^{234} _{90} Th + ^{4} _{2} He + Energy\\)\n\\(^{234} _{90} Th \\to ^{234} _{91} Pa + ^{0} _{-1} e + Energy\\)\n\\(^{234} _{91} \\to ^{0} _{-1} e + ^{234} _{92} + Energy\\)\n(d) Uses of radioisotopes:\n(1) Radio therapy (2) Radio dating (3) Used as tracers and preservatives in agriculture.\n\n', '(a) Fission: Nuclear reaction in which a heavy nucleus is bombarded with neutrons and split into approximate equal parts emitting neutrons and releasing large amounts of energy.\nFusion: Nuclear reaction between two light nuclei forming a heavier nucleus with the release of a large amount of energy.~(b) Advantages of Fusion over Fission:\n(1)By - products are non-radioactive. (2) Easily achieved with lightest elements so that nuclear repulsion is easily overcome as nuclei approach each other. (3) Raw materials are cheaply available.\nWhy fusion is not normally used : Very high temperature is required to overcome coulomb repulsive forces between the two light nuclei. It is difficult to get materials that will withstand such high temperature. Mathematically,\n\\(T^{\\frac{1}{2}} = \\frac{0.693}{\\lambda}\\), where \\(\\lambda\\) is the decay constant.\nDecay constant = \\(\\frac{\\text{no of atoms disintegrating per sec}}{\\text{no of atoms in the source at that instant}}\\)\nIsotopes: Isotopes of an element are atoms which have the same proton numbers/atoms number but different nucleon numbers/mass number.\nDecay scheme: \\(^{238} _{92} U \\to ^{234} _{90} Th + ^{4} _{2} He + Energy\\)\n\\(^{234} _{90} Th \\to ^{234} _{91} Pa + ^{0} _{-1} e + Energy\\)\n\\(^{234} _{91} \\to ^{0} _{-1} e + ^{234} _{92} + Energy\\)\n(d) Uses of radioisotopes:\n(1) Radio therapy (2) Radio dating (3) Used as tracers and preservatives in agriculture.\n~(d) Uses of radioisotopes:\n(1) Radio therapy (2) Radio dating (3) Used as tracers and preservatives in agriculture.\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
(293, '2019', '\n\nQuestion\n\n(a) Explain what is meant by acceleration of free fall due to gravity, g.\n(b) State two reasons why g varies on the surface of the earth\n(c) A stone is projected upwards at an angle of 30° to the horizontal from the top of a tower of height 100 m and it hits the ground at a point Q. If the initial velocity of projection is 100ms\\(^{-1}\\), calculate the\n(i) maximum height of the stone above the ground;\n(ii) time it takes to reach this height;\n(iii) time of flight\n(iv) horizontal distance from the foot of the tower to the point Q. (Neglect air resistance and take g as 10m\\(^{-2}\\))\n\n', '', '(a) Explain what is meant by acceleration of free fall due to gravity, g.\n(b) State two reasons why g varies on the surface of the earth\n(c) A stone is projected upwards at an angle of 30° to the horizontal from the top of a tower of height 100 m and it hits the ground at a point Q. If the initial velocity of projection is 100ms\\(^{-1}\\), calculate the\n(i) maximum height of the stone above the ground;\n(ii) time it takes to reach this height;\n(iii) time of flight\n(iv) horizontal distance from the foot of the tower to the point Q. (Neglect air resistance and take g as 10m\\(^{-2}\\))', '(a) Explain what is meant by acceleration of free fall due to gravity, g.~(b) State two reasons why g varies on the surface of the earth\n(c) A stone is projected upwards at an angle of 30° to the horizontal from the top of a tower of height 100 m and it hits the ground at a point Q. If the initial velocity of projection is 100ms\\(^{-1}\\), calculate the\n(i) maximum height of the stone above the ground;\n(ii) time it takes to reach this height;\n(iii) time of flight\n(iv) horizontal distance from the foot of the tower to the point Q. (Neglect air resistance and take g as 10m\\(^{-2}\\))~(c) A stone is projected upwards at an angle of 30° to the horizontal from the top of a tower of height 100 m and it hits the ground at a point Q. If the initial velocity of projection is 100ms\\(^{-1}\\), calculate the\n(i) maximum height of the stone above the ground;\n(ii) time it takes to reach this height;\n(iii) time of flight\n(iv) horizontal distance from the foot of the tower to the point Q. (Neglect air resistance and take g as 10m\\(^{-2}\\))', NULL, '\n\nExplanation\n(a) This is the force of attraction of the Earth on a mass\nIt is a vector quantity and is measured in ms\\(^{-2}\\)\nIt pulls the acceleration when a body is in vacuum or without resistance. OR/F = g\n(all symbols explained) where F = force of gravity, M = mass of the object\n\n(b) — (Shape of the earth). The earth is not a perfect sphere (1) or the shape of the earth is elliptical - Rotation of the Earth about its polar axis.\n(c)(i) Let H = height of tower and h = vertical difference covered by the stone from the point of projection.\n(i) V\\(_2\\) = (Usin\\(\\theta\\))\\(^2\\) - 2gh\n\\(\\theta\\) = 100 x 100 x (\\(\\frac{1}{2}\\))\\(^2\\) - 20h\nh = \\(\\frac{2500}{20}\\)\n= 125m\n\nheight above the ground\n(i) S = H + 125 S = 100 + 125 S = 225m\n(ii) V = U sin \\(\\theta\\) – gt\n0 = 100 sin 30° – 10t\nt = \\(\\frac{100 \\times 0.5}{10}\\)\nt = 5s\n(iii) Using S = ut + 1/2 gt\\(^2\\)\n225 = 0 + 1/2 x 10t\\(^2\\)\nt = \\(\\sqrt{45}\\) = 6.71\nTime of flight, T= 5 + 6.\nT = 11.71s\n\n(iv) Range R = UTcos\\(\\theta\\)\nOr S = ut – 1/2 gt\\(^2\\)\ng = 0\n=100 x 0.8660 x 11.71\n= 1014.08 (C14)\nRange \\(\\frac{U^2 sin20}{ g}\\)\n= 1013.22m\n\n', '', 'Explanation\n(a) This is the force of attraction of the Earth on a mass\nIt is a vector quantity and is measured in ms\\(^{-2}\\)\nIt pulls the acceleration when a body is in vacuum or without resistance. OR/F = g\n(all symbols explained) where F = force of gravity, M = mass of the object\n\n(b) — (Shape of the earth). The earth is not a perfect sphere (1) or the shape of the earth is elliptical - Rotation of the Earth about its polar axis.\n(c)(i) Let H = height of tower and h = vertical difference covered by the stone from the point of projection.\n(i) V\\(_2\\) = (Usin\\(\\theta\\))\\(^2\\) - 2gh\n\\(\\theta\\) = 100 x 100 x (\\(\\frac{1}{2}\\))\\(^2\\) - 20h\nh = \\(\\frac{2500}{20}\\)\n= 125m\n\nheight above the ground\n(i) S = H + 125 S = 100 + 125 S = 225m\n(ii) V = U sin \\(\\theta\\) – gt\n0 = 100 sin 30° – 10t\nt = \\(\\frac{100 \\times 0.5}{10}\\)\nt = 5s\n(iii) Using S = ut + 1/2 gt\\(^2\\)\n225 = 0 + 1/2 x 10t\\(^2\\)\nt = \\(\\sqrt{45}\\) = 6.71\nTime of flight, T= 5 + 6.\nT = 11.71s\n\n(iv) Range R = UTcos\\(\\theta\\)\nOr S = ut – 1/2 gt\\(^2\\)\ng = 0\n=100 x 0.8660 x 11.71\n= 1014.08 (C14)\nRange \\(\\frac{U^2 sin20}{ g}\\)\n= 1013.22m\n\n', '(a) This is the force of attraction of the Earth on a mass\nIt is a vector quantity and is measured in ms\\(^{-2}\\)\nIt pulls the acceleration when a body is in vacuum or without resistance. OR/F = g\n(all symbols explained) where F = force of gravity, M = mass of the object\n~(b) — (Shape of the earth). The earth is not a perfect sphere (1) or the shape of the earth is elliptical - Rotation of the Earth about its polar axis.\n(c)(i) Let H = height of tower and h = vertical difference covered by the stone from the point of projection.\n(i) V\\(_2\\) = (Usin\\(\\theta\\))\\(^2\\) - 2gh\n\\(\\theta\\) = 100 x 100 x (\\(\\frac{1}{2}\\))\\(^2\\) - 20h\nh = \\(\\frac{2500}{20}\\)\n= 125m\n\nheight above the ground\n(i) S = H + 125 S = 100 + 125 S = 225m\n(ii) V = U sin \\(\\theta\\) – gt\n0 = 100 sin 30° – 10t\nt = \\(\\frac{100 \\times 0.5}{10}\\)\nt = 5s\n(iii) Using S = ut + 1/2 gt\\(^2\\)\n225 = 0 + 1/2 x 10t\\(^2\\)\nt = \\(\\sqrt{45}\\) = 6.71\nTime of flight, T= 5 + 6.\nT = 11.71s\n\n(iv) Range R = UTcos\\(\\theta\\)\nOr S = ut – 1/2 gt\\(^2\\)\ng = 0\n=100 x 0.8660 x 11.71\n= 1014.08 (C14)\nRange \\(\\frac{U^2 sin20}{ g}\\)\n= 1013.22m\n~(c)(i) Let H = height of tower and h = vertical difference covered by the stone from the point of projection.\n(i) V\\(_2\\) = (Usin\\(\\theta\\))\\(^2\\) - 2gh\n\\(\\theta\\) = 100 x 100 x (\\(\\frac{1}{2}\\))\\(^2\\) - 20h\nh = \\(\\frac{2500}{20}\\)\n= 125m\n\nheight above the ground\n(i) S = H + 125 S = 100 + 125 S = 225m\n(ii) V = U sin \\(\\theta\\) – gt\n0 = 100 sin 30° – 10t\nt = \\(\\frac{100 \\times 0.5}{10}\\)\nt = 5s\n(iii) Using S = ut + 1/2 gt\\(^2\\)\n225 = 0 + 1/2 x 10t\\(^2\\)\nt = \\(\\sqrt{45}\\) = 6.71\nTime of flight, T= 5 + 6.\nT = 11.71s\n\n(iv) Range R = UTcos\\(\\theta\\)\nOr S = ut – 1/2 gt\\(^2\\)\ng = 0\n=100 x 0.8660 x 11.71\n= 1014.08 (C14)\nRange \\(\\frac{U^2 sin20}{ g}\\)\n= 1013.22m\n~d) where F = force of gravity, M = mass of the object\n\n(b) — (Shape of the earth). The earth is not a perfect sphere (1) or the shape of the earth is elliptical - Rotation of the Earth about its polar axis.\n(c)(i) Let H = height of tower and h = vertical difference covered by the stone from the point of projection.\n(i) V\\(_2\\) = (Usin\\(\\theta\\))\\(^2\\) - 2gh\n\\(\\theta\\) = 100 x 100 x (\\(\\frac{1}{2}\\))\\(^2\\) - 20h\nh = \\(\\frac{2500}{20}\\)\n= 125m\n\nheight above the ground\n(i) S = H + 125 S = 100 + 125 S = 225m\n(ii) V = U sin \\(\\theta\\) – gt\n0 = 100 sin 30° – 10t\nt = \\(\\frac{100 \\times 0.5}{10}\\)\nt = 5s\n(iii) Using S = ut + 1/2 gt\\(^2\\)\n225 = 0 + 1/2 x 10t\\(^2\\)\nt = \\(\\sqrt{45}\\) = 6.71\nTime of flight, T= 5 + 6.\nT = 11.71s\n\n(iv) Range R = UTcos\\(\\theta\\)\nOr S = ut – 1/2 gt\\(^2\\)\ng = 0\n=100 x 0.8660 x 11.71\n= 1014.08 (C14)\nRange \\(\\frac{U^2 sin20}{ g}\\)\n= 1013.22m\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(294, '2019', '\n\nQuestion\n\n(a) State Faraday\'s law of electromagnetic induction.\n(b) Draw a labelled diagram of an induction coil and explain how it works.\n(c) How is the effect of eddy currents minimized in the coil?\n(d) State two reasons why a capacitor should be included in the primary circuit of the coil.\n(e) State three uses of an induction coil.\n\n\n', '', '(a) State Faraday\'s law of electromagnetic induction.\n(b) Draw a labelled diagram of an induction coil and explain how it works.\n(c) How is the effect of eddy currents minimized in the coil?\n(d) State two reasons why a capacitor should be included in the primary circuit of the coil.\n(e) State three uses of an induction coil.\n', '(a) State Faraday\'s law of electromagnetic induction.~(b) Draw a labelled diagram of an induction coil and explain how it works.\n(c) How is the effect of eddy currents minimized in the coil?\n(d) State two reasons why a capacitor should be included in the primary circuit of the coil.\n(e) State three uses of an induction coil.\n~(c) How is the effect of eddy currents minimized in the coil?\n(d) State two reasons why a capacitor should be included in the primary circuit of the coil.\n(e) State three uses of an induction coil.\n~(d) State two reasons why a capacitor should be included in the primary circuit of the coil.\n(e) State three uses of an induction coil.\n~(e) State three uses of an induction coil.\n', NULL, '\n\nExplanation\n(a) Whenever there is a change in the magnetic flux linked with a circuit, an electromotive force is induced, (1\\(\\frac{1}{2}\\)) the strength of which is proportional to the rate of change of the flux linked with the circuit (1\\(\\frac{1}{2}\\)).\nE \\(\\alpha\\) \\(\\frac{d \\theta}{dt} or E = \\frac{Nd \\theta}{dt}\\)\n\nWhere E = induced e.m.f.\n\\(\\theta\\) = magnetic flux.\n\n(b)\n\nWorkability: All except key and capacitor. When the platinum contacts touch ar,d the key k is closed, current passes through the primary coil and the core is magnetised. The soft iron armature is then attracted so the, the platinum contacts separate and the circuit is broken. When this happens, the core loses its magnetism and the armature is pulled back by the spring so that contact is re-established. As the contact is repeatedly made and broken in the primary circuit, changes of magnetic flux take place and induces an e.m.f. in the secondary coil. Because of the much larger number of turns in the secondary compared with the primary, a very high e.m.f is induced in the secondary coil which is sufficient to cause a spark to jump across the spark gap.\n(c) The core of the coil is made of bundle of soft-iron wires OR that are of the coil is laminated. Or use of particle of iron dust. Or lamination of iron sheet or iron core not metallic sheet\n\n(d) – To control sparking at the platinum contacts (which otherwise wears off the contact points)\n-- It promotes a rapid decay of the magnetic flux at the break of the primary circuit and this helps to increase the secondary e.m.f.\n\n(e) It is used in the\n– coil\n– ignition system of motor cars\n– inveitigation of high voltages\n– study of electric discharge through gases\n– operation of x-ray tubes\n– merse radio transmitter.\n\n\n\n\n', '', 'Explanation\n(a) Whenever there is a change in the magnetic flux linked with a circuit, an electromotive force is induced, (1\\(\\frac{1}{2}\\)) the strength of which is proportional to the rate of change of the flux linked with the circuit (1\\(\\frac{1}{2}\\)).\nE \\(\\alpha\\) \\(\\frac{d \\theta}{dt} or E = \\frac{Nd \\theta}{dt}\\)\n\nWhere E = induced e.m.f.\n\\(\\theta\\) = magnetic flux.\n\n(b)\n\nWorkability: All except key and capacitor. When the platinum contacts touch ar,d the key k is closed, current passes through the primary coil and the core is magnetised. The soft iron armature is then attracted so the, the platinum contacts separate and the circuit is broken. When this happens, the core loses its magnetism and the armature is pulled back by the spring so that contact is re-established. As the contact is repeatedly made and broken in the primary circuit, changes of magnetic flux take place and induces an e.m.f. in the secondary coil. Because of the much larger number of turns in the secondary compared with the primary, a very high e.m.f is induced in the secondary coil which is sufficient to cause a spark to jump across the spark gap.\n(c) The core of the coil is made of bundle of soft-iron wires OR that are of the coil is laminated. Or use of particle of iron dust. Or lamination of iron sheet or iron core not metallic sheet\n\n(d) – To control sparking at the platinum contacts (which otherwise wears off the contact points)\n-- It promotes a rapid decay of the magnetic flux at the break of the primary circuit and this helps to increase the secondary e.m.f.\n\n(e) It is used in the\n– coil\n– ignition system of motor cars\n– inveitigation of high voltages\n– study of electric discharge through gases\n– operation of x-ray tubes\n– merse radio transmitter.\n\n\n\n\n', '(a) Whenever there is a change in the magnetic flux linked with a circuit, an electromotive force is induced, (1\\(\\frac{1}{2}\\)) the strength of which is proportional to the rate of change of the flux linked with the circuit (1\\(\\frac{1}{2}\\)).\nE \\(\\alpha\\) \\(\\frac{d \\theta}{dt} or E = \\frac{Nd \\theta}{dt}\\)\n\nWhere E = induced e.m.f.\n\\(\\theta\\) = magnetic flux.\n~(b)\n\nWorkability: All except key and capacitor. When the platinum contacts touch ar,d the key k is closed, current passes through the primary coil and the core is magnetised. The soft iron armature is then attracted so the, the platinum contacts separate and the circuit is broken. When this happens, the core loses its magnetism and the armature is pulled back by the spring so that contact is re-established. As the contact is repeatedly made and broken in the primary circuit, changes of magnetic flux take place and induces an e.m.f. in the secondary coil. Because of the much larger number of turns in the secondary compared with the primary, a very high e.m.f is induced in the secondary coil which is sufficient to cause a spark to jump across the spark gap.\n(c) The core of the coil is made of bundle of soft-iron wires OR that are of the coil is laminated. Or use of particle of iron dust. Or lamination of iron sheet or iron core not metallic sheet\n\n(d) – To control sparking at the platinum contacts (which otherwise wears off the contact points)\n-- It promotes a rapid decay of the magnetic flux at the break of the primary circuit and this helps to increase the secondary e.m.f.\n\n(e) It is used in the\n– coil\n– ignition system of motor cars\n– inveitigation of high voltages\n– study of electric discharge through gases\n– operation of x-ray tubes\n– merse radio transmitter.\n\n\n\n~(c) The core of the coil is made of bundle of soft-iron wires OR that are of the coil is laminated. Or use of particle of iron dust. Or lamination of iron sheet or iron core not metallic sheet\n\n(d) – To control sparking at the platinum contacts (which otherwise wears off the contact points)\n-- It promotes a rapid decay of the magnetic flux at the break of the primary circuit and this helps to increase the secondary e.m.f.\n\n(e) It is used in the\n– coil\n– ignition system of motor cars\n– inveitigation of high voltages\n– study of electric discharge through gases\n– operation of x-ray tubes\n– merse radio transmitter.\n\n\n\n~(d) – To control sparking at the platinum contacts (which otherwise wears off the contact points)\n-- It promotes a rapid decay of the magnetic flux at the break of the primary circuit and this helps to increase the secondary e.m.f.\n\n(e) It is used in the\n– coil\n– ignition system of motor cars\n– inveitigation of high voltages\n– study of electric discharge through gases\n– operation of x-ray tubes\n– merse radio transmitter.\n\n\n\n~(e) It is used in the\n– coil\n– ignition system of motor cars\n– inveitigation of high voltages\n– study of electric discharge through gases\n– operation of x-ray tubes\n– merse radio transmitter.\n\n\n\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(295, '2019', '\n\nQuestion\n\n(a) What is surface tension? Explain the phenomenon in terms of intermolecular forces\n(b) Describe a simple experiment to demonstrate the surface tension of a liquid\n(c) State three examples to illustrate the effects of surface tension\n(d) Why does water wet a clean glass surface whereas mercury does not?