INSERT INTO `exam_simulation` (`id`, `examYear`, `questions`, `options`, `answer`, `explaination`, `imageDir`, `topic`, `package_id`, `subject_id`, `status`) VALUES
(1311, '2007', 'Find the LCM of 23×3×52,2×32×5and22×32×52^{3}\\times 3\\times 5^{2}, 2\\times 3^{2}\\times 5 \\hspace{1mm}and \\hspace{1mm}2^{2}\\times 3^{2}\\times 5', ' 22×33×522^{2}\\times 3^{3}\\times 5^{2}~ 23×323×52^{3}\\times 3^{23}\\times 5~ 22×32×522^{2}\\times 3^{2}\\times 5^{2}~ 23×33×522^{3}\\times 3^{3}\\times 5^{2}~', ' 23×33×522^{3}\\times 3^{3}\\times 5^{2}', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1312, '2007', 'Simplify (0.3×105)', ' 7.5×10−47.5\\times 10^{-4}~ 7.5×10−37.5\\times 10^{-3}~ 7.5×10−27.5\\times 10^{-2}~ 7.5×10−17.5\\times 10^{-1}~', ' 7.5×10−37.5\\times 10^{-3}', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1313, '2007', 'If p', ' ~~~~', '', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1314, '2007', 'Simplify 3√45−12√5+16√203\\sqrt{45}-12\\sqrt{5}+16\\sqrt{20}leaving your answer in surd form', ' 29√529\\sqrt{5}~ 14√1514\\sqrt{15}~ 12√1512\\sqrt{15}~ 11√511\\sqrt{5}~', ' 29√529\\sqrt{5}', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1315, '2007', 'Simplify3x3(3x)3\\frac{3x^{3}}{(3x)^{3}}', '1~ 13\\frac{1}{3}~ 19\\frac{1}{9}~ 127\\frac{1}{27}~', ' 19\\frac{1}{9}', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1316, '2007', 'N140,000 is shared between ABU, Kayode and Uche. Abu has twice as much as Kayode, and Kayode has twice as much as Uche. What is Kayode\'s share?', ' N80,000~ N40,000~ N20,000~ N10,000~', ' N40,000', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1317, '2007', 'I am x years old and my brother is 3 years older how old was my brother last year', ' (x - 4) years~ (x + 2) years~ (3x - 1) years~ (3x + 1) years~', ' (3x - 1) years', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1318, '2007', 'For what value of x is the expression 2x−1x+32x−1x+3\\frac{2x-1}{x+3}not defined?', '3~2~ 1/2~-3~', '-3', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1319, '2007', 'If x =3, y = 2 and z = 4 what is the value of 3x2−2y+z3x2−2y+z3x^{2}-2y+z', '17~27~35~71~', '27', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1320, '2007', 'Solve the simultaneous equation: x+y=2 and 3x-2y=1', ' x=2 and y=1~ x=1 and y=1~ x=1 and y=2~ x=-1 and y=1~', ' x=1 and y=1', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1321, '2007', 'If 8x+1=148x+1=148^{x+1}=\\frac{1}{4}, find x', ' −53−53-\\frac{5}{3}~-1~ −13−13-\\frac{1}{3}~ 1313\\frac{1}{3}~', ' −53−53-\\frac{5}{3}', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1322, '2007', 'The area of a square field is 110.25m2. Find the cost of fencing it round at N75.00 per meter square', ' N1,575.00~ N3,150.00~ N4,734.30~ N8,268.75~', ' N8,268.75', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1323, '2007', 'A chord of length 6cm is drawn in a circle of radius 5cm. Find the distance of the chord from the center of the circle', ' 2.5cm~ 3.0cm~ 3.5cm~ 4.0cm~', ' 4.0cm', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1324, '2007', 'A sector of a circle of radius 14cm containing an angle 60o is folded to form a cone. Calculate the radius of the base of the cone.', ' 512cm512cm5\\frac{1}{2}cm~ 423cm423cm4\\frac{2}{3}cm~ 312cm312cm3\\frac{1}{2}cm~ 213cm213cm2\\frac{1}{3}cm~', ' 213cm213cm2\\frac{1}{3}cm', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1325, '2007', 'IN the diagram, |LN| = 4cm, LNM = 90o and tan y = 2/3. What is the area of the ', ' 24cm2~ 12cm2~ 10cm2~ 6cm2~', ' 12cm2', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1326, '2007', 'Evaluate(111two)2 and leave your answer in base 2.', ' 111001two~ 110001two~ 101001two~ 10010two~', ' 110001two', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1327, '2007', 'Find the LCM of 23 x 3 x 52, 2 x 32 x 5 and 23 x 33 x 5', ' 22 x 33 x 52~ 23 x 33 x 5~ 22 x 32 x 52~ 23 x 33 x 52~', ' 23 x 33 x 52', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1328, '2007', 'Simplify(0.3 x 105) ', ' 7.5 x 10-4~ 7.5 x 10-3~ 7.5 x 10-2~ 7.5 x 10-1~', ' 7.5 x 10-3', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1329, '2007', 'If p ', ' q ~~~~', '', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1330, '2007', 'Simplify 3√45−12√15+16√2045−1215+1620\\sqrt{45} - 12\\sqrt{15} + 16\\sqrt{20}, leaving your answer in surd form', ' 29√55\\sqrt{5}~ 14√1515\\sqrt{15}~ 12√1515\\sqrt{15}~ 11√1515\\sqrt{15}~', ' 29√55\\sqrt{5}', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1331, '2007', 'Simplify 3x3(3x)33x3(3x)3\\frac{3x^3}{(3x)^3}', '1~ 1313\\frac{1}{3}~ 1919\\frac{1}{9}~ 127127\\frac{1}{27}~', ' 1919\\frac{1}{9}', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1332, '2007', 'N140,000 is shared between Abu, Kayode and Uche. Abu has twice as much as Kayode and Kayode has twice as much as Uche. What is Kayode\'sshare?', ' N80,000~ N40,000~ N2,000~10000~', ' N40,000', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1333, '2007', 'I am x years old and my brother is 3 years older. How old was my brother last year?', ' (x - 4)years~ (x + 2) years~ (3x - 10years~ (3x + 1)years~', ' (x + 2) years', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1334, '2007', 'For what value of x is the expression 2x−1x+32x−1x+3\\frac{2x - 1}{x + 3} not defined?', '3~2~ 1212\\frac{1}{2}~-3~', '-3', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1335, '2007', 'If x = 3, Y = 2 and z = 4, what is the value of 3x2 - 2y + z?', '17~27~35~71~', '27', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1336, '2007', 'Solve the simultaneous equations x + y = 2 and 3x - 2y = 1', ' x = 2 and y = 1~ x = 1 and y = 1~ x = 1 and y = 2~ x = -1 and y = 1~', ' x = 1 and y = 1', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1337, '2007', 'If 8x-1 = 1414\\frac{1}{4}, find x', ' -−53−53\\frac{-5}{3}~-1~ -1313\\frac{1}{3}~ 1313\\frac{1}{3}~', ' 1313\\frac{1}{3}', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1338, '2007', 'A chord of length 6cm is drawn in a circle of radius 5cm. Find the distance of the chord from the centre of the circle.', ' 2.5cm~ 3.0cm~ 3.5cm~ 4.0cm~', ' 4.0cm', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1339, '2007', 'The area of a square field is 110.25m2. Find the cost of fencing it round at N75.00 per metre', ' N1,575.00~ N3,150.00~ N4,734.30~ N8,268.75~', ' N3,150.00', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1340, '2007', 'A sector of a circle of radius 14cm containing an angle 60o is folded to form a cone. Calculate the radius of the base of the cone', ' 512cm12cm\\frac{1}{2}cm~ 423cm23cm\\frac{2}{3}cm~ 312cm12cm\\frac{1}{2}cm~ 213cm13cm\\frac{1}{3}cm~', ' 213cm13cm\\frac{1}{3}cm', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1341, '2007', 'Find the volume of a solid cylinder with base radius 10cm and height 14cm.', '2203~ 880cm3>~ 1400cm3~ 4400cm3~', ' 4400cm3', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1342, '2007', 'In a , < PQR = < PRQ = 45o. which of the following statements is/are correct?  i. ', ' ii only~ i and ii only~ ii and iii only~ i and ii only~', ' ii and iii only', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1343, '2007', 'The table above gives the distribution of the marks of a number of students in a test.Mark123456Frequency536425Mark123456Frequency536425\\begin{array}{c|c} Mark ', '2~3~5~6~', '3', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1344, '2007', 'If the probability of an event occurring is x, what is the probability of the event not occurring?', ' 1 - x~ x - 1~0~ 1x1x\\frac{1}{x}~', ' 1 - x', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1345, '2007', 'The table shows the ages (in years) of twenty children chosen at random from a community. What is the mean age?Age (years)12345Number of children23564Age (years)12345Number of children23564\\begin{array}{c|c} \\text{Age (years)} ', ' 4.46 