\n(e) State two methods by which the surface tension of a liquid may reduced.\n\n', '', '(a) What is surface tension? Explain the phenomenon in terms of intermolecular forces\n(b) Describe a simple experiment to demonstrate the surface tension of a liquid\n(c) State three examples to illustrate the effects of surface tension\n(d) Why does water wet a clean glass surface whereas mercury does not?\n(e) State two methods by which the surface tension of a liquid may reduced.', '(a) What is surface tension? Explain the phenomenon in terms of intermolecular forces~(b) Describe a simple experiment to demonstrate the surface tension of a liquid\n(c) State three examples to illustrate the effects of surface tension\n(d) Why does water wet a clean glass surface whereas mercury does not?\n(e) State two methods by which the surface tension of a liquid may reduced.~(c) State three examples to illustrate the effects of surface tension\n(d) Why does water wet a clean glass surface whereas mercury does not?\n(e) State two methods by which the surface tension of a liquid may reduced.~(d) Why does water wet a clean glass surface whereas mercury does not?\n(e) State two methods by which the surface tension of a liquid may reduced.~(e) State two methods by which the surface tension of a liquid may reduced.', NULL, '\n\nExplanation\n(a) Surface tension is the property of a liquid which makes its surface behave as though covered with an elastic sink. Or it is the tangential force on the surface of a liquid. Surface tension is force per unit length on the surface of a liquid\n\n\nConsider two molecules A and B as shown above. The molecule A will be attracted uniformly in all directions trot pulled in any one particular. direction. On the other hand, B is attracted more by molecules within the body of the liquid and it therefore experiences a resultant inward force at right angles to the surface. There will therefore be a tendency for the surface of the liquid to contact and have as little an area as possible. The surface is thus in a tension.\nDiagram or description link\n\n\nNB: Water and needle shown but not labelled\nA beaker is filled to the brim with water as shown above. A dry needle is placed on a small piece of filter paper, which is then gently placed on the water surface. It would be observed that the filter paper sinks to the bottom of the beaker while the needle is left floating. The surface thereby acts like an elastic skin covering the liquid which is able to support the weight of the needle\n\n(c) – When water is allowed to drip slowly from a water tap, the shape of each small drop is spherical.\n- When mercury spills on a glass sheet, it forms spherical droplets.\n- Soap bubbles are spherical in shape. - Insects (water - skaker) area able to walk across a water surface etc.\n- Toy duck on liquid\n- A flowing or floating needle on water.\n\n(d) The adhesion of water molecules to glass molecules is stronger than the cohesion between water molecules. Hence, when water spills on a clean glass surface it spreads and wets the glass. On the other hand, the cohesion between mercury molecules is greater than the adhesion between mercury and glass molecules. Therefore when mercury spills on glass it forms small spherical droplets on larger flattened drops\n(e) By adding\n- detergent/soap\n- alcohol\n- oil\n- By heating the liquid/increasing the temperature of the liquid\n- adding camphor.\n\n', '', 'Explanation\n(a) Surface tension is the property of a liquid which makes its surface behave as though covered with an elastic sink. Or it is the tangential force on the surface of a liquid. Surface tension is force per unit length on the surface of a liquid\n\n\nConsider two molecules A and B as shown above. The molecule A will be attracted uniformly in all directions trot pulled in any one particular. direction. On the other hand, B is attracted more by molecules within the body of the liquid and it therefore experiences a resultant inward force at right angles to the surface. There will therefore be a tendency for the surface of the liquid to contact and have as little an area as possible. The surface is thus in a tension.\nDiagram or description link\n\n\nNB: Water and needle shown but not labelled\nA beaker is filled to the brim with water as shown above. A dry needle is placed on a small piece of filter paper, which is then gently placed on the water surface. It would be observed that the filter paper sinks to the bottom of the beaker while the needle is left floating. The surface thereby acts like an elastic skin covering the liquid which is able to support the weight of the needle\n\n(c) – When water is allowed to drip slowly from a water tap, the shape of each small drop is spherical.\n- When mercury spills on a glass sheet, it forms spherical droplets.\n- Soap bubbles are spherical in shape. - Insects (water - skaker) area able to walk across a water surface etc.\n- Toy duck on liquid\n- A flowing or floating needle on water.\n\n(d) The adhesion of water molecules to glass molecules is stronger than the cohesion between water molecules. Hence, when water spills on a clean glass surface it spreads and wets the glass. On the other hand, the cohesion between mercury molecules is greater than the adhesion between mercury and glass molecules. Therefore when mercury spills on glass it forms small spherical droplets on larger flattened drops\n(e) By adding\n- detergent/soap\n- alcohol\n- oil\n- By heating the liquid/increasing the temperature of the liquid\n- adding camphor.\n\n', '~(c) – When water is allowed to drip slowly from a water tap, the shape of each small drop is spherical.\n- When mercury spills on a glass sheet, it forms spherical droplets.\n- Soap bubbles are spherical in shape. - Insects (water - skaker) area able to walk across a water surface etc.\n- Toy duck on liquid\n- A flowing or floating needle on water.\n\n(d) The adhesion of water molecules to glass molecules is stronger than the cohesion between water molecules. Hence, when water spills on a clean glass surface it spreads and wets the glass. On the other hand, the cohesion between mercury molecules is greater than the adhesion between mercury and glass molecules. Therefore when mercury spills on glass it forms small spherical droplets on larger flattened drops\n(e) By adding\n- detergent/soap\n- alcohol\n- oil\n- By heating the liquid/increasing the temperature of the liquid\n- adding camphor.\n~(d) The adhesion of water molecules to glass molecules is stronger than the cohesion between water molecules. Hence, when water spills on a clean glass surface it spreads and wets the glass. On the other hand, the cohesion between mercury molecules is greater than the adhesion between mercury and glass molecules. Therefore when mercury spills on glass it forms small spherical droplets on larger flattened drops\n(e) By adding\n- detergent/soap\n- alcohol\n- oil\n- By heating the liquid/increasing the temperature of the liquid\n- adding camphor.\n~(e) By adding\n- detergent/soap\n- alcohol\n- oil\n- By heating the liquid/increasing the temperature of the liquid\n- adding camphor.\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(296, '2019', '\n\nQuestion\n\n(a) Define the boiling it of a liquid.\n(b) Describe with the a d. labelled diagram, an experiment to determire the boiling point of a small quantity of a liquid.\n(c) factors that may affect the boiling point of a liquid\n(d) Using the kinetic theory of matter, explain why pure water changes to steam at S.T.P. without any change in temperature, although heat is being supplied to the water.\n\n\n', '', '(a) Define the boiling it of a liquid.\n(b) Describe with the a d. labelled diagram, an experiment to determire the boiling point of a small quantity of a liquid.\n(c) factors that may affect the boiling point of a liquid\n(d) Using the kinetic theory of matter, explain why pure water changes to steam at S.T.P. without any change in temperature, although heat is being supplied to the water.\n', '(a) Define the boiling it of a liquid.~(b) Describe with the a d. labelled diagram, an experiment to determire the boiling point of a small quantity of a liquid.\n(c) factors that may affect the boiling point of a liquid\n(d) Using the kinetic theory of matter, explain why pure water changes to steam at S.T.P. without any change in temperature, although heat is being supplied to the water.\n~(c) factors that may affect the boiling point of a liquid\n(d) Using the kinetic theory of matter, explain why pure water changes to steam at S.T.P. without any change in temperature, although heat is being supplied to the water.\n~(d) Using the kinetic theory of matter, explain why pure water changes to steam at S.T.P. without any change in temperature, although heat is being supplied to the water.\n', NULL, '\n\nExplanation\na) The boiling pointer of a liquid is the temperature at which its s.v.p. is equal to the external atmospheric pressure OR The temperature at which a (pure) liquid boils when the atmospheric pressure on it is 760 mm Hg. OR The temperature at which the liquid changes from the liquid to the gaseous state without change in temperature when the atmospheric pressure is normal (760mm Hg).\n\n\nMethod: Some mercury is introduced into J - tube and the entrapped air removed. Some liquid is introduced into the shorter length of the J - tube above the mercury. The beaker and its contents are heated and the water stirred continuously. At a stage when the levels of mercury in the two limbs of the J-tube are the same the temperature is read. At this temperature the s.v.p of the liquid is equal to the atmospheric pressure, and this temperature is the boiling point of the liquid. The process is repeated with the beaker and its contents cooling. The temperature at which the levels of mercury in the two limbs are equal is read and the mean of the two taken.\n\n(c) Pressure on the liquid, impurities in the liquid.\n(d) The extra heat supplied during boiling is used in breaking the intermolecular attractions and in mabling the steam expand or enabling the molecules move further apart against the atmospheric pressure. Hence, there is no change in temperature during the change of state.\n\n', '', 'Explanation\na) The boiling pointer of a liquid is the temperature at which its s.v.p. is equal to the external atmospheric pressure OR The temperature at which a (pure) liquid boils when the atmospheric pressure on it is 760 mm Hg. OR The temperature at which the liquid changes from the liquid to the gaseous state without change in temperature when the atmospheric pressure is normal (760mm Hg).\n\n\nMethod: Some mercury is introduced into J - tube and the entrapped air removed. Some liquid is introduced into the shorter length of the J - tube above the mercury. The beaker and its contents are heated and the water stirred continuously. At a stage when the levels of mercury in the two limbs of the J-tube are the same the temperature is read. At this temperature the s.v.p of the liquid is equal to the atmospheric pressure, and this temperature is the boiling point of the liquid. The process is repeated with the beaker and its contents cooling. The temperature at which the levels of mercury in the two limbs are equal is read and the mean of the two taken.\n\n(c) Pressure on the liquid, impurities in the liquid.\n(d) The extra heat supplied during boiling is used in breaking the intermolecular attractions and in mabling the steam expand or enabling the molecules move further apart against the atmospheric pressure. Hence, there is no change in temperature during the change of state.\n\n', '~(c) Pressure on the liquid, impurities in the liquid.\n(d) The extra heat supplied during boiling is used in breaking the intermolecular attractions and in mabling the steam expand or enabling the molecules move further apart against the atmospheric pressure. Hence, there is no change in temperature during the change of state.\n~(d) The extra heat supplied during boiling is used in breaking the intermolecular attractions and in mabling the steam expand or enabling the molecules move further apart against the atmospheric pressure. Hence, there is no change in temperature during the change of state.\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(297, '2019', '\n\nQuestion\n\n(a) Explain the terms:- uniform acceleration and average speed.\n(b) A body at rest is given an initial uniform acceleration of 8.0ms\\(^{-2}\\) for 30s after which the acceleration is reduced to 5.0ms\\(^{-2}\\) for the next 20s. The body maintains the speed attained for 60s after which it is brought to rest in 20s. Draw the velocity-timegraph of the motion using the information given above.\n(c) Using the graph, calculate the: (i) maximum speed attained during the motion; (ii) average retardation as the body is being brought to rest; (iii) total distance travelled during the first 50s; (iv) average speed during the same interval as in (ii).\n\n', '', '(a) Explain the terms:- uniform acceleration and average speed.\n(b) A body at rest is given an initial uniform acceleration of 8.0ms\\(^{-2}\\) for 30s after which the acceleration is reduced to 5.0ms\\(^{-2}\\) for the next 20s. The body maintains the speed attained for 60s after which it is brought to rest in 20s. Draw the velocity-timegraph of the motion using the information given above.\n(c) Using the graph, calculate the: (i) maximum speed attained during the motion; (ii) average retardation as the body is being brought to rest; (iii) total distance travelled during the first 50s; (iv) average speed during the same interval as in (ii).', '(a) Explain the terms:- uniform acceleration and average speed.~(b) A body at rest is given an initial uniform acceleration of 8.0ms\\(^{-2}\\) for 30s after which the acceleration is reduced to 5.0ms\\(^{-2}\\) for the next 20s. The body maintains the speed attained for 60s after which it is brought to rest in 20s. Draw the velocity-timegraph of the motion using the information given above.\n(c) Using the graph, calculate the: (i) maximum speed attained during the motion; (ii) average retardation as the body is being brought to rest; (iii) total distance travelled during the first 50s; (iv) average speed during the same interval as in (ii).~(c) Using the graph, calculate the: (i) maximum speed attained during the motion; (ii) average retardation as the body is being brought to rest; (iii) total distance travelled during the first 50s; (iv) average speed during the same interval as in (ii).', NULL, '\n\nExplanation\n(a) Uniform acceleration occurs when the change in velocity per unit time is constant. The S.I. unit is ms\\(^{-2}\\).\n\\(a = \\frac{v - u}{t}\\);a - uniform acceleration, v - final velocity, u - initial velocity, t - time.\nGraphically,\n\nacceleration is the same acceleration at x is different every point from that at Y.\nNote - acceleration is given by slope in both graphs.\n(b)\n(i) Maximum speed = 240 m/s.\n(ii) \\(a = \\frac{v - u}{t} = \\frac{0 - 200}{20} = -10 ms^{-2}\\)\n(iii) Area of I = \\(\\frac{1}{2} \\times 30 \\times 240 = 3600m\\)\nArea of II = \\(20 \\times 200 = 4000m\\)\nArea of III = \\(\\frac{1}{2} \\times 20 \\times 40 = 400m\\)\nTotal distance in the first 50s : \\(3600m + 4000m + 400m = 8000m\\)\n(iv) Average speed during the first 50s = \\(\\frac{8000m}{50s} = 160 ms^{-1}\\)\n\n', '', 'Explanation\n(a) Uniform acceleration occurs when the change in velocity per unit time is constant. The S.I. unit is ms\\(^{-2}\\).\n\\(a = \\frac{v - u}{t}\\);a - uniform acceleration, v - final velocity, u - initial velocity, t - time.\nGraphically,\n\nacceleration is the same acceleration at x is different every point from that at Y.\nNote - acceleration is given by slope in both graphs.\n(b)\n(i) Maximum speed = 240 m/s.\n(ii) \\(a = \\frac{v - u}{t} = \\frac{0 - 200}{20} = -10 ms^{-2}\\)\n(iii) Area of I = \\(\\frac{1}{2} \\times 30 \\times 240 = 3600m\\)\nArea of II = \\(20 \\times 200 = 4000m\\)\nArea of III = \\(\\frac{1}{2} \\times 20 \\times 40 = 400m\\)\nTotal distance in the first 50s : \\(3600m + 4000m + 400m = 8000m\\)\n(iv) Average speed during the first 50s = \\(\\frac{8000m}{50s} = 160 ms^{-1}\\)\n\n', '(a) Uniform acceleration occurs when the change in velocity per unit time is constant. The S.I. unit is ms\\(^{-2}\\).\n\\(a = \\frac{v - u}{t}\\);a - uniform acceleration, v - final velocity, u - initial velocity, t - time.\nGraphically,\n\nacceleration is the same acceleration at x is different every point from that at Y.\nNote - acceleration is given by slope in both graphs.~(b)\n(i) Maximum speed = 240 m/s.\n(ii) \\(a = \\frac{v - u}{t} = \\frac{0 - 200}{20} = -10 ms^{-2}\\)\n(iii) Area of I = \\(\\frac{1}{2} \\times 30 \\times 240 = 3600m\\)\nArea of II = \\(20 \\times 200 = 4000m\\)\nArea of III = \\(\\frac{1}{2} \\times 20 \\times 40 = 400m\\)\nTotal distance in the first 50s : \\(3600m + 4000m + 400m = 8000m\\)\n(iv) Average speed during the first 50s = \\(\\frac{8000m}{50s} = 160 ms^{-1}\\)\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(298, '2019', '\n\nQuestion\n\n(a) Distinguish between temperature and heat. State the units in which they are measured\n(ii) State two physical properties used for measuring temperature.\n(b) (i) Describe, with the aid of a diagram, howthe upper fixed point is determined for a mecury-in-glass thermometer. State one precaution to ensure accurate results\n(ii) State one advantage which a constant-volume gas thermometer has over other thermometers and one reason why it is seldom used as an everyday laboratory instrument.\n(c) Using the kinetic theory of matter explain why evaporation causes cooling.\n\n', '', '(a) Distinguish between temperature and heat. State the units in which they are measured\n(ii) State two physical properties used for measuring temperature.\n(b) (i) Describe, with the aid of a diagram, howthe upper fixed point is determined for a mecury-in-glass thermometer. State one precaution to ensure accurate results\n(ii) State one advantage which a constant-volume gas thermometer has over other thermometers and one reason why it is seldom used as an everyday laboratory instrument.\n(c) Using the kinetic theory of matter explain why evaporation causes cooling.', '(a) Distinguish between temperature and heat. State the units in which they are measured\n(ii) State two physical properties used for measuring temperature.~(b) (i) Describe, with the aid of a diagram, howthe upper fixed point is determined for a mecury-in-glass thermometer. State one precaution to ensure accurate results\n(ii) State one advantage which a constant-volume gas thermometer has over other thermometers and one reason why it is seldom used as an everyday laboratory instrument.\n(c) Using the kinetic theory of matter explain why evaporation causes cooling.~(c) Using the kinetic theory of matter explain why evaporation causes cooling.', NULL, '\n\nExplanation\n(a)(i) Temperature is a number which expresses how hot or cold a body is on scale such as centigrade orkelvin. S.I. unit is kelvin (K). Temperature decides which direction heat will flow while, heat is a form of energy S.I. is Joules (J).\n(ii) Two physical properties used for measuring temperature are:- (1) Change in volume of liquid with temperature (2) Change in resistance. Others are: (3) Expansion of solids (4) Expansion of volumeof gas.\n(b)(i) Determination of the upper - fixed point for a mercury-in-glass thermometer.\n\nTo determine the upper fixed point, a hypsometer is used. The experiment is set up as shown above. The manometer is used to measure the pressure of the steam.\nIf it is not 76cm Hg, an adjustment is made to the upper fixed point. When the mercury level is constant, a mark is made on its glass to fix the point. One precaution that must be observed in this experiment is that the thermometer must not touch the water, because its temperature is affected by dissolved impurities in the water.\n(ii) The constant - volume gas thermometer measures temperature more accurately than other thermometers because its glass expansion is negligible compared to the expansion of the gas. It is seldom used as everyday laboratory thermometer because it is not easily handled unlike the laboratory thermometer.\n(c) In the kinetic theory, molecules are in random motion. They all have different velocities. Near the surface, some liquid molecules with high velocities in an upward direction may have sufficient kinetic energy to break through the surface. Then they become molecules of vapour. The average kinetic energy ofthe liquid remaining is therefore reduced, and so its temperature is lowered.\n\n', '', 'Explanation\n(a)(i) Temperature is a number which expresses how hot or cold a body is on scale such as centigrade orkelvin. S.I. unit is kelvin (K). Temperature decides which direction heat will flow while, heat is a form of energy S.I. is Joules (J).\n(ii) Two physical properties used for measuring temperature are:- (1) Change in volume of liquid with temperature (2) Change in resistance. Others are: (3) Expansion of solids (4) Expansion of volumeof gas.\n(b)(i) Determination of the upper - fixed point for a mercury-in-glass thermometer.\n\nTo determine the upper fixed point, a hypsometer is used. The experiment is set up as shown above. The manometer is used to measure the pressure of the steam.\nIf it is not 76cm Hg, an adjustment is made to the upper fixed point. When the mercury level is constant, a mark is made on its glass to fix the point. One precaution that must be observed in this experiment is that the thermometer must not touch the water, because its temperature is affected by dissolved impurities in the water.\n(ii) The constant - volume gas thermometer measures temperature more accurately than other thermometers because its glass expansion is negligible compared to the expansion of the gas. It is seldom used as everyday laboratory thermometer because it is not easily handled unlike the laboratory thermometer.\n(c) In the kinetic theory, molecules are in random motion. They all have different velocities. Near the surface, some liquid molecules with high velocities in an upward direction may have sufficient kinetic energy to break through the surface. Then they become molecules of vapour. The average kinetic energy ofthe liquid remaining is therefore reduced, and so its temperature is lowered.\n\n', '(a)(i) Temperature is a number which expresses how hot or cold a body is on scale such as centigrade orkelvin. S.I. unit is kelvin (K). Temperature decides which direction heat will flow while, heat is a form of energy S.I. is Joules (J).\n(ii) Two physical properties used for measuring temperature are:- (1) Change in volume of liquid with temperature (2) Change in resistance. Others are: (3) Expansion of solids (4) Expansion of volumeof gas.~(b)(i) Determination of the upper - fixed point for a mercury-in-glass thermometer.\n\nTo determine the upper fixed point, a hypsometer is used. The experiment is set up as shown above. The manometer is used to measure the pressure of the steam.\nIf it is not 76cm Hg, an adjustment is made to the upper fixed point. When the mercury level is constant, a mark is made on its glass to fix the point. One precaution that must be observed in this experiment is that the thermometer must not touch the water, because its temperature is affected by dissolved impurities in the water.\n(ii) The constant - volume gas thermometer measures temperature more accurately than other thermometers because its glass expansion is negligible compared to the expansion of the gas. It is seldom used as everyday laboratory thermometer because it is not easily handled unlike the laboratory thermometer.\n(c) In the kinetic theory, molecules are in random motion. They all have different velocities. Near the surface, some liquid molecules with high velocities in an upward direction may have sufficient kinetic energy to break through the surface. Then they become molecules of vapour. The average kinetic energy ofthe liquid remaining is therefore reduced, and so its temperature is lowered.\n~(c) In the kinetic theory, molecules are in random motion. They all have different velocities. Near the surface, some liquid molecules with high velocities in an upward direction may have sufficient kinetic energy to break through the surface. Then they become molecules of vapour. The average kinetic energy ofthe liquid remaining is therefore reduced, and so its temperature is lowered.\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(299, '2019', '\n\nQuestion\n\n(a) Explain the terms reactance and impedance in an a.c circuit.\n(b) A source of e.m.f. 240 v and frequency 50 Hz is connected to a resistor, an inductor and a capacitor in series. When the current in the capacitor is 10A, the potential difference across the resistor is 140V and that across the inductor is 50V. Draw the vector diagram of the potential difference across the inductor, the capacitor and the resistor.\nCalculate the (i) potential difference across the capacitor; (ii) capacitance of the capacitor; (iii) inductance of the inductor.\n\n', '', '(a) Explain the terms reactance and impedance in an a.c circuit.\n(b) A source of e.m.f. 240 v and frequency 50 Hz is connected to a resistor, an inductor and a capacitor in series. When the current in the capacitor is 10A, the potential difference across the resistor is 140V and that across the inductor is 50V. Draw the vector diagram of the potential difference across the inductor, the capacitor and the resistor.\nCalculate the (i) potential difference across the capacitor; (ii) capacitance of the capacitor; (iii) inductance of the inductor.', '(a) Explain the terms reactance and impedance in an a.c circuit.~(b) A source of e.m.f. 240 v and frequency 50 Hz is connected to a resistor, an inductor and a capacitor in series. When the current in the capacitor is 10A, the potential difference across the resistor is 140V and that across the inductor is 50V. Draw the vector diagram of the potential difference across the inductor, the capacitor and the resistor.\nCalculate the (i) potential difference across the capacitor; (ii) capacitance of the capacitor; (iii) inductance of the inductor.', NULL, '\n\nExplanation\n(a) Reactance is the opposition in ohms to th alternating current by a capacitor or inductor. It is defined by\n\\(X_{C} = \\frac{V}{I}\\) or \\(X_{L} = \\frac{V}{I}\\)\n\\(X_{C} = \\frac{1}{2 \\pi f c}\\) or \\(X_{L} = 2 \\pi fl\\)\nV - voltage (V), c - current, f - frequency (Hz), I - inductance (H), c - capacitance (F).\nImpedance is the opposition in ohms to the alternating current by combination of resistor, inductor or capacitor. It is defined as:\n\\(Z= \\sqrt{R^{2} + X_{L} ^{2}}\\)\n\\(Z= \\sqrt{R^{2} + X^{2}}\\)\n\\(Z= \\sqrt{R^{2} + (X_{C} - X_{L})^{2}}\\)\n(b)\nHence using,\n\\(V = \\sqrt{V_{R} ^{2} + (V_{L} - V_{C})^{2}}\\)\n\\(240 = \\sqrt{140^{2} + (50 - V_{C})^{2}}\\)\nSquare both sides,\n\\(57600 = 19600 + (50 - V_{C})^{2} \\implies 57600 - 19600 = (50 - V_{C})^{2}\\)\n\\(38000 = (50 - V_{C})^{2}\\)\n\\(\\sqrt{38000} = 50 - V_{C}\\)\n\\(\\pm 194.936 = 50 - V_{C} \\implies V_{C} = \\pm 194.936 +50\\)\n\\(V_{C} = -144.936 ; V_{C} = 244.936v\\)\nThe voltage cannot be negative, hence, \\(V_{C} = 244.936v\\)\n(ii) Using \\(X_{C} = \\frac{1}{2 \\pi fc} = \\frac{V}{I}\\)\n\\(\\frac{244.936}{10} = \\frac{1}{2\\pi \\times 50 \\times c}\\)\n\\(24.4936 = \\frac{1}{2\\pi \\times 50 \\times c} \\implies c = \\frac{1}{2\\pi \\times 50 \\times 24.4936}\\)\n\\(c = \\frac{1}{7694.89} = 0.00013F = 130\\mu F\\)\n\n', '', 'Explanation\n(a) Reactance is the opposition in ohms to th alternating current by a capacitor or inductor. It is defined by\n\\(X_{C} = \\frac{V}{I}\\) or \\(X_{L} = \\frac{V}{I}\\)\n\\(X_{C} = \\frac{1}{2 \\pi f c}\\) or \\(X_{L} = 2 \\pi fl\\)\nV - voltage (V), c - current, f - frequency (Hz), I - inductance (H), c - capacitance (F).\nImpedance is the opposition in ohms to the alternating current by combination of resistor, inductor or capacitor. It is defined as:\n\\(Z= \\sqrt{R^{2} + X_{L} ^{2}}\\)\n\\(Z= \\sqrt{R^{2} + X^{2}}\\)\n\\(Z= \\sqrt{R^{2} + (X_{C} - X_{L})^{2}}\\)\n(b)\nHence using,\n\\(V = \\sqrt{V_{R} ^{2} + (V_{L} - V_{C})^{2}}\\)\n\\(240 = \\sqrt{140^{2} + (50 - V_{C})^{2}}\\)\nSquare both sides,\n\\(57600 = 19600 + (50 - V_{C})^{2} \\implies 57600 - 19600 = (50 - V_{C})^{2}\\)\n\\(38000 = (50 - V_{C})^{2}\\)\n\\(\\sqrt{38000} = 50 - V_{C}\\)\n\\(\\pm 194.936 = 50 - V_{C} \\implies V_{C} = \\pm 194.936 +50\\)\n\\(V_{C} = -144.936 ; V_{C} = 244.936v\\)\nThe voltage cannot be negative, hence, \\(V_{C} = 244.936v\\)\n(ii) Using \\(X_{C} = \\frac{1}{2 \\pi fc} = \\frac{V}{I}\\)\n\\(\\frac{244.936}{10} = \\frac{1}{2\\pi \\times 50 \\times c}\\)\n\\(24.4936 = \\frac{1}{2\\pi \\times 50 \\times c} \\implies c = \\frac{1}{2\\pi \\times 50 \\times 24.4936}\\)\n\\(c = \\frac{1}{7694.89} = 0.00013F = 130\\mu F\\)\n\n', '(a) Reactance is the opposition in ohms to th alternating current by a capacitor or inductor. It is defined by\n\\(X_{C} = \\frac{V}{I}\\) or \\(X_{L} = \\frac{V}{I}\\)\n\\(X_{C} = \\frac{1}{2 \\pi f c}\\) or \\(X_{L} = 2 \\pi fl\\)\nV - voltage (V), c - current, f - frequency (Hz), I - inductance (H), c - capacitance (F).\nImpedance is the opposition in ohms to the alternating current by combination of resistor, inductor or capacitor. It is defined as:\n\\(Z= \\sqrt{R^{2} + X_{L} ^{2}}\\)\n\\(Z= \\sqrt{R^{2} + X^{2}}\\)\n\\(Z= \\sqrt{R^{2} + (X_{C} - X_{L})^{2}}\\)~(b)\nHence using,\n\\(V = \\sqrt{V_{R} ^{2} + (V_{L} - V_{C})^{2}}\\)\n\\(240 = \\sqrt{140^{2} + (50 - V_{C})^{2}}\\)\nSquare both sides,\n\\(57600 = 19600 + (50 - V_{C})^{2} \\implies 57600 - 19600 = (50 - V_{C})^{2}\\)\n\\(38000 = (50 - V_{C})^{2}\\)\n\\(\\sqrt{38000} = 50 - V_{C}\\)\n\\(\\pm 194.936 = 50 - V_{C} \\implies V_{C} = \\pm 194.936 +50\\)\n\\(V_{C} = -144.936 ; V_{C} = 244.936v\\)\nThe voltage cannot be negative, hence, \\(V_{C} = 244.936v\\)\n(ii) Using \\(X_{C} = \\frac{1}{2 \\pi fc} = \\frac{V}{I}\\)\n\\(\\frac{244.936}{10} = \\frac{1}{2\\pi \\times 50 \\times c}\\)\n\\(24.4936 = \\frac{1}{2\\pi \\times 50 \\times c} \\implies c = \\frac{1}{2\\pi \\times 50 \\times 24.4936}\\)\n\\(c = \\frac{1}{7694.89} = 0.00013F = 130\\mu F\\)\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
(300, '2019', '\n\nQuestion\n\n(a)(i) By means of a labelled diagram, describe the mode uf operation of a modern X-ray tube.\n(ii) State the energy transformation which takes place during the operation.\n(b) Explain the terms hardness and intensity as applied to X-rays\n(c)(i) State three uses of X-rays\n(ii) State one hazard of over-exposure to X-rays in a radiological laboratory, indicating two safety precautions.\n\n', '', '(a)(i) By means of a labelled diagram, describe the mode uf operation of a modern X-ray tube.\n(ii) State the energy transformation which takes place during the operation.\n(b) Explain the terms hardness and intensity as applied to X-rays\n(c)(i) State three uses of X-rays\n(ii) State one hazard of over-exposure to X-rays in a radiological laboratory, indicating two safety precautions.', '(a)(i) By means of a labelled diagram, describe the mode uf operation of a modern X-ray tube.\n(ii) State the energy transformation which takes place during the operation.~(b) Explain the terms hardness and intensity as applied to X-rays\n(c)(i) State three uses of X-rays\n(ii) State one hazard of over-exposure to X-rays in a radiological laboratory, indicating two safety precautions.~(c)(i) State three uses of X-rays\n(ii) State one hazard of over-exposure to X-rays in a radiological laboratory, indicating two safety precautions.', NULL, '\n\nExplanation\n(a)(i)\nMode of Operation:\nInthe X-ray tube a Low voltage high p.d isapplied between the hot cathode and the anode. Electrons are emitted from the cathode and are accelerated to an extremely high speed. They are suddenly decelerated when they strike the anode causing the emission of high - energy radiation of short wavelength - X-ray. In this process the anode becomes very hot, and requires cooling fins on the outside of the tube.\nEnergy Transformation: The electrons have kinetic energy which is partly changed to internal energy (potential) and finally changed to heat energy.\n(b) Hardness: Refers to X-rays with high penetrating power, short wavelength but high frequency. They are produced by increasing the p.d. between the cathode and anode. Such X-rays are called hard X-rays. Less penetrating or lower frequency ones are called soft X-rays. They are produced at lower p.d.\nIntensity: High intensity X-rays beam results from the increase of current through the filament which leads to increase in the number of electrons.\n(c)(i) Uses of X-rays:\n(1) Investigating the structure of crystals ; (2) X-ray examination of the body for broken bones; (3) As invincible bullets for the treatment of tumours ; (4) Examination of metals for defects and fractures.\n(ii) They may cause damage to living tissue and to cells. Precaution to be taken includes lead screening and regular health checks.\n\n', '', 'Explanation\n(a)(i)\nMode of Operation:\nInthe X-ray tube a Low voltage high p.d isapplied between the hot cathode and the anode. Electrons are emitted from the cathode and are accelerated to an extremely high speed. They are suddenly decelerated when they strike the anode causing the emission of high - energy radiation of short wavelength - X-ray. In this process the anode becomes very hot, and requires cooling fins on the outside of the tube.\nEnergy Transformation: The electrons have kinetic energy which is partly changed to internal energy (potential) and finally changed to heat energy.\n(b) Hardness: Refers to X-rays with high penetrating power, short wavelength but high frequency. They are produced by increasing the p.d. between the cathode and anode. Such X-rays are called hard X-rays. Less penetrating or lower frequency ones are called soft X-rays. They are produced at lower p.d.\nIntensity: High intensity X-rays beam results from the increase of current through the filament which leads to increase in the number of electrons.\n(c)(i) Uses of X-rays:\n(1) Investigating the structure of crystals ; (2) X-ray examination of the body for broken bones; (3) As invincible bullets for the treatment of tumours ; (4) Examination of metals for defects and fractures.\n(ii) They may cause damage to living tissue and to cells. Precaution to be taken includes lead screening and regular health checks.\n\n', '(a)(i)\nMode of Operation:\nInthe X-ray tube a Low voltage high p.d isapplied between the hot cathode and the anode. Electrons are emitted from the cathode and are accelerated to an extremely high speed. They are suddenly decelerated when they strike the anode causing the emission of high - energy radiation of short wavelength - X-ray. In this process the anode becomes very hot, and requires cooling fins on the outside of the tube.\nEnergy Transformation: The electrons have kinetic energy which is partly changed to internal energy (potential) and finally changed to heat energy.~(b) Hardness: Refers to X-rays with high penetrating power, short wavelength but high frequency. They are produced by increasing the p.d. between the cathode and anode. Such X-rays are called hard X-rays. Less penetrating or lower frequency ones are called soft X-rays. They are produced at lower p.d.\nIntensity: High intensity X-rays beam results from the increase of current through the filament which leads to increase in the number of electrons.\n(c)(i) Uses of X-rays:\n(1) Investigating the structure of crystals ; (2) X-ray examination of the body for broken bones; (3) As invincible bullets for the treatment of tumours ; (4) Examination of metals for defects and fractures.\n(ii) They may cause damage to living tissue and to cells. Precaution to be taken includes lead screening and regular health checks.