years~ 3.35 years~ 3.30years~ 3.00 years~', ' 3.35 years', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1346, '2007', 'The table shows the ages (in years) of twenty children chosen at random from a community. What is the median of the distribution?Age (years)12345Number of children23564Age (years)12345Number of children23564\\begin{array}{c|c} \\text{Age (years)} ', ' 3.5 years~ 3.0 years~ 2.5 years~ 2.0 years~', ' 3.5 years', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1347, '2007', 'Divide the sum of 8, 6, 7, 2, 0, 4, 7, 2, 3, by their mean', '9~8~7~6~', '9', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1348, '2007', 'Q is 32 km away from P on a bearing 042o and R is 25km from P on a bearing of 132o. Calculate the bearing of R from Q.', ' 122o~ 184o~ 190o~ 226o~', ' 184o', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1349, '2007', 'The bearing of a point P from another point Q is 310o. If |PQ| = 200m, how far west of Q is P?', ' 128.6m~ 153.2m~ 167.8m~ 187.9m~', ' 153.2m', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1350, '2007', 'Find the value of a if log10 a2 + log10a2 = 0.9030', '4~2~1.6~0~', '1.6', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1351, '2007', 'A messenger was paid N2.50 an hour during the normal working hours and 4.00 n hour during overtime. If he received N31.00 for 10 hours work, how many hours are for overtime?', '2~3~4~6~', '4', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1352, '2007', 'At what rate per cent per annum will N520.00 yield a simple interest of N39.00 in three years?', '0.04~ 31212\\frac{1}{2}%~0.03~ 21212\\frac{1}{2}%~', ' 21212\\frac{1}{2}%', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1353, '2007', 'PQR is an equilateral triangle with sides 2√3cm√3cm\\sqrt{3cm}. calculate its height', ' 1.7cm~ 3.0cm~ 3.2cm~ 3.9cm~', ' 3.0cm', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1354, '2007', 'Which of the following is a factor of 2 - x - x2?', ' 1 - x~ 1 + x~ x - 1~ 2 - x~', ' 1 - x', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1355, '2007', 'make w the subject of the relation a+bcwd+fa+bcwd+f\\frac{a + bc}{wd + f} = g', ' a+bc−fgdga+bc−fgdg\\frac{a + bc - fg}{dg}~ a−bc+fgdga−bc+fgdg\\frac{a - bc + fg}{dg}~ a+bc−fdga+bc−fdg\\frac{a + bc - f}{dg}~ a+bc−dgdga+bc−dgdg\\frac{a + bc - dg}{dg}~', ' a+bc−fgdga+bc−fgdg\\frac{a + bc - fg}{dg}', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1356, '2007', 'For what range of values of x is 4x - 3(2x - 1) > 1?', ' x > -1~ x > 1~~~', '', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1357, '2007', 'Find the smallest value of k such that 22 x 32 5 x k is a perfect square', '3~5~15~30~', '15', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1358, '2007', 'Expand the expression(3a - xy)(3a + xy)', ' 9a2 - x2y2~ 9a2 + x2y2~ 9a2 - xy~ 9a2 + x2y~', ' 9a2 - x2y2', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1359, '2007', 'Simplify 42x−2x+xx42x−2x+xx\\frac{4}{2x} - \\frac{2x + x}{x}', '-1~ -2x~ 2x~ 2−x2x2−x2x\\frac{2 - x}{2x}~', '-1', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1360, '2007', 'Two circles have radii 16cm and 23cm. What is the difference between their circumference? take ', ' 422.92cm~ 149.92cm~ 44.00cm~ 43.96cm~', ' 44.00cm', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1361, '2007', 'The volume of a cylinder is 1200cm3 and the area of its base is 1500cm2. Find the height of the cylinder.', ' 80.00cm~ 8.00cm~ 0.80cm~ 0.08cm~', ' 8.00cm', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1362, '2007', 'In the diagram |LN| = 4cm, LNM = 90o and tan y = 2323\\frac{2}{3}. What is the area of the ', ' 24cm2~ 12cm2~ 10cm2~ 6cm2~', ' 12cm2', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1363, '2007', 'From the diagram, find the value of x in the diagram.', ' 80o~ 70o~ 55o~ 35o~', ' 70o', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1364, '2007', 'In the diagram, |SP| = |SR