\n~(c)(i) Uses of X-rays:\n(1) Investigating the structure of crystals ; (2) X-ray examination of the body for broken bones; (3) As invincible bullets for the treatment of tumours ; (4) Examination of metals for defects and fractures.\n(ii) They may cause damage to living tissue and to cells. Precaution to be taken includes lead screening and regular health checks.\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(301, '2020', '\n\nQuestion\n\n(a) Explain what is meant by the following statement. The specific latent heat of fusion of ice is \\(3.4 \\times10^{5}Jkg^{-1}\\).\n(b) Describe an experiment to determine the specific latent heat of fusion of ice. State two precautions necessary to obtain an accurate result.\n(c) Using the kinetic theory of matter, explain why ice can change to water at 0°C without any change in temperature.\n\n', '', '(a) Explain what is meant by the following statement. The specific latent heat of fusion of ice is \\(3.4 \\times10^{5}Jkg^{-1}\\).\n(b) Describe an experiment to determine the specific latent heat of fusion of ice. State two precautions necessary to obtain an accurate result.\n(c) Using the kinetic theory of matter, explain why ice can change to water at 0°C without any change in temperature.', '(a) Explain what is meant by the following statement. The specific latent heat of fusion of ice is \\(3.4 \\times10^{5}Jkg^{-1}\\).~(b) Describe an experiment to determine the specific latent heat of fusion of ice. State two precautions necessary to obtain an accurate result.\n(c) Using the kinetic theory of matter, explain why ice can change to water at 0°C without any change in temperature.~(c) Using the kinetic theory of matter, explain why ice can change to water at 0°C without any change in temperature.', NULL, '\n\nExplanation\n(a) The specific latent heat of fusion of ice is \\(3.4 \\times10^{5}Jkg^{-1}\\) means that \\(3.4\\times10^{5} J\\) of hoat energy is required to change 1kg mass of ice from solid state to the liquid state without a change in temperature.\n\nWeigh a clean, dry copper calorimeter empty, together with stirer. Half fill the calorimeter with water previously warmedto about 8°C above room temperature and reweigh it to find the mass of water. Record the temperature of the water when it is say about 5°C above room temperature and then immediately add small pieces of ice, to the water. Stir the water until all the ice is melted before adding another piece. Continue until final temperature is 5°C below the room temperature. Record this temperature. Finally, reweigh the calorimeter to find the massof ice melted.\nCalculations: Heat gained byice in forming water = Heat lost by water andcalorimeter\n\\(M_{1} L + M_{1} C_{w} \\theta_{1} = (M_{w} C_{w} + M_{c} C_{c}) \\Delta \\theta\\)\n\\(\\therefore L = \\frac{(M_{w} C_{w} + M_{c} C_{c}) \\Delta \\theta - M_{1} C_{w} \\theta_{f}}{M_{1}}\\)\nwhere \\(M_{1} - \\text{mass of ice}\\)\n\\(M_{c} - \\text{mass of calorimeter}\\)\n\\(C_{w} - \\text{specific heat capacity of water}\\)\n\\(C_{c} - \\text{specific heat capacity of calorimeter}\\)\n\\(\\theta_{f} - \\text{final temperature}\\)\n\\(\\Delta \\theta - \\text{change in temperature of calorimeter and water}\\)\n(c)In a solid, such as ice, the molecules, which are held together by inter molecular forces, vibrate about a mean position. As ice absorbs more heat, the vibration of the molecules have enough energy to overcome the force binding them together. They are now free to move about in a random fashion as in a liquid.\n\n', '', 'Explanation\n(a) The specific latent heat of fusion of ice is \\(3.4 \\times10^{5}Jkg^{-1}\\) means that \\(3.4\\times10^{5} J\\) of hoat energy is required to change 1kg mass of ice from solid state to the liquid state without a change in temperature.\n\nWeigh a clean, dry copper calorimeter empty, together with stirer. Half fill the calorimeter with water previously warmedto about 8°C above room temperature and reweigh it to find the mass of water. Record the temperature of the water when it is say about 5°C above room temperature and then immediately add small pieces of ice, to the water. Stir the water until all the ice is melted before adding another piece. Continue until final temperature is 5°C below the room temperature. Record this temperature. Finally, reweigh the calorimeter to find the massof ice melted.\nCalculations: Heat gained byice in forming water = Heat lost by water andcalorimeter\n\\(M_{1} L + M_{1} C_{w} \\theta_{1} = (M_{w} C_{w} + M_{c} C_{c}) \\Delta \\theta\\)\n\\(\\therefore L = \\frac{(M_{w} C_{w} + M_{c} C_{c}) \\Delta \\theta - M_{1} C_{w} \\theta_{f}}{M_{1}}\\)\nwhere \\(M_{1} - \\text{mass of ice}\\)\n\\(M_{c} - \\text{mass of calorimeter}\\)\n\\(C_{w} - \\text{specific heat capacity of water}\\)\n\\(C_{c} - \\text{specific heat capacity of calorimeter}\\)\n\\(\\theta_{f} - \\text{final temperature}\\)\n\\(\\Delta \\theta - \\text{change in temperature of calorimeter and water}\\)\n(c)In a solid, such as ice, the molecules, which are held together by inter molecular forces, vibrate about a mean position. As ice absorbs more heat, the vibration of the molecules have enough energy to overcome the force binding them together. They are now free to move about in a random fashion as in a liquid.\n\n', '~(c)In a solid, such as ice, the molecules, which are held together by inter molecular forces, vibrate about a mean position. As ice absorbs more heat, the vibration of the molecules have enough energy to overcome the force binding them together. They are now free to move about in a random fashion as in a liquid.\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(302, '2020', '\n\nQuestion\n\n(a)(i) What is meant by resonance?\n(ii) Outline the necessary steps taken in a simple experiment to illustrate top resonance\n(iii) Explain why a vibrating tuning fork sounds louder when its stem is pressed against a table top than when held in air.\n(b) Explain with the aid of diagrams, how a concave mirror could be used to: (i) Ignite a piece of carbon paper; (ii) produce an exact copy of a picture on a screen.\n\n', '', '(a)(i) What is meant by resonance?\n(ii) Outline the necessary steps taken in a simple experiment to illustrate top resonance\n(iii) Explain why a vibrating tuning fork sounds louder when its stem is pressed against a table top than when held in air.\n(b) Explain with the aid of diagrams, how a concave mirror could be used to: (i) Ignite a piece of carbon paper; (ii) produce an exact copy of a picture on a screen.', '(a)(i) What is meant by resonance?\n(ii) Outline the necessary steps taken in a simple experiment to illustrate top resonance\n(iii) Explain why a vibrating tuning fork sounds louder when its stem is pressed against a table top than when held in air.~(b) Explain with the aid of diagrams, how a concave mirror could be used to: (i) Ignite a piece of carbon paper; (ii) produce an exact copy of a picture on a screen.', NULL, '\n\nExplanation\n(a) Resonance occurs when one vibrating body sets other body into vibration at its own natural frequency as a result of impulse received from the first body. (ii) Simple experiment to illustrate resonance.\n\nThe apparatus is arranged as shown. A cylinder is 3/4 filled with water and a tube is dipped into the water so that the length I can be varied. A tuning fork f is set vibrating just above the tube but not touching it. As the fork is set vibrating the length I is gradually increased. A point is reached when a large sound is heard. At this point resonance has occurred because the vibrating tuning fork has set the air in the tube vibrating at its own frequency.\n(b)\nParallel light rays are reflected by a concave mirror on a piece of carbon paper. The converging of the rays on the carbon paper will cause it to ignite.\n\nThe object Ois kept on the centre of curvature of a concave mirror. An exact but inverted image I is formed at the centre of curvature where the screen should be placed.\n\n', '', 'Explanation\n(a) Resonance occurs when one vibrating body sets other body into vibration at its own natural frequency as a result of impulse received from the first body. (ii) Simple experiment to illustrate resonance.\n\nThe apparatus is arranged as shown. A cylinder is 3/4 filled with water and a tube is dipped into the water so that the length I can be varied. A tuning fork f is set vibrating just above the tube but not touching it. As the fork is set vibrating the length I is gradually increased. A point is reached when a large sound is heard. At this point resonance has occurred because the vibrating tuning fork has set the air in the tube vibrating at its own frequency.\n(b)\nParallel light rays are reflected by a concave mirror on a piece of carbon paper. The converging of the rays on the carbon paper will cause it to ignite.\n\nThe object Ois kept on the centre of curvature of a concave mirror. An exact but inverted image I is formed at the centre of curvature where the screen should be placed.\n\n', '(a) Resonance occurs when one vibrating body sets other body into vibration at its own natural frequency as a result of impulse received from the first body. (ii) Simple experiment to illustrate resonance.\n\nThe apparatus is arranged as shown. A cylinder is 3/4 filled with water and a tube is dipped into the water so that the length I can be varied. A tuning fork f is set vibrating just above the tube but not touching it. As the fork is set vibrating the length I is gradually increased. A point is reached when a large sound is heard. At this point resonance has occurred because the vibrating tuning fork has set the air in the tube vibrating at its own frequency.~(b)\nParallel light rays are reflected by a concave mirror on a piece of carbon paper. The converging of the rays on the carbon paper will cause it to ignite.\n\nThe object Ois kept on the centre of curvature of a concave mirror. An exact but inverted image I is formed at the centre of curvature where the screen should be placed.\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(303, '2020', '\n\nQuestion\n\n(a) Draw a simple labelled diagram illustrating the principle of a step-down transformer and explain how it works\n(b) State three ways by which energy is lost in a transformer and how they can be minimized.\n(c) If a transformer is used to light a lamp rated at 60W, 220V from a 4400V a.c. supply, calculate the; (i) ratio of the number of turns of the primary coil to the secondary coil in the transformer (ii) current taken from the main circuit if the efficiency of the transforme is 95%.\n\n', '', '(a) Draw a simple labelled diagram illustrating the principle of a step-down transformer and explain how it works\n(b) State three ways by which energy is lost in a transformer and how they can be minimized.\n(c) If a transformer is used to light a lamp rated at 60W, 220V from a 4400V a.c. supply, calculate the; (i) ratio of the number of turns of the primary coil to the secondary coil in the transformer (ii) current taken from the main circuit if the efficiency of the transforme is 95%.', '(a) Draw a simple labelled diagram illustrating the principle of a step-down transformer and explain how it works~(b) State three ways by which energy is lost in a transformer and how they can be minimized.\n(c) If a transformer is used to light a lamp rated at 60W, 220V from a 4400V a.c. supply, calculate the; (i) ratio of the number of turns of the primary coil to the secondary coil in the transformer (ii) current taken from the main circuit if the efficiency of the transforme is 95%.~(c) If a transformer is used to light a lamp rated at 60W, 220V from a 4400V a.c. supply, calculate the; (i) ratio of the number of turns of the primary coil to the secondary coil in the transformer (ii) current taken from the main circuit if the efficiency of the transforme is 95%.', NULL, '\n\nExplanation\n(a)\nThe step-down transformer has more turns in the primary turns than in the secondary turns. It works by electromagnetic induction, the magnetic flux from the primary coil in the iron core links the secondary coil. When an alternating e.m.f \\(E_{p}\\) is connected to the primary coil, an alternating current flows. This produces corresponding variation of magnetic flux in the iron core which links the secondary coil. An induced e.m.f \\(E_{s}\\) of the same frequency as \\(E_{p}\\) is obtained. From Faraday\'s law, the e.m.f induced in a coil proportional to the rate of change of the magnetic flux linking the coil. So,\n\\(\\frac{E_{s}}{E_{p}} = \\frac{N_{s}}{N_{p}}\\). \\(N_{s}\\) - Number of turns in secondary coil; \\(N_{p}\\) - number of turns in primary coil.\n(b)Three ways by which energy is lost and how they can be minimised are:\n(1) Energy is lost through Eddy current produced in the soft — iron. This is reduced by laminating the core in thin strips. (2) Heat generation in the copper coils. This is reduced by using copper coils with low resistance. (3). Loss of energy due to reversal of iroes of force. This is reduced by using soft magnetic materials for the core.\n(c)(i) Using \\(\\frac{E_{p}}{E_{s}} = \\frac{N_{p}}{N_{s}}\\)\n\\(\\frac{4400}{220} = \\frac{N_{p}}{N_{s}}\\)\nThe ratio of the primary turns to the secondary turns is 20 : 1.\n(ii) \\(Efficiency = \\frac{\\text{power output}}{\\text{power input}}\\)\n\\(\\frac{95}{100} = \\frac{60}{input}\\)\n\\(\\text{Power input} = \\frac{6000}{95} = 63.2W\\)\n\\(\\therefore Current I = \\frac{P}{V} = \\frac{63.2}{4400}\\)\n\\(I = 0.0144A = 14.4 mA\\)\n\n', '', 'Explanation\n(a)\nThe step-down transformer has more turns in the primary turns than in the secondary turns. It works by electromagnetic induction, the magnetic flux from the primary coil in the iron core links the secondary coil. When an alternating e.m.f \\(E_{p}\\) is connected to the primary coil, an alternating current flows. This produces corresponding variation of magnetic flux in the iron core which links the secondary coil. An induced e.m.f \\(E_{s}\\) of the same frequency as \\(E_{p}\\) is obtained. From Faraday\'s law, the e.m.f induced in a coil proportional to the rate of change of the magnetic flux linking the coil. So,\n\\(\\frac{E_{s}}{E_{p}} = \\frac{N_{s}}{N_{p}}\\). \\(N_{s}\\) - Number of turns in secondary coil; \\(N_{p}\\) - number of turns in primary coil.\n(b)Three ways by which energy is lost and how they can be minimised are:\n(1) Energy is lost through Eddy current produced in the soft — iron. This is reduced by laminating the core in thin strips. (2) Heat generation in the copper coils. This is reduced by using copper coils with low resistance. (3). Loss of energy due to reversal of iroes of force. This is reduced by using soft magnetic materials for the core.\n(c)(i) Using \\(\\frac{E_{p}}{E_{s}} = \\frac{N_{p}}{N_{s}}\\)\n\\(\\frac{4400}{220} = \\frac{N_{p}}{N_{s}}\\)\nThe ratio of the primary turns to the secondary turns is 20 : 1.\n(ii) \\(Efficiency = \\frac{\\text{power output}}{\\text{power input}}\\)\n\\(\\frac{95}{100} = \\frac{60}{input}\\)\n\\(\\text{Power input} = \\frac{6000}{95} = 63.2W\\)\n\\(\\therefore Current I = \\frac{P}{V} = \\frac{63.2}{4400}\\)\n\\(I = 0.0144A = 14.4 mA\\)\n\n', '(a)\nThe step-down transformer has more turns in the primary turns than in the secondary turns. It works by electromagnetic induction, the magnetic flux from the primary coil in the iron core links the secondary coil. When an alternating e.m.f \\(E_{p}\\) is connected to the primary coil, an alternating current flows. This produces corresponding variation of magnetic flux in the iron core which links the secondary coil. An induced e.m.f \\(E_{s}\\) of the same frequency as \\(E_{p}\\) is obtained. From Faraday\'s law, the e.m.f induced in a coil proportional to the rate of change of the magnetic flux linking the coil. So,\n\\(\\frac{E_{s}}{E_{p}} = \\frac{N_{s}}{N_{p}}\\). \\(N_{s}\\) - Number of turns in secondary coil; \\(N_{p}\\) - number of turns in primary coil.~(b)Three ways by which energy is lost and how they can be minimised are:\n(1) Energy is lost through Eddy current produced in the soft — iron. This is reduced by laminating the core in thin strips. (2) Heat generation in the copper coils. This is reduced by using copper coils with low resistance. (3). Loss of energy due to reversal of iroes of force. This is reduced by using soft magnetic materials for the core.\n(c)(i) Using \\(\\frac{E_{p}}{E_{s}} = \\frac{N_{p}}{N_{s}}\\)\n\\(\\frac{4400}{220} = \\frac{N_{p}}{N_{s}}\\)\nThe ratio of the primary turns to the secondary turns is 20 : 1.\n(ii) \\(Efficiency = \\frac{\\text{power output}}{\\text{power input}}\\)\n\\(\\frac{95}{100} = \\frac{60}{input}\\)\n\\(\\text{Power input} = \\frac{6000}{95} = 63.2W\\)\n\\(\\therefore Current I = \\frac{P}{V} = \\frac{63.2}{4400}\\)\n\\(I = 0.0144A = 14.4 mA\\)\n~(c)(i) Using \\(\\frac{E_{p}}{E_{s}} = \\frac{N_{p}}{N_{s}}\\)\n\\(\\frac{4400}{220} = \\frac{N_{p}}{N_{s}}\\)\nThe ratio of the primary turns to the secondary turns is 20 : 1.\n(ii) \\(Efficiency = \\frac{\\text{power output}}{\\text{power input}}\\)\n\\(\\frac{95}{100} = \\frac{60}{input}\\)\n\\(\\text{Power input} = \\frac{6000}{95} = 63.2W\\)\n\\(\\therefore Current I = \\frac{P}{V} = \\frac{63.2}{4400}\\)\n\\(I = 0.0144A = 14.4 mA\\)\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(304, '2020', '\n\nQuestion\n\n(a) Explain what is meant by photoelectric emission\n(b) Draw a labelled diagram showing the structure of a simple type of a photocell and explain its mode of operation.\n(c) State four applications of photoelectric emission.\n(d) In a photocell, no electrons are emitted until the threshold frequency of light is reached. Explain what happens to the energy of the light before emission of electrons begins. State one factor that may affect the number of emitted electrons.\n\n', '', '(a) Explain what is meant by photoelectric emission\n(b) Draw a labelled diagram showing the structure of a simple type of a photocell and explain its mode of operation.\n(c) State four applications of photoelectric emission.\n(d) In a photocell, no electrons are emitted until the threshold frequency of light is reached. Explain what happens to the energy of the light before emission of electrons begins. State one factor that may affect the number of emitted electrons.', '(a) Explain what is meant by photoelectric emission~(b) Draw a labelled diagram showing the structure of a simple type of a photocell and explain its mode of operation.\n(c) State four applications of photoelectric emission.\n(d) In a photocell, no electrons are emitted until the threshold frequency of light is reached. Explain what happens to the energy of the light before emission of electrons begins. State one factor that may affect the number of emitted electrons.~(c) State four applications of photoelectric emission.\n(d) In a photocell, no electrons are emitted until the threshold frequency of light is reached. Explain what happens to the energy of the light before emission of electrons begins. State one factor that may affect the number of emitted electrons.~(d) In a photocell, no electrons are emitted until the threshold frequency of light is reached. Explain what happens to the energy of the light before emission of electrons begins. State one factor that may affect the number of emitted electrons.', NULL, '\n\nExplanation\n(a) When ultraviolet rays fall on zinc plate, electrons are liberated from the zinc plate. This phenomenon is called photoelectric emission. In general the emission of electrons from matter as a result of electromagnetic wave falling on the matter is referred to as photoelectric effect.Emission of electrons depends only on the threshold frequency. But the rate of emission depends on the intensity of light. The threshold frequency is not the same for all metals. The energy of emitted electrons varies from zero to a maximum value.\n(b) photo cell\n\nWhen light photons fall on the caesium metal, it emits electrons which move to the collector to generate current flow. As the intensity of Iight varies, the rate of emission of electrons varies and thus, the current flow varies.\n(c) Applications of photoelectric emission.\n(1) Production of television cameras (2) Burglary/ alarms (3) Automatic switches for putting on lights at dusk. (4) Industrial controls and counting operations.\n(d) The energy of the light is absorbed by the photocell and are used to overcome the force that brids the electrons but may not be sufficient to remove the electron until light with sufficient threshold frequency. One factor that may affect the number of emitted electrons is the intensity of the light.\n\n', '', 'Explanation\n(a) When ultraviolet rays fall on zinc plate, electrons are liberated from the zinc plate. This phenomenon is called photoelectric emission. In general the emission of electrons from matter as a result of electromagnetic wave falling on the matter is referred to as photoelectric effect.Emission of electrons depends only on the threshold frequency. But the rate of emission depends on the intensity of light. The threshold frequency is not the same for all metals. The energy of emitted electrons varies from zero to a maximum value.\n(b) photo cell\n\nWhen light photons fall on the caesium metal, it emits electrons which move to the collector to generate current flow. As the intensity of Iight varies, the rate of emission of electrons varies and thus, the current flow varies.\n(c) Applications of photoelectric emission.\n(1) Production of television cameras (2) Burglary/ alarms (3) Automatic switches for putting on lights at dusk. (4) Industrial controls and counting operations.\n(d) The energy of the light is absorbed by the photocell and are used to overcome the force that brids the electrons but may not be sufficient to remove the electron until light with sufficient threshold frequency. One factor that may affect the number of emitted electrons is the intensity of the light.\n\n', '(a) When ultraviolet rays fall on zinc plate, electrons are liberated from the zinc plate. This phenomenon is called photoelectric emission. In general the emission of electrons from matter as a result of electromagnetic wave falling on the matter is referred to as photoelectric effect.Emission of electrons depends only on the threshold frequency. But the rate of emission depends on the intensity of light. The threshold frequency is not the same for all metals. The energy of emitted electrons varies from zero to a maximum value.~(b) photo cell\n\nWhen light photons fall on the caesium metal, it emits electrons which move to the collector to generate current flow. As the intensity of Iight varies, the rate of emission of electrons varies and thus, the current flow varies.\n(c) Applications of photoelectric emission.\n(1) Production of television cameras (2) Burglary/ alarms (3) Automatic switches for putting on lights at dusk. (4) Industrial controls and counting operations.\n(d) The energy of the light is absorbed by the photocell and are used to overcome the force that brids the electrons but may not be sufficient to remove the electron until light with sufficient threshold frequency. One factor that may affect the number of emitted electrons is the intensity of the light.\n~(c) Applications of photoelectric emission.\n(1) Production of television cameras (2) Burglary/ alarms (3) Automatic switches for putting on lights at dusk. (4) Industrial controls and counting operations.\n(d) The energy of the light is absorbed by the photocell and are used to overcome the force that brids the electrons but may not be sufficient to remove the electron until light with sufficient threshold frequency. One factor that may affect the number of emitted electrons is the intensity of the light.\n~(d) The energy of the light is absorbed by the photocell and are used to overcome the force that brids the electrons but may not be sufficient to remove the electron until light with sufficient threshold frequency. One factor that may affect the number of emitted electrons is the intensity of the light.\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(305, '2020', '\n\nQuestion\n\n(a)(i) Explain what is meant by a machine (ii) Define the terms: mechanical advantage, velocity ratio and efficiency as applied to a machine. Derive the equation connecting the three terms.\n(b) Explain why the efficiency of a machine is usually less than 100%\n(c) A screw jack whose pitch is 4.4mm is used to raise a body of mass 8000 kg through a height of 20cm. The length of the tommy bar of the jack is 70cm. If the efficiency of the jack is 80%, calcuate the: (i) velocity ratio of the jack; (ii) mechanical advantage of the jack (iii) effort required in raising the body, (iv) work done by the effort in raising the body \\((g = 10ms^{-2}; \\pi= \\frac{22}{7}\\))\n\n', '', '(a)(i) Explain what is meant by a machine (ii) Define the terms: mechanical advantage, velocity ratio and efficiency as applied to a machine. Derive the equation connecting the three terms.\n(b) Explain why the efficiency of a machine is usually less than 100%\n(c) A screw jack whose pitch is 4.4mm is used to raise a body of mass 8000 kg through a height of 20cm. The length of the tommy bar of the jack is 70cm. If the efficiency of the jack is 80%, calcuate the: (i) velocity ratio of the jack; (ii) mechanical advantage of the jack (iii) effort required in raising the body, (iv) work done by the effort in raising the body \\((g = 10ms^{-2}; \\pi= \\frac{22}{7}\\))', '(a)(i) Explain what is meant by a machine (ii) Define the terms: mechanical advantage, velocity ratio and efficiency as applied to a machine. Derive the equation connecting the three terms.~(b) Explain why the efficiency of a machine is usually less than 100%\n(c) A screw jack whose pitch is 4.4mm is used to raise a body of mass 8000 kg through a height of 20cm. The length of the tommy bar of the jack is 70cm. If the efficiency of the jack is 80%, calcuate the: (i) velocity ratio of the jack; (ii) mechanical advantage of the jack (iii) effort required in raising the body, (iv) work done by the effort in raising the body \\((g = 10ms^{-2}; \\pi= \\frac{22}{7}\\))~(c) A screw jack whose pitch is 4.4mm is used to raise a body of mass 8000 kg through a height of 20cm. The length of the tommy bar of the jack is 70cm. If the efficiency of the jack is 80%, calcuate the: (i) velocity ratio of the jack; (ii) mechanical advantage of the jack (iii) effort required in raising the body, (iv) work done by the effort in raising the body \\((g = 10ms^{-2}; \\pi= \\frac{22}{7}\\))', NULL, '\n\nExplanation\n(a)(i) A machine is a device used to overcome a load at one end by the application of an effort at another end. Examples of simple machine include the screw jack, can opener, pulley system. (ii) M.A. is the ratio of the load (L) to the Effort (E).\n\\(M.A = \\frac{Load}{Effort}\\)\nV.R. is the ratio of the distance moved by the effort tc the distarice moved by the load (at the same time).\n\\(V.R = \\frac{\\text{distance moved by effort}}{\\text{distance moved by load}}\\)\nEfficiency (E) is the ratio of the useful work done by the machine (work output) to the total work put into the machine (work input).\n\\(Efficiency = \\frac{\\text{work done on load}}{\\text{work done by effort}} \\times 100%\\)\n\n\\(Efficiency = \\frac{M.A}{V.R} \\times 100%\\);\n\\(Work = Force \\times Distance\\)\n\\(Efficiency = \\frac{\\text{work output}}{\\text{work input}} \\)\n= \\(\\frac{Load \\times \\text{distance moved by load}}{Effort \\times \\text{distance moved by effort}}\\)\n= \\(M.A \\times \\frac{1}{V.R} \\times 100%\\)\n(b) The efficiency of a simple machine is not 100% because of the work done in overcoming friction.\n(c) length of tommy bar. L = 70cm = 0.7m; Pitch = 4.4mm = 0.0044m.\n(i) \\(V.R = \\frac{2\\pi L}{pitch} = \\frac{2 \\pi \\times 0.7}{0.0044}\\)\n= \\(2 \\times \\frac{22}{7} \\times \\frac{0.7}{44 \\times 10^{-4}}\\)\n\\(V.R = 1000\\)\n(ii) \\(Efficiency = \\frac{M.A}{V.R} \\times 100%\\)\n\\(\\frac{80}{100} = \\frac{M.A}{1000} \\implies M.A = 800\\)\n(iii) \\(M.A = \\frac{Load}{Effort}\\)\n\\(Effort = \\frac{8000 \\times 10}{800} = 100N\\)\n(iv) Work done = \\(\\frac{load \\times \\text{distance moved by load}}{effort \\times \\text{distance moved by effort}}\\)\n= \\(\\frac{8000 \\times 10 \\times 0.2}{0.8} = 20,000J\\)\n\n', '', 'Explanation\n(a)(i) A machine is a device used to overcome a load at one end by the application of an effort at another end. Examples of simple machine include the screw jack, can opener, pulley system. (ii) M.A. is the ratio of the load (L) to the Effort (E).\n\\(M.A = \\frac{Load}{Effort}\\)\nV.R. is the ratio of the distance moved by the effort tc the distarice moved by the load (at the same time).\n\\(V.R = \\frac{\\text{distance moved by effort}}{\\text{distance moved by load}}\\)\nEfficiency (E) is the ratio of the useful work done by the machine (work output) to the total work put into the machine (work input).\n\\(Efficiency = \\frac{\\text{work done on load}}{\\text{work done by effort}} \\times 100%\\)\n\n\\(Efficiency = \\frac{M.A}{V.R} \\times 100%\\);\n\\(Work = Force \\times Distance\\)\n\\(Efficiency = \\frac{\\text{work output}}{\\text{work input}} \\)\n= \\(\\frac{Load \\times \\text{distance moved by load}}{Effort \\times \\text{distance moved by effort}}\\)\n= \\(M.A \\times \\frac{1}{V.R} \\times 100%\\)\n(b) The efficiency of a simple machine is not 100% because of the work done in overcoming friction.\n(c) length of tommy bar. L = 70cm = 0.7m; Pitch = 4.4mm = 0.0044m.\n(i) \\(V.R = \\frac{2\\pi L}{pitch} = \\frac{2 \\pi \\times 0.7}{0.0044}\\)\n= \\(2 \\times \\frac{22}{7} \\times \\frac{0.7}{44 \\times 10^{-4}}\\)\n\\(V.R = 1000\\)\n(ii) \\(Efficiency = \\frac{M.A}{V.R} \\times 100%\\)\n\\(\\frac{80}{100} = \\frac{M.A}{1000} \\implies M.A = 800\\)\n(iii) \\(M.A = \\frac{Load}{Effort}\\)\n\\(Effort = \\frac{8000 \\times 10}{800} = 100N\\)\n(iv) Work done = \\(\\frac{load \\times \\text{distance moved by load}}{effort \\times \\text{distance moved by effort}}\\)\n= \\(\\frac{8000 \\times 10 \\times 0.2}{0.8} = 20,000J\\)\n\n', '(a)(i) A machine is a device used to overcome a load at one end by the application of an effort at another end. Examples of simple machine include the screw jack, can opener, pulley system. (ii) M.A. is the ratio of the load (L) to the Effort (E).\n\\(M.A = \\frac{Load}{Effort}\\)\nV.R. is the ratio of the distance moved by the effort tc the distarice moved by the load (at the same time).\n\\(V.R = \\frac{\\text{distance moved by effort}}{\\text{distance moved by load}}\\)\nEfficiency (E) is the ratio of the useful work done by the machine (work output) to the total work put into the machine (work input).\n\\(Efficiency = \\frac{\\text{work done on load}}{\\text{work done by effort}} \\times 100%\\)\n\n\\(Efficiency = \\frac{M.A}{V.R} \\times 100%\\);\n\\(Work = Force \\times Distance\\)\n\\(Efficiency = \\frac{\\text{work output}}{\\text{work input}} \\)\n= \\(\\frac{Load \\times \\text{distance moved by load}}{Effort \\times \\text{distance moved by effort}}\\)\n= \\(M.A \\times \\frac{1}{V.R} \\times 100%\\)~(b) The efficiency of a simple machine is not 100% because of the work done in overcoming friction.\n(c) length of tommy bar. L = 70cm = 0.7m; Pitch = 4.4mm = 0.0044m.\n(i) \\(V.R = \\frac{2\\pi L}{pitch} = \\frac{2 \\pi \\times 0.7}{0.0044}\\)\n= \\(2 \\times \\frac{22}{7} \\times \\frac{0.7}{44 \\times 10^{-4}}\\)\n\\(V.R = 1000\\)\n(ii) \\(Efficiency = \\frac{M.A}{V.R} \\times 100%\\)\n\\(\\frac{80}{100} = \\frac{M.A}{1000} \\implies M.A = 800\\)\n(iii) \\(M.A = \\frac{Load}{Effort}\\)\n\\(Effort = \\frac{8000 \\times 10}{800} = 100N\\)\n(iv) Work done = \\(\\frac{load \\times \\text{distance moved by load}}{effort \\times \\text{distance moved by effort}}\\)\n= \\(\\frac{8000 \\times 10 \\times 0.2}{0.8} = 20,000J\\)\n~(c) length of tommy bar. L = 70cm = 0.7m; Pitch = 4.4mm = 0.0044m.\n(i) \\(V.R = \\frac{2\\pi L}{pitch} = \\frac{2 \\pi \\times 0.7}{0.0044}\\)\n= \\(2 \\times \\frac{22}{7} \\times \\frac{0.7}{44 \\times 10^{-4}}\\)\n\\(V.R = 1000\\)\n(ii) \\(Efficiency = \\frac{M.A}{V.R} \\times 100%\\)\n\\(\\frac{80}{100} = \\frac{M.A}{1000} \\implies M.A = 800\\)\n(iii) \\(M.A = \\frac{Load}{Effort}\\)\n\\(Effort = \\frac{8000 \\times 10}{800} = 100N\\)\n(iv) Work done = \\(\\frac{load \\times \\text{distance moved by load}}{effort \\times \\text{distance moved by effort}}\\)\n= \\(\\frac{8000 \\times 10 \\times 0.2}{0.8} = 20,000J\\)\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(306, '2020', '\n\nQuestion\n\n(a) With the aid of a labelled diagram, describe an experiment to illustrate the relationship between the volume and the temperature of a given mass of air at constant pressure.\n(b) A uniform capillary tube of negligible expansivity sealed at one end, contains air trapped by a pellet of mercury. The trapped air column is 13.7cm long at 0°C and 18.7cm long at 100°C. Calculate the cubical expansivity of the air at constant pressure.\n(c) Using the kinetic theory of gases, explain why the volume of a fixed mass of gas at constant pressure increases with increase in temperature.\n\n\n', '', '(a) With the aid of a labelled diagram, describe an experiment to illustrate the relationship between the volume and the temperature of a given mass of air at constant pressure.\n(b) A uniform capillary tube of negligible expansivity sealed at one end, contains air trapped by a pellet of mercury. The trapped air column is 13.7cm long at 0°C and 18.7cm long at 100°C. Calculate the cubical expansivity of the air at constant pressure.\n(c) Using the kinetic theory of gases, explain why the volume of a fixed mass of gas at constant pressure increases with increase in temperature.\n', '(a) With the aid of a labelled diagram, describe an experiment to illustrate the relationship between the volume and the temperature of a given mass of air at constant pressure.~(b) A uniform capillary tube of negligible expansivity sealed at one end, contains air trapped by a pellet of mercury. The trapped air column is 13.7cm long at 0°C and 18.7cm long at 100°C. Calculate the cubical expansivity of the air at constant pressure.\n(c) Using the kinetic theory of gases, explain why the volume of a fixed mass of gas at constant pressure increases with increase in temperature.\n~(c) Using the kinetic theory of gases, explain why the volume of a fixed mass of gas at constant pressure increases with increase in temperature.\n', NULL, '\n\nExplanation\n(a)\nA mass of dry air is entrappedin a uniform capillary tube, sealed at one end, by means of a mercury pellet. Both the tube and thermometer are then firmly attacheda half-meter rule and placed in a water bathas shown above. The water is stirred and its temperature \\(\\theta_{1}\\)at a steady state is noted. The corresponding length, \\(L_{1}\\) of the entrapped air is also noted and recorded. The water is gradually heated and stirred uniformly until another steady temperature, \\(\\theta_{2}\\)is attained and recorded. The corresponding length, \\(L_{2}\\) of the air column in the capillary tube is again read and recorded. The experiment is repeated to obtain more sets of readings. A graph of L against \\(\\theta\\)is plotted as shown.\n\nDeduction: Since the volume V of the tube is proportional to its length, L the graph shows that the volume of the entrapped air increases (directly) with increase in temperature at constant pressure.\nPrecautions: Dry air must be used. Also, error ofparallax in reading thermometer and meter rule avoided.\n(b) \\(S = \\frac{V_{1} - V_{0}}{V_{0} \\theta}\\)\n= \\(\\frac{18.7 - 13.7}{13.7 \\times 100}\\)\n= \\(\\frac{5}{1370}\\)\n= \\(3.65 \\times 10^{-3} K^{-1}\\)\n(c) As temperature increases the kinetic energies the molecules increase and they bombard the wall the container at a faster rate since their velocities hasincreased. To keep the rate of change of momentum per unit area (i.e pressure) the same, the molecules will have to travel longer distances betweon collision This implies increase in volume.\n\n', '', 'Explanation\n(a)\nA mass of dry air is entrappedin a uniform capillary tube, sealed at one end, by means of a mercury pellet. Both the tube and thermometer are then firmly attacheda half-meter rule and placed in a water bathas shown above. The water is stirred and its temperature \\(\\theta_{1}\\)at a steady state is noted. The corresponding length, \\(L_{1}\\) of the entrapped air is also noted and recorded. The water is gradually heated and stirred uniformly until another steady temperature, \\(\\theta_{2}\\)is attained and recorded. The corresponding length, \\(L_{2}\\) of the air column in the capillary tube is again read and recorded. The experiment is repeated to obtain more sets of readings. A graph of L against \\(\\theta\\)is plotted as shown.\n\nDeduction: Since the volume V of the tube is proportional to its length, L the graph shows that the volume of the entrapped air increases (directly) with increase in temperature at constant pressure.\nPrecautions: Dry air must be used. Also, error ofparallax in reading thermometer and meter rule avoided.\n(b) \\(S = \\frac{V_{1} - V_{0}}{V_{0} \\theta}\\)\n= \\(\\frac{18.7 - 13.7}{13.7 \\times 100}\\)\n= \\(\\frac{5}{1370}\\)\n= \\(3.65 \\times 10^{-3} K^{-1}\\)\n(c) As temperature increases the kinetic energies the molecules increase and they bombard the wall the container at a faster rate since their velocities hasincreased. To keep the rate of change of momentum per unit area (i.e pressure) the same, the molecules will have to travel longer distances betweon collision This implies increase in volume.\n\n', '(a)\nA mass of dry air is entrappedin a uniform capillary tube, sealed at one end, by means of a mercury pellet. Both the tube and thermometer are then firmly attacheda half-meter rule and placed in a water bathas shown above. The water is stirred and its temperature \\(\\theta_{1}\\)at a steady state is noted. The corresponding length, \\(L_{1}\\) of the entrapped air is also noted and recorded. The water is gradually heated and stirred uniformly until another steady temperature, \\(\\theta_{2}\\)is attained and recorded. The corresponding length, \\(L_{2}\\) of the air column in the capillary tube is again read and recorded. The experiment is repeated to obtain more sets of readings. A graph of L against \\(\\theta\\)is plotted as shown.\n\nDeduction: Since the volume V of the tube is proportional to its length, L the graph shows that the volume of the entrapped air increases (directly) with increase in temperature at constant pressure.\nPrecautions: Dry air must be used. Also, error ofparallax in reading thermometer and meter rule avoided.~(b) \\(S = \\frac{V_{1} - V_{0}}{V_{0} \\theta}\\)\n= \\(\\frac{18.7 - 13.7}{13.7 \\times 100}\\)\n= \\(\\frac{5}{1370}\\)\n= \\(3.65 \\times 10^{-3} K^{-1}\\)\n(c) As temperature increases the kinetic energies the molecules increase and they bombard the wall the container at a faster rate since their velocities hasincreased. To keep the rate of change of momentum per unit area (i.e pressure) the same, the molecules will have to travel longer distances betweon collision This implies increase in volume.\n~(c) As temperature increases the kinetic energies the molecules increase and they bombard the wall the container at a faster rate since their velocities hasincreased. To keep the rate of change of momentum per unit area (i.e pressure) the same, the molecules will have to travel longer distances betweon collision This implies increase in volume.\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(307, '2020', '\n\nQuestion\n\n(a) State the laws of electromagnetic induction.\n(b) Draw a labelled diagram of a simple d.c. generator and explain how it works.\n(c) State three methods by which higher e.m.f. could be obtained from the generator.\n\n', '', '(a) State the laws of electromagnetic induction.\n(b) Draw a labelled diagram of a simple d.c. generator and explain how it works.\n(c) State three methods by which higher e.m.f. could be obtained from the generator.', '(a) State the laws of electromagnetic induction.~(b) Draw a labelled diagram of a simple d.c. generator and explain how it works.\n(c) State three methods by which higher e.m.f. could be obtained from the generator.~(c) State three methods by which higher e.m.f. could be obtained from the generator.', NULL, '\n\nExplanation\n(a) Whenever there is a change in the magnetic flux linked with a circuit, an e.m.f. is induced, the strength of which is proportional to the rate of the change ofthe flux linked with the circuit. The direction of the induced current is always such as to oppose thechange producing it.\n\nAs coil isrotating in the magnetic field, an e.m.f whichis due to the changing flux induces between its terminals. The e.m.f. varies from maximum when the coil is horizontal to minimum when the coil is vertical at which position connections between brushes and terminals of coil are changed by commutator. This ensures that current continues to flow in the same direction.\n(c) – By increasing the number of turns – Winding the coil on the soft iron core – Increasing the strength of the magnet – turning the coil faster.\n\n', '', 'Explanation\n(a) Whenever there is a change in the magnetic flux linked with a circuit, an e.m.f. is induced, the strength of which is proportional to the rate of the change ofthe flux linked with the circuit. The direction of the induced current is always such as to oppose thechange producing it.\n\nAs coil isrotating in the magnetic field, an e.m.f whichis due to the changing flux induces between its terminals. The e.m.f. varies from maximum when the coil is horizontal to minimum when the coil is vertical at which position connections between brushes and terminals of coil are changed by commutator. This ensures that current continues to flow in the same direction.\n(c) – By increasing the number of turns – Winding the coil on the soft iron core – Increasing the strength of the magnet – turning the coil faster.\n\n', '~(c) – By increasing the number of turns – Winding the coil on the soft iron core – Increasing the strength of the magnet – turning the coil faster.\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
(308, '2020', '\n\nQuestion\n\n(a) Explain the following, illustrating your answer with one example in each case: (i) nuclear fusion: (ii) nuclear fission: (iii) radiation hazards.\n(b) State two advantages of fusion over fission and explain briefly why, in spite of these advantages, fusion is not normally used for the generation of power.\n(c) The current, I in an a.c. circuit is given by the equation: \\(I = 30 sin 100\\pi t\\), where t is the time in seconds. Deduce the following from this equation: (i) frequency of the current (ii) peak value of the current, (iii) r.m.s valueof the current.\n\n', '', '(a) Explain the following, illustrating your answer with one example in each case: (i) nuclear fusion: (ii) nuclear fission: (iii) radiation hazards.\n(b) State two advantages of fusion over fission and explain briefly why, in spite of these advantages, fusion is not normally used for the generation of power.\n(c) The current, I in an a.c. circuit is given by the equation: \\(I = 30 sin 100\\pi t\\), where t is the time in seconds. Deduce the following from this equation: (i) frequency of the current (ii) peak value of the current, (iii) r.m.s valueof the current.', '(a) Explain the following, illustrating your answer with one example in each case: (i) nuclear fusion: (ii) nuclear fission: (iii) radiation hazards.~(b) State two advantages of fusion over fission and explain briefly why, in spite of these advantages, fusion is not normally used for the generation of power.\n(c) The current, I in an a.c. circuit is given by the equation: \\(I = 30 sin 100\\pi t\\), where t is the time in seconds. Deduce the following from this equation: (i) frequency of the current (ii) peak value of the current, (iii) r.m.s valueof the current.~(c) The current, I in an a.c. circuit is given by the equation: \\(I = 30 sin 100\\pi t\\), where t is the time in seconds. Deduce the following from this equation: (i) frequency of the current (ii) peak value of the current, (iii) r.m.s valueof the current.', NULL, '\n\nExplanation\n(a)(i) Fusion:Fusion is the coming together of two light nuclei to form a heavier nucleus with the release of energy.\nExample: \\(^{2} _{1} H + ^{2} _{1} H \\to ^{4} _{2} He + Energy\\)\n(ii) Fission:Fission is the breaking up of heavy nucleus into two nuclei of comparable masses when bombarded with fast neutrons or energetic particles releasing energy in the process.\nExample: \\(^{235} _{92} U + ^{1} _{0} N \\to ^{90} _{39} Kr + ^{144} _{50} Ba + ^{1} _{0} N + ^{1} _{0} N\\)\n(iii) Radiation Hazards. Radioactive substances can be dangerous if one is exposed to their radiations for a long time. Some of the radiations are highly penetrating and if the body is over-exposed to them, they can destroy the cells in tissues and upset the natural chemistry of the body. Examples include: – Leukaemia – Skin burns\n(b) Advantages of fusion over fission include: (1) Easily achieved with lightest element so that nuclear repulsion is easily overcome as nuclei approach each other (2) Raw materials are cheaply available. (3) Produces less dangerous by products. (4) There is no upper limit to the mass of hydrogen that can be obtained. Very high temperature is required to overcome the coulomb repulsive forces between the two light nuclei. This poses severe technological problem due to the fact that materials to withstand this high temperature are difficult to come by.\n(c) \\(I = I_{m} \\sin 2 \\pi ft ... (1)\\)\nI - Instantaneous value of current\n\\(I_{m}\\) - Peak value of current\nF - frequency of current\n\\(I = 30 \\sin 100 \\pi t ... (2)\\)\nComparing (1) and (2),\n(i) \\(2 \\pi f = 100\\pi \\therefore f = 50 Hz\\)\n(ii) \\(I_{m} = 30A\\)\n(iii) \\(I_{r.m.s} = \\frac{I_{m}}{\\sqrt{2}}\\)\n= \\(\\frac{30}{\\sqrt{2}} = 21.2A\\)\n\n', '', 'Explanation\n(a)(i) Fusion:Fusion is the coming together of two light nuclei to form a heavier nucleus with the release of energy.\nExample: \\(^{2} _{1} H + ^{2} _{1} H \\to ^{4} _{2} He + Energy\\)\n(ii) Fission:Fission is the breaking up of heavy nucleus into two nuclei of comparable masses when bombarded with fast neutrons or energetic particles releasing energy in the process.\nExample: \\(^{235} _{92} U + ^{1} _{0} N \\to ^{90} _{39} Kr + ^{144} _{50} Ba + ^{1} _{0} N + ^{1} _{0} N\\)\n(iii) Radiation Hazards. Radioactive substances can be dangerous if one is exposed to their radiations for a long time. Some of the radiations are highly penetrating and if the body is over-exposed to them, they can destroy the cells in tissues and upset the natural chemistry of the body. Examples include: – Leukaemia – Skin burns\n(b) Advantages of fusion over fission include: (1) Easily achieved with lightest element so that nuclear repulsion is easily overcome as nuclei approach each other (2) Raw materials are cheaply available. (3) Produces less dangerous by products. (4) There is no upper limit to the mass of hydrogen that can be obtained. Very high temperature is required to overcome the coulomb repulsive forces between the two light nuclei. This poses severe technological problem due to the fact that materials to withstand this high temperature are difficult to come by.\n(c) \\(I = I_{m} \\sin 2 \\pi ft ... (1)\\)\nI - Instantaneous value of current\n\\(I_{m}\\) - Peak value of current\nF - frequency of current\n\\(I = 30 \\sin 100 \\pi t ... (2)\\)\nComparing (1) and (2),\n(i) \\(2 \\pi f = 100\\pi \\therefore f = 50 Hz\\)\n(ii) \\(I_{m} = 30A\\)\n(iii) \\(I_{r.m.s} = \\frac{I_{m}}{\\sqrt{2}}\\)\n= \\(\\frac{30}{\\sqrt{2}} = 21.2A\\)\n\n', '(a)(i) Fusion:Fusion is the coming together of two light nuclei to form a heavier nucleus with the release of energy.\nExample: \\(^{2} _{1} H + ^{2} _{1} H \\to ^{4} _{2} He + Energy\\)\n(ii) Fission:Fission is the breaking up of heavy nucleus into two nuclei of comparable masses when bombarded with fast neutrons or energetic particles releasing energy in the process.\nExample: \\(^{235} _{92} U + ^{1} _{0} N \\to ^{90} _{39} Kr + ^{144} _{50} Ba + ^{1} _{0} N + ^{1} _{0} N\\)\n(iii) Radiation Hazards. Radioactive substances can be dangerous if one is exposed to their radiations for a long time. Some of the radiations are highly penetrating and if the body is over-exposed to them, they can destroy the cells in tissues and upset the natural chemistry of the body. Examples include: – Leukaemia – Skin burns~(b) Advantages of fusion over fission include: (1) Easily achieved with lightest element so that nuclear repulsion is easily overcome as nuclei approach each other (2) Raw materials are cheaply available. (3) Produces less dangerous by products. (4) There is no upper limit to the mass of hydrogen that can be obtained. Very high temperature is required to overcome the coulomb repulsive forces between the two light nuclei. This poses severe technological problem due to the fact that materials to withstand this high temperature are difficult to come by.\n(c) \\(I = I_{m} \\sin 2 \\pi ft ... (1)\\)\nI - Instantaneous value of current\n\\(I_{m}\\) - Peak value of current\nF - frequency of current\n\\(I = 30 \\sin 100 \\pi t ... (2)\\)\nComparing (1) and (2),\n(i) \\(2 \\pi f = 100\\pi \\therefore f = 50 Hz\\)\n(ii) \\(I_{m} = 30A\\)\n(iii) \\(I_{r.m.s} = \\frac{I_{m}}{\\sqrt{2}}\\)\n= \\(\\frac{30}{\\sqrt{2}} = 21.2A\\)\n~(c) \\(I = I_{m} \\sin 2 \\pi ft ... (1)\\)\nI - Instantaneous value of current\n\\(I_{m}\\) - Peak value of current\nF - frequency of current\n\\(I = 30 \\sin 100 \\pi t ... (2)\\)\nComparing (1) and (2),\n(i) \\(2 \\pi f = 100\\pi \\therefore f = 50 Hz\\)\n(ii) \\(I_{m} = 30A\\)\n(iii) \\(I_{r.m.s} = \\frac{I_{m}}{\\sqrt{2}}\\)\n= \\(\\frac{30}{\\sqrt{2}} = 21.2A\\)\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(309, '2020', '\n\nQuestion\n\n(a) What is meant by the statement:The specific heat capacity of copper is \\(400 J kg^{-1}K^{-1}\\)?\n(b)(i) Describe an experiment to determine the specific heat capacity of copper using a copper ball.\n(ii) State two precautions necessary to obtain accurate results\n(iii) A piece of copper ball of mass 20 g at 200°C is placed in a copper calorimeter of mass 60 g containing 50 g of water at 30°C, ignoring heat losses, calculate the final steady temperature of the mixture (Specific heat capacity of water = \\(4.2 J g^{-1}K^{-1}\\)) (Specific heat capacity of copper = \\(0.4 Jg^{-1}K^{-1}\\)).\n\n', '', '(a) What is meant by the statement:The specific heat capacity of copper is \\(400 J kg^{-1}K^{-1}\\)?\n(b)(i) Describe an experiment to determine the specific heat capacity of copper using a copper ball.\n(ii) State two precautions necessary to obtain accurate results\n(iii) A piece of copper ball of mass 20 g at 200°C is placed in a copper calorimeter of mass 60 g containing 50 g of water at 30°C, ignoring heat losses, calculate the final steady temperature of the mixture (Specific heat capacity of water = \\(4.2 J g^{-1}K^{-1}\\)) (Specific heat capacity of copper = \\(0.4 Jg^{-1}K^{-1}\\)).', '(a) What is meant by the statement:The specific heat capacity of copper is \\(400 J kg^{-1}K^{-1}\\)?~(b)(i) Describe an experiment to determine the specific heat capacity of copper using a copper ball.\n(ii) State two precautions necessary to obtain accurate results\n(iii) A piece of copper ball of mass 20 g at 200°C is placed in a copper calorimeter of mass 60 g containing 50 g of water at 30°C, ignoring heat losses, calculate the final steady temperature of the mixture (Specific heat capacity of water = \\(4.2 J g^{-1}K^{-1}\\)) (Specific heat capacity of copper = \\(0.4 Jg^{-1}K^{-1}\\)).', NULL, '\n\nExplanation\n(a) The specific heat capacity of copper is 400 \\(JKg^{-1}K^{-1}\\) means that 400 J of heat energy is required to raise (or lower) the temperature of 1 kg of a piece of copper by 1 kelvin.\n(b) Experiment to determine the specific heat capacity of copper using a copper ball. The piece of copper ball is suspended by a thread and placed in a beaker of boiling water. While the temperature of the copper is being raised to boiling point of water, a copper calorimeter is half filled with water. The initial temperature of the water is measured and recorded and the hot copper ball is lifted by the thread and transferred to the calorimeter when the water has reached boiling point. The calorimeter cover is put on, the mixture is well stirred and the final steady temperature read from the themometer. The copper ball is weighed either before or after the experiment. The result is tabulated as shown below.\nMass of copper ball - m (kg)\nTemperature of boiling water - \\(\\theta\\) (K)\nMass of copper calorimeter - \\(m_{c}\\) (kg)\nMass of calorimeter + water - \\(m_{x}\\) (kg)\nMass of water \\((m_{x} - m_{c})\\) - \\(m_{w}\\) (kg)\nSpecific heat capacity - C \\((Jkg^{-1} K^{-1})\\)\nInitial temperature of water = \\(\\theta_{w}\\) (K)\nFinal temperature of water = \\(\\theta_{f}\\) (K)\nUsing Heat lost by copper ball = Heat gained by calorimeter + Heat gained by water\n\\(mc(\\theta - \\theta_{f}) = m_{c} c(\\theta_{f} - \\theta_{w}) + m_{w} c_{w} (\\theta_{f} - \\theta_{w})\\)\n\\(c = \\frac{m_{w} c_{w} (\\theta_{f} - \\theta_{w})}{m (\\theta - \\theta_{f}) - m_{c} (\\theta_{f} - \\theta_{w})}\\)\nPrecautions: (i) The copper ball should be transferred quickly into the calorimeter without splashing. (ii) Just before transfer, the copper should be shaken slightly to remove adhering hot water.\n(iii) Heat lost by copper ball = heat gained by copper calorimeter + Hoat gained by water.\n\\(m_{c} (\\theta - \\theta_{f}) = m c (\\theta_{f} - \\theta_{w}) + m_{w} c_{w} (\\theta_{f} - \\theta_{w})\\)\n\\(20 \\times 0.4 \\times (200 - \\theta_{f}) = 60 \\times 0.4 \\times (\\theta_{f} - 30) + 50 \\times 4.2 \\times (\\theta_{f} - 30)\\)\n\\(8(200 - \\theta_{f}) = 24(\\theta_{f} - 30) + 210 (\\theta_{f} - 30)\\)\n\\(\\theta_{f} = \\frac{8620}{242} = 35.6K\\)\n\n', '', 'Explanation\n(a) The specific heat capacity of copper is 400 \\(JKg^{-1}K^{-1}\\) means that 400 J of heat energy is required to raise (or lower) the temperature of 1 kg of a piece of copper by 1 kelvin.\n(b) Experiment to determine the specific heat capacity of copper using a copper ball. The piece of copper ball is suspended by a thread and placed in a beaker of boiling water. While the temperature of the copper is being raised to boiling point of water, a copper calorimeter is half filled with water. The initial temperature of the water is measured and recorded and the hot copper ball is lifted by the thread and transferred to the calorimeter when the water has reached boiling point. The calorimeter cover is put on, the mixture is well stirred and the final steady temperature read from the themometer. The copper ball is weighed either before or after the experiment. The result is tabulated as shown below.\nMass of copper ball - m (kg)\nTemperature of boiling water - \\(\\theta\\) (K)\nMass of copper calorimeter - \\(m_{c}\\) (kg)\nMass of calorimeter + water - \\(m_{x}\\) (kg)\nMass of water \\((m_{x} - m_{c})\\) - \\(m_{w}\\) (kg)\nSpecific heat capacity - C \\((Jkg^{-1} K^{-1})\\)\nInitial temperature of water = \\(\\theta_{w}\\) (K)\nFinal temperature of water = \\(\\theta_{f}\\) (K)\nUsing Heat lost by copper ball = Heat gained by calorimeter + Heat gained by water\n\\(mc(\\theta - \\theta_{f}) = m_{c} c(\\theta_{f} - \\theta_{w}) + m_{w} c_{w} (\\theta_{f} - \\theta_{w})\\)\n\\(c = \\frac{m_{w} c_{w} (\\theta_{f} - \\theta_{w})}{m (\\theta - \\theta_{f}) - m_{c} (\\theta_{f} - \\theta_{w})}\\)\nPrecautions: (i) The copper ball should be transferred quickly into the calorimeter without splashing. (ii) Just before transfer, the copper should be shaken slightly to remove adhering hot water.\n(iii) Heat lost by copper ball = heat gained by copper calorimeter + Hoat gained by water.\n\\(m_{c} (\\theta - \\theta_{f}) = m c (\\theta_{f} - \\theta_{w}) + m_{w} c_{w} (\\theta_{f} - \\theta_{w})\\)\n\\(20 \\times 0.4 \\times (200 - \\theta_{f}) = 60 \\times 0.4 \\times (\\theta_{f} - 30) + 50 \\times 4.2 \\times (\\theta_{f} - 30)\\)\n\\(8(200 - \\theta_{f}) = 24(\\theta_{f} - 30) + 210 (\\theta_{f} - 30)\\)\n\\(\\theta_{f} = \\frac{8620}{242} = 35.6K\\)\n\n', '(a) The specific heat capacity of copper is 400 \\(JKg^{-1}K^{-1}\\) means that 400 J of heat energy is required to raise (or lower) the temperature of 1 kg of a piece of copper by 1 kelvin.~(b) Experiment to determine the specific heat capacity of copper using a copper ball. The piece of copper ball is suspended by a thread and placed in a beaker of boiling water. While the temperature of the copper is being raised to boiling point of water, a copper calorimeter is half filled with water. The initial temperature of the water is measured and recorded and the hot copper ball is lifted by the thread and transferred to the calorimeter when the water has reached boiling point. The calorimeter cover is put on, the mixture is well stirred and the final steady temperature read from the themometer. The copper ball is weighed either before or after the experiment. The result is tabulated as shown below.\nMass of copper ball - m (kg)\nTemperature of boiling water - \\(\\theta\\) (K)\nMass of copper calorimeter - \\(m_{c}\\) (kg)\nMass of calorimeter + water - \\(m_{x}\\) (kg)\nMass of water \\((m_{x} - m_{c})\\) - \\(m_{w}\\) (kg)\nSpecific heat capacity - C \\((Jkg^{-1} K^{-1})\\)\nInitial temperature of water = \\(\\theta_{w}\\) (K)\nFinal temperature of water = \\(\\theta_{f}\\) (K)\nUsing Heat lost by copper ball = Heat gained by calorimeter + Heat gained by water\n\\(mc(\\theta - \\theta_{f}) = m_{c} c(\\theta_{f} - \\theta_{w}) + m_{w} c_{w} (\\theta_{f} - \\theta_{w})\\)\n\\(c = \\frac{m_{w} c_{w} (\\theta_{f} - \\theta_{w})}{m (\\theta - \\theta_{f}) - m_{c} (\\theta_{f} - \\theta_{w})}\\)\nPrecautions: (i) The copper ball should be transferred quickly into the calorimeter without splashing. (ii) Just before transfer, the copper should be shaken slightly to remove adhering hot water.\n(iii) Heat lost by copper ball = heat gained by copper calorimeter + Hoat gained by water.\n\\(m_{c} (\\theta - \\theta_{f}) = m c (\\theta_{f} - \\theta_{w}) + m_{w} c_{w} (\\theta_{f} - \\theta_{w})\\)\n\\(20 \\times 0.4 \\times (200 - \\theta_{f}) = 60 \\times 0.4 \\times (\\theta_{f} - 30) + 50 \\times 4.2 \\times (\\theta_{f} - 30)\\)\n\\(8(200 - \\theta_{f}) = 24(\\theta_{f} - 30) + 210 (\\theta_{f} - 30)\\)\n\\(\\theta_{f} = \\frac{8620}{242} = 35.6K\\)\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(310, '2020', '\n\nQuestion\n\n(a) Describe an experiment to show how the frequency of the note emitted by a vibrating string depends on the tension in the string\n(b) Draw diagrams showing a vibrating string fixed at both ends emitting (i) fundamental frequency (ii) second overtone indicate the nodes and antinodes on the diagrams\n(c) With the aid of a ray diagram show how a virtual image of an object is formed by a (i) concavemirror (ii) converging lens\n\n', '', '(a) Describe an experiment to show how the frequency of the note emitted by a vibrating string depends on the tension in the string\n(b) Draw diagrams showing a vibrating string fixed at both ends emitting (i) fundamental frequency (ii) second overtone indicate the nodes and antinodes on the diagrams\n(c) With the aid of a ray diagram show how a virtual image of an object is formed by a (i) concavemirror (ii) converging lens', '(a) Describe an experiment to show how the frequency of the note emitted by a vibrating string depends on the tension in the string~(b) Draw diagrams showing a vibrating string fixed at both ends emitting (i) fundamental frequency (ii) second overtone indicate the nodes and antinodes on the diagrams\n(c) With the aid of a ray diagram show how a virtual image of an object is formed by a (i) concavemirror (ii) converging lens~(c) With the aid of a ray diagram show how a virtual image of an object is formed by a (i) concavemirror (ii) converging lens', NULL, '\n\nExplanation\n(a)\nThe experiment is set up as shown in the diagram. The wooden box effective length, I of the string remains constant throughout the experiment. Also the string used is of uniform thickness and must not be changed to ensure that the mass per unit length is constant throughout the experiment. The length of string is tied to the stake at one free end while a known mass is added to the other end. This creates a tension in the string which is equal to the weight of the attached mass. A tuning fork of known frequency is set vibrating and the string is tuned in unison with the fork. The frequency of the fork is recorded. The procedure is repeated by increasing the masses and using different tuning forks four more times. The corresponding tension (weights) and frequency are recorded in a table. Experiment to show that frequency is proportional to the square root of the tension, T. Mathematically, the graph of F against \\(\\sqrt{T}\\) is a straight line through the origin.\n(b)(i)\n(ii) Second overtone\n\n(c)(i)\n(ii)\n\n', '', 'Explanation\n(a)\nThe experiment is set up as shown in the diagram. The wooden box effective length, I of the string remains constant throughout the experiment. Also the string used is of uniform thickness and must not be changed to ensure that the mass per unit length is constant throughout the experiment. The length of string is tied to the stake at one free end while a known mass is added to the other end. This creates a tension in the string which is equal to the weight of the attached mass. A tuning fork of known frequency is set vibrating and the string is tuned in unison with the fork. The frequency of the fork is recorded. The procedure is repeated by increasing the masses and using different tuning forks four more times. The corresponding tension (weights) and frequency are recorded in a table. Experiment to show that frequency is proportional to the square root of the tension, T. Mathematically, the graph of F against \\(\\sqrt{T}\\) is a straight line through the origin.\n(b)(i)\n(ii) Second overtone\n\n(c)(i)\n(ii)\n\n', '(a)\nThe experiment is set up as shown in the diagram. The wooden box effective length, I of the string remains constant throughout the experiment. Also the string used is of uniform thickness and must not be changed to ensure that the mass per unit length is constant throughout the experiment. The length of string is tied to the stake at one free end while a known mass is added to the other end. This creates a tension in the string which is equal to the weight of the attached mass. A tuning fork of known frequency is set vibrating and the string is tuned in unison with the fork. The frequency of the fork is recorded. The procedure is repeated by increasing the masses and using different tuning forks four more times. The corresponding tension (weights) and frequency are recorded in a table. Experiment to show that frequency is proportional to the square root of the tension, T. Mathematically, the graph of F against \\(\\sqrt{T}\\) is a straight line through the origin.~(b)(i)\n(ii) Second overtone\n\n(c)(i)\n(ii)\n~(c)(i)\n(ii)\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(311, '2020', '\n\nQuestion\n\n(a) Explain what is meant by a magnetic field\n(b)(i) Describe an experiment to show that a magnetic field exists around a straight wire carrying current\n(ii) Draw a labelled diagram showing the pattern and direction of the magnetic field rroduced around the wire. (Neglect the earth\'s magnetic field).\n(c) Sketch the magnetic field due to two straight parallel wires carrying current in the same direction. Indicate the neutral point in the field\n(d) Explain, with the aid of a labelled diagram, how a delicate magnetic material could be protected from the earth\'s magnetic field.\n\n', '', '(a) Explain what is meant by a magnetic field\n(b)(i) Describe an experiment to show that a magnetic field exists around a straight wire carrying current\n(ii) Draw a labelled diagram showing the pattern and direction of the magnetic field rroduced around the wire. (Neglect the earth\'s magnetic field).\n(c) Sketch the magnetic field due to two straight parallel wires carrying current in the same direction. Indicate the neutral point in the field\n(d) Explain, with the aid of a labelled diagram, how a delicate magnetic material could be protected from the earth\'s magnetic field.', '(a) Explain what is meant by a magnetic field~(b)(i) Describe an experiment to show that a magnetic field exists around a straight wire carrying current\n(ii) Draw a labelled diagram showing the pattern and direction of the magnetic field rroduced around the wire. (Neglect the earth\'s magnetic field).\n(c) Sketch the magnetic field due to two straight parallel wires carrying current in the same direction. Indicate the neutral point in the field\n(d) Explain, with the aid of a labelled diagram, how a delicate magnetic material could be protected from the earth\'s magnetic field.~(c) Sketch the magnetic field due to two straight parallel wires carrying current in the same direction. Indicate the neutral point in the field\n(d) Explain, with the aid of a labelled diagram, how a delicate magnetic material could be protected from the earth\'s magnetic field.~(d) Explain, with the aid of a labelled diagram, how a delicate magnetic material could be protected from the earth\'s magnetic field.', NULL, '\n\nExplanation\n(a)A magnetic field is a region in which a force of magnetic origin is experienced. It is usually due to the presence of magnetic flux which is a vector quantity.\n(b)\nA straight wire carrying current is filled through a hole in a cardboard as shown. Compasses are placed round the conductor or the cardboard. The needles of the compass will be noticed to deflect ina particular direction. They all conform to a circular pattern indicating that the magnetic field around the conductor carrying current is circular or concentric in shape. When the current is reversed, the direction of the compass needle reverses as well.\n(c)\n(d) Current in the same direction\n\nThe material is surrounded by a ring of iron. This shields the flux of the earth away from the material.\n\n\n', '', 'Explanation\n(a)A magnetic field is a region in which a force of magnetic origin is experienced. It is usually due to the presence of magnetic flux which is a vector quantity.\n(b)\nA straight wire carrying current is filled through a hole in a cardboard as shown. Compasses are placed round the conductor or the cardboard. The needles of the compass will be noticed to deflect ina particular direction. They all conform to a circular pattern indicating that the magnetic field around the conductor carrying current is circular or concentric in shape. When the current is reversed, the direction of the compass needle reverses as well.\n(c)\n(d) Current in the same direction\n\nThe material is surrounded by a ring of iron. This shields the flux of the earth away from the material.\n\n\n', '(a)A magnetic field is a region in which a force of magnetic origin is experienced. It is usually due to the presence of magnetic flux which is a vector quantity.~(b)\nA straight wire carrying current is filled through a hole in a cardboard as shown. Compasses are placed round the conductor or the cardboard. The needles of the compass will be noticed to deflect ina particular direction. They all conform to a circular pattern indicating that the magnetic field around the conductor carrying current is circular or concentric in shape. When the current is reversed, the direction of the compass needle reverses as well.\n(c)\n(d) Current in the same direction\n\nThe material is surrounded by a ring of iron. This shields the flux of the earth away from the material.\n\n~(c)\n(d) Current in the same direction\n\nThe material is surrounded by a ring of iron. This shields the flux of the earth away from the material.\n\n~(d) Current in the same direction\n\nThe material is surrounded by a ring of iron. This shields the flux of the earth away from the material.\n\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(312, '2020', '\n\nQuestion\n\n(a) State the laws of electromagnetic induction\n(b) Explain how one of the laws illustrates the principle of conservation of energy\n(c)(i) Draw a labelled diagram of a simple d.c. electric motor and explain how it works.\n(ii) State two reasons why the efficiency of an electric motor is less than 100%.\n\n', '', '(a) State the laws of electromagnetic induction\n(b) Explain how one of the laws illustrates the principle of conservation of energy\n(c)(i) Draw a labelled diagram of a simple d.c. electric motor and explain how it works.\n(ii) State two reasons why the efficiency of an electric motor is less than 100%.', '(a) State the laws of electromagnetic induction~(b) Explain how one of the laws illustrates the principle of conservation of energy\n(c)(i) Draw a labelled diagram of a simple d.c. electric motor and explain how it works.\n(ii) State two reasons why the efficiency of an electric motor is less than 100%.~(c)(i) Draw a labelled diagram of a simple d.c. electric motor and explain how it works.\n(ii) State two reasons why the efficiency of an electric motor is less than 100%.', NULL, '\n\nExplanation\n(a) Whenever there is a change in the magnetic flux linked with a circuit an electromotive force is induced, the strength of which is proportional to the rate of change of the flux linked with the circuit. Lenz\'s law states that the direction of the induced current is always such as to oppose the charge producing it.\n(b) Lenz\'s law illustrates the law of conservation of energy. In this case mechanical energy i.e kinetic energy, due to the motion of the bar magnet is changed to electrical energy in the form of induced current.\n(c)\nA simple d.c. electric motor can supply direct current (d.c.) by using a split - ring commutator.The brush (-) touchesone half of the split ring X while the other brush (+) touches the other half Y.As the coil rotates it cuts through the magnetic flux linking the magnets and current is induced which flows through the coil as shown in the diagram. Asthe coil rotates past the vertical Y moves round to touch (-) while X moves round to touch (+). This action reverses the connections to the external resistance connected to (-) and (+). The reverse happens when the current is just zero (coil vertical) so instead of the current reversing in the external resistance, it continues to flow in the same direction.\nc(ii) (1) Part of the current is wasted in form of heat due to the resistance of the coil and this reduces efficiency;\n(2) Friction between the moveable parts of the motor makes the efficiency to reduce.\n\n', '', 'Explanation\n(a) Whenever there is a change in the magnetic flux linked with a circuit an electromotive force is induced, the strength of which is proportional to the rate of change of the flux linked with the circuit. Lenz\'s law states that the direction of the induced current is always such as to oppose the charge producing it.\n(b) Lenz\'s law illustrates the law of conservation of energy. In this case mechanical energy i.e kinetic energy, due to the motion of the bar magnet is changed to electrical energy in the form of induced current.\n(c)\nA simple d.c. electric motor can supply direct current (d.c.) by using a split - ring commutator.The brush (-) touchesone half of the split ring X while the other brush (+) touches the other half Y.As the coil rotates it cuts through the magnetic flux linking the magnets and current is induced which flows through the coil as shown in the diagram. Asthe coil rotates past the vertical Y moves round to touch (-) while X moves round to touch (+). This action reverses the connections to the external resistance connected to (-) and (+). The reverse happens when the current is just zero (coil vertical) so instead of the current reversing in the external resistance, it continues to flow in the same direction.\nc(ii) (1) Part of the current is wasted in form of heat due to the resistance of the coil and this reduces efficiency;\n(2) Friction between the moveable parts of the motor makes the efficiency to reduce.\n\n', '(a) Whenever there is a change in the magnetic flux linked with a circuit an electromotive force is induced, the strength of which is proportional to the rate of change of the flux linked with the circuit. Lenz\'s law states that the direction of the induced current is always such as to oppose the charge producing it.~(b) Lenz\'s law illustrates the law of conservation of energy. In this case mechanical energy i.e kinetic energy, due to the motion of the bar magnet is changed to electrical energy in the form of induced current.\n(c)\nA simple d.c. electric motor can supply direct current (d.c.) by using a split - ring commutator.The brush (-) touchesone half of the split ring X while the other brush (+) touches the other half Y.As the coil rotates it cuts through the magnetic flux linking the magnets and current is induced which flows through the coil as shown in the diagram. Asthe coil rotates past the vertical Y moves round to touch (-) while X moves round to touch (+). This action reverses the connections to the external resistance connected to (-) and (+). The reverse happens when the current is just zero (coil vertical) so instead of the current reversing in the external resistance, it continues to flow in the same direction.\nc(ii) (1) Part of the current is wasted in form of heat due to the resistance of the coil and this reduces efficiency;\n(2) Friction between the moveable parts of the motor makes the efficiency to reduce.\n~(c)\nA simple d.c. electric motor can supply direct current (d.c.) by using a split - ring commutator.The brush (-) touchesone half of the split ring X while the other brush (+) touches the other half Y.As the coil rotates it cuts through the magnetic flux linking the magnets and current is induced which flows through the coil as shown in the diagram. Asthe coil rotates past the vertical Y moves round to touch (-) while X moves round to touch (+). This action reverses the connections to the external resistance connected to (-) and (+). The reverse happens when the current is just zero (coil vertical) so instead of the current reversing in the external resistance, it continues to flow in the same direction.\nc(ii) (1) Part of the current is wasted in form of heat due to the resistance of the coil and this reduces efficiency;\n(2) Friction between the moveable parts of the motor makes the efficiency to reduce.\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(313, '2020', '\n\nQuestion\n\nA stone of mass 20g is released from a catapult whose rubber is stretched through 5cm. If the force constant of the rubber is 200Nm\\(^{-1}\\), calculate the speed with which the stone leaves the catapult.\n\n\n', 'A stone of mass 20g is released from a catapult whose rubber is stretched through 5cm. If the force constant of the rubber is 200Nm\\(^{-1}\\), calculate the speed with which the stone leaves the catapult.\n', '', '', NULL, '\n\nExplanation\n½Ke\\(^{2}\\) =½mv\\(^{6}\\)\n½ x 200 x (0.05)\\(^{2}\\) =½ x 0.02 x v\\(^{2}\\)\nv\\(^{2}\\) = 25\nv = 5 m s\\(^{-1}\\)\n\n', '½Ke\\(^{2}\\) =½mv\\(^{6}\\)\n½ x 200 x (0.05)\\(^{2}\\) =½ x 0.02 x v\\(^{2}\\)\nv\\(^{2}\\) = 25\nv = 5 m s\\(^{-1}\\)\n\n', '', '', 'WAEC-PK-1', 'WAEC-PHY', 1),
(314, '2020', '\n\nQuestion\n\n(a)The diagram above illustrates a projectile motion.Identify each of the physical quantities labeled P, β, H and R.\n\n(b)Write an equation to show the relationship between P, g and Rmax’ where g is the acceleration due to gravity and Rmax is maximum R.\n\n\n', '', '(a)The diagram above illustrates a projectile motion.Identify each of the physical quantities labeled P, β, H and R.\n\n(b)Write an equation to show the relationship between P, g and Rmax’ where g is the acceleration due to gravity and Rmax is maximum R.\n', '(a)The diagram above illustrates a projectile motion.Identify each of the physical quantities labeled P, β, H and R.\n~(b)Write an equation to show the relationship between P, g and Rmax’ where g is the acceleration due to gravity and Rmax is maximum R.\n', NULL, '\n\nExplanation\nThe expected answer is\n(a)Identification of labelled physical quantities\nP- initial velocity/ velocity of projection\nb - angle of projection \nH - maximum height reached\nR-range/horizontal distance (covered)\n\n(b)Equation showing relationship between P, g and Rmax\nRmax = \\(\\frac{p^2}{g}\\)\n\n', '', 'Explanation\nThe expected answer is\n(a)Identification of labelled physical quantities\nP- initial velocity/ velocity of projection\nb - angle of projection \nH - maximum height reached\nR-range/horizontal distance (covered)\n\n(b)Equation showing relationship between P, g and Rmax\nRmax = \\(\\frac{p^2}{g}\\)\n\n', '(a)Identification of labelled physical quantities\nP- initial velocity/ velocity of projection\nb - angle of projection \nH - maximum height reached\nR-range/horizontal distance (covered)\n~(b)Equation showing relationship between P, g and Rmax\nRmax = \\(\\frac{p^2}{g}\\)\n~d)\n\n(b)Equation showing relationship between P, g and Rmax\nRmax = \\(\\frac{p^2}{g}\\)\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(315, '2020', '\n\nQuestion\n\nState three observable phenomena in which waves behave like a particle.\n\n', 'State three observable phenomena in which waves behave like a particle.', '', '', NULL, '\n\nExplanation\nObservable phenomena in which waves behave like a particle\n-reflection;\n-refraction;\n-diffraction;\n-polarization;\n-interference;\n-Compton effect;\n-blackbody radiation\n\n', 'Observable phenomena in which waves behave like a particle\n-reflection;\n-refraction;\n-diffraction;\n-polarization;\n-interference;\n-Compton effect;\n-blackbody radiation\n\n', '', '', 'WAEC-PK-1', 'WAEC-PHY', 1),
(316, '2020', '\n\nQuestion\n\nList three magnetic elements that determine the earth’s magnetic field at a point.\n\n', 'List three magnetic elements that determine the earth’s magnetic field at a point.', '', '', NULL, '\n\nExplanation\nMagnetic elements that determine the earth’s magnetic field at a point\n-Angle of dip/inclination\n-angle of declination/variation\n-horizontal component of the earth’s magnetic field\n\n', 'Magnetic elements that determine the earth’s magnetic field at a point\n-Angle of dip/inclination\n-angle of declination/variation\n-horizontal component of the earth’s magnetic field\n\n', '', '', 'WAEC-PK-1', 'WAEC-PHY', 1),
(317, '2020', '\n\nQuestion\n\nExplain each of the following terms as used in Electronics.\n(a) free electrons;\n(b) holes.\n\n', '', 'Explain each of the following terms as used in Electronics.\n(a) free electrons;\n(b) holes.', '(a) free electrons;~(b) holes.', NULL, '\n\nExplanation\nExplanation of terms\n(a)Free electrons; Negative charge carriers \n(b) Holes; Positive charge carriers\n\n', '', 'Explanation\nExplanation of terms\n(a)Free electrons; Negative charge carriers \n(b) Holes; Positive charge carriers\n\n', '(a)Free electrons; Negative charge carriers ~(b) Holes; Positive charge carriers\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(318, '2013', 'Question\n\nA used car was purchased at N900,000.00. Its value depreciated by 30% in the first year. In each subsequent year, the depreciation was 22% of its value at the beginning of the year. If the car was bought on the 1st of March, 2011, calculate, correct to the nearest hundred naira, the value of the car on the 28th of February, 2015.\n\n', 'A used car was purchased at N900,000.00. Its value depreciated by 30% in the first year. In each subsequent year, the depreciation was 22% of its value at the beginning of the year. If the car was bought on the 1st of March, 2011, calculate, correct to the nearest hundred naira, the value of the car on the 28th of February, 2015.', '', '', NULL, 'Explanation\nBeginning of 1st year (March 2011 - February 2012)- N900,000\nDepreciation (30%) = \\(\\frac{30}{100} \\times N900,000\\)\n= N270,000\nPrice at the end of 1st year = N(900,000 - 270,000)\n= N630,000.\nBeginning of 2nd year (March 2012 - February 2013)- N630,000\nDepreciation (22%) = \\(\\frac{22}{100} \\times N630,000\\)\n= N138,600\nPrice at the end of 2nd year = N(630,000 - 138,600)\n= N491,400.\nBeginning of 3rd year (March 2013 - February 2014)- N491,400\nDepreciation (22%) = \\(\\frac{22}{100} \\times N491,400\\)\n= N108,108\nPrice at the end of the 3rd year = N(491,400 - 108,108)\n= N383,292.\nBeginning of 4th year (March 2014 - February 2015)- N383,292\nDepreciation (22%) = \\(\\frac{22}{100} \\times N383,292\\)\n= N84,324.24\nPrice at the end of the 4th year = N(383,292 - 84,324.24)\n= N298,967.76 \\(\\approxeq\\) N299,000 (to the nearest hundred naira).\n\n', 'Beginning of 1st year (March 2011 - February 2012)- N900,000\nDepreciation (30%) = \\(\\frac{30}{100} \\times N900,000\\)\n= N270,000\nPrice at the end of 1st year = N(900,000 - 270,000)\n= N630,000.\nBeginning of 2nd year (March 2012 - February 2013)- N630,000\nDepreciation (22%) = \\(\\frac{22}{100} \\times N630,000\\)\n= N138,600\nPrice at the end of 2nd year = N(630,000 - 138,600)\n= N491,400.\nBeginning of 3rd year (March 2013 - February 2014)- N491,400\nDepreciation (22%) = \\(\\frac{22}{100} \\times N491,400\\)\n= N108,108\nPrice at the end of the 3rd year = N(491,400 - 108,108)\n= N383,292.\nBeginning of 4th year (March 2014 - February 2015)- N383,292\nDepreciation (22%) = \\(\\frac{22}{100} \\times N383,292\\)\n= N84,324.24\nPrice at the end of the 4th year = N(383,292 - 84,324.24)\n= N298,967.76 \\(\\approxeq\\) N299,000 (to the nearest hundred naira).\n\n', '', '', 'WAEC-PK-1', 'WAEC-MATH', 1),
(319, '2013', '\n\nQuestion\n\n(a) The diagonals of a rhombus are 10.2 cm and 9.3 cm long. Calculate, correct to one decimal place, the perimeter of the rhombus.\n(b) Given that \\(\\sin x = \\frac{3}{5}, 0° < x < 90°\\), find the value of \\(5\\cos x - 4\\tan x\\).\n\n', '', '(a) The diagonals of a rhombus are 10.2 cm and 9.3 cm long. Calculate, correct to one decimal place, the perimeter of the rhombus.\n(b) Given that \\(\\sin x = \\frac{3}{5}, 0° < x < 90°\\), find the value of \\(5\\cos x - 4\\tan x\\).', '(a) The diagonals of a rhombus are 10.2 cm and 9.3 cm long. Calculate, correct to one decimal place, the perimeter of the rhombus.~(b) Given that \\(\\sin x = \\frac{3}{5}, 0° < x < 90°\\), find the value of \\(5\\cos x - 4\\tan x\\).', NULL, '\n\nExplanation\n(a)\nLet a side of the rhombus be n.\nThen \\(n^{2} = 4.65^{2} + 5.1^{2}\\)\n= \\(21.6225 + 26.01 \\)\n\\(n^{2} = 47.6325\\)\n\\(n = \\sqrt{47.6325} = 6.902 cm\\)\nHence, the perimeter of the rhombus = 4n\n= \\(4 \\times 6.902\\)\n= \\(27.608 cm \\approxeq 27.6 cm\\) (to 1 d.p)\n(b) Right- angled triangle ABC\n\n\\(\\sin x = \\frac{3}{5}\\), Thus, \\(m^{2} = 5^{2} - 3^{2} = 25 - 9 = 16\\)\n\\(m = 4\\).\n\\(\\cos x = \\frac{4}{5} \\implies 5\\cos x = 5\\times \\frac{4}{5} = 4\\)\n\\(\\tan x = \\frac{3}{4} \\implies 4\\tan x = 4 \\times \\frac{3}{4} = 3\\)\n\\(5\\cos x - 4\\tan x = 4 - 3 = 1\\).\n\n', '', 'Explanation\n(a)\nLet a side of the rhombus be n.\nThen \\(n^{2} = 4.65^{2} + 5.1^{2}\\)\n= \\(21.6225 + 26.01 \\)\n\\(n^{2} = 47.6325\\)\n\\(n = \\sqrt{47.6325} = 6.902 cm\\)\nHence, the perimeter of the rhombus = 4n\n= \\(4 \\times 6.902\\)\n= \\(27.608 cm \\approxeq 27.6 cm\\) (to 1 d.p)\n(b) Right- angled triangle ABC\n\n\\(\\sin x = \\frac{3}{5}\\), Thus, \\(m^{2} = 5^{2} - 3^{2} = 25 - 9 = 16\\)\n\\(m = 4\\).\n\\(\\cos x = \\frac{4}{5} \\implies 5\\cos x = 5\\times \\frac{4}{5} = 4\\)\n\\(\\tan x = \\frac{3}{4} \\implies 4\\tan x = 4 \\times \\frac{3}{4} = 3\\)\n\\(5\\cos x - 4\\tan x = 4 - 3 = 1\\).\n\n', '(a)\nLet a side of the rhombus be n.\nThen \\(n^{2} = 4.65^{2} + 5.1^{2}\\)\n= \\(21.6225 + 26.01 \\)\n\\(n^{2} = 47.6325\\)\n\\(n = \\sqrt{47.6325} = 6.902 cm\\)\nHence, the perimeter of the rhombus = 4n\n= \\(4 \\times 6.902\\)\n= \\(27.608 cm \\approxeq 27.6 cm\\) (to 1 d.p)~(b) Right- angled triangle ABC\n\n\\(\\sin x = \\frac{3}{5}\\), Thus, \\(m^{2} = 5^{2} - 3^{2} = 25 - 9 = 16\\)\n\\(m = 4\\).\n\\(\\cos x = \\frac{4}{5} \\implies 5\\cos x = 5\\times \\frac{4}{5} = 4\\)\n\\(\\tan x = \\frac{3}{4} \\implies 4\\tan x = 4 \\times \\frac{3}{4} = 3\\)\n\\(5\\cos x - 4\\tan x = 4 - 3 = 1\\).\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(320, '2013', '\n\nQuestion\n\n(a)\nIn the diagram\n(i) The value of x; (ii)\n(b) If \\(2N4_{seven} = 15N_{nine}\\), find the value of N.\n\n', '', '(a)\nIn the diagram\n(i) The value of x; (ii)\n(b) If \\(2N4_{seven} = 15N_{nine}\\), find the value of N.', '(a)\nIn the diagram\n(i) The value of x; (ii)~(b) If \\(2N4_{seven} = 15N_{nine}\\), find the value of N.', NULL, '\n\nExplanation\n(a)\n(i) In the diagram,\n\\(x° + 90° = (3x + 15)°\\) (Sum of opposite interior angles)\nThus, \\(90° - 15° = 3x - x \\implies 75° = 2x\\)\n\\(x = 37.5°\\)\n(ii)\n\n< RSQ = 90° - 37.5° = 52.5°\n(b) \\(2N4_{seven} = 15N_{nine}\\)\n\\(2N4_{seven} = (2 \\times 7^{2}) + (N \\times 7^{1}) + (4 \\times 7^{0})\\)\n= \\(98 + 7N + 4\\)\n= \\(102 + 7N\\)\n\\(15N_{nine} = (1 \\times 9^{2}) + (5 \\times 9^{1}) + (N \\times 9^{0})\\)\n= \\(81 + 45 + N\\)\n= \\(126 + N\\)\n\\(\\implies 102 + 7N = 126 + N\\)\n\\(7N - N = 126 - 102 = 24\\)\n\\(6N = 24 \\implies N = 4\\)\nTherefore, \\(244_{seven} = 154_{nine}\\).\n\n\n', '', 'Explanation\n(a)\n(i) In the diagram,\n\\(x° + 90° = (3x + 15)°\\) (Sum of opposite interior angles)\nThus, \\(90° - 15° = 3x - x \\implies 75° = 2x\\)\n\\(x = 37.5°\\)\n(ii)\n\n< RSQ = 90° - 37.5° = 52.5°\n(b) \\(2N4_{seven} = 15N_{nine}\\)\n\\(2N4_{seven} = (2 \\times 7^{2}) + (N \\times 7^{1}) + (4 \\times 7^{0})\\)\n= \\(98 + 7N + 4\\)\n= \\(102 + 7N\\)\n\\(15N_{nine} = (1 \\times 9^{2}) + (5 \\times 9^{1}) + (N \\times 9^{0})\\)\n= \\(81 + 45 + N\\)\n= \\(126 + N\\)\n\\(\\implies 102 + 7N = 126 + N\\)\n\\(7N - N = 126 - 102 = 24\\)\n\\(6N = 24 \\implies N = 4\\)\nTherefore, \\(244_{seven} = 154_{nine}\\).\n\n\n', '(a)\n(i) In the diagram,\n\\(x° + 90° = (3x + 15)°\\) (Sum of opposite interior angles)\nThus, \\(90° - 15° = 3x - x \\implies 75° = 2x\\)\n\\(x = 37.5°\\)\n(ii)\n\n< RSQ = 90° - 37.5° = 52.5°~(b) \\(2N4_{seven} = 15N_{nine}\\)\n\\(2N4_{seven} = (2 \\times 7^{2}) + (N \\times 7^{1}) + (4 \\times 7^{0})\\)\n= \\(98 + 7N + 4\\)\n= \\(102 + 7N\\)\n\\(15N_{nine} = (1 \\times 9^{2}) + (5 \\times 9^{1}) + (N \\times 9^{0})\\)\n= \\(81 + 45 + N\\)\n= \\(126 + N\\)\n\\(\\implies 102 + 7N = 126 + N\\)\n\\(7N - N = 126 - 102 = 24\\)\n\\(6N = 24 \\implies N = 4\\)\nTherefore, \\(244_{seven} = 154_{nine}\\).\n\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(321, '2013', '\n\nQuestion\n\n(a) If the mean of m, n, s, p and q is 12, calculate the mean of (m + 4), (n - 3), (s + 6), (p - 2) and (q + 8).\n(b) In a community of 500 people, the 75th percentile age is 65 years while the 25th percentile age is 15 years. How many of the people are between 15 and 65 years?\n\n', '', '(a) If the mean of m, n, s, p and q is 12, calculate the mean of (m + 4), (n - 3), (s + 6), (p - 2) and (q + 8).\n(b) In a community of 500 people, the 75th percentile age is 65 years while the 25th percentile age is 15 years. How many of the people are between 15 and 65 years?', '(a) If the mean of m, n, s, p and q is 12, calculate the mean of (m + 4), (n - 3), (s + 6), (p - 2) and (q + 8).~(b) In a community of 500 people, the 75th percentile age is 65 years while the 25th percentile age is 15 years. How many of the people are between 15 and 65 years?', NULL, '\n\nExplanation\n(a) \\(\\frac{m + n + s + p + q}{5} = 12 \\)\n\\(\\implies m + n + s + p + q = 12 \\times 5 = 60\\)\n\\(\\frac{(m + 4) + (n - 3) + (s + 6) + (p - 2) + (q + 8)}{5} = \\frac{m + n + s + p + q + 4 - 3 + 6 - 2 + 8}{5}\\)\n= \\(\\frac{60 + 13}{5} = \\frac{73}{5}\\)\n= 14.6.\n(b) In the community, there are 500 people.\nThe 75th percentile age = 65 years\nNumber of people in the 75th percentile = \\(\\frac{75 \\times 500}{100} = 375\\)\nThe 25th percentile = 15 years\nNumber of people in the 25th percentile = \\(\\frac{25 \\times 500}{100} = 125\\)\nNumber of people between 15 years and 65 years = 375 - 125 = 250 people.\n\n', '', 'Explanation\n(a) \\(\\frac{m + n + s + p + q}{5} = 12 \\)\n\\(\\implies m + n + s + p + q = 12 \\times 5 = 60\\)\n\\(\\frac{(m + 4) + (n - 3) + (s + 6) + (p - 2) + (q + 8)}{5} = \\frac{m + n + s + p + q + 4 - 3 + 6 - 2 + 8}{5}\\)\n= \\(\\frac{60 + 13}{5} = \\frac{73}{5}\\)\n= 14.6.\n(b) In the community, there are 500 people.\nThe 75th percentile age = 65 years\nNumber of people in the 75th percentile = \\(\\frac{75 \\times 500}{100} = 375\\)\nThe 25th percentile = 15 years\nNumber of people in the 25th percentile = \\(\\frac{25 \\times 500}{100} = 125\\)\nNumber of people between 15 years and 65 years = 375 - 125 = 250 people.\n\n', '(a) \\(\\frac{m + n + s + p + q}{5} = 12 \\)\n\\(\\implies m + n + s + p + q = 12 \\times 5 = 60\\)\n\\(\\frac{(m + 4) + (n - 3) + (s + 6) + (p - 2) + (q + 8)}{5} = \\frac{m + n + s + p + q + 4 - 3 + 6 - 2 + 8}{5}\\)\n= \\(\\frac{60 + 13}{5} = \\frac{73}{5}\\)\n= 14.6.~(b) In the community, there are 500 people.\nThe 75th percentile age = 65 years\nNumber of people in the 75th percentile = \\(\\frac{75 \\times 500}{100} = 375\\)\nThe 25th percentile = 15 years\nNumber of people in the 25th percentile = \\(\\frac{25 \\times 500}{100} = 125\\)\nNumber of people between 15 years and 65 years = 375 - 125 = 250 people.\n', 'WAEC-PK-1', 'WAEC-MATH', 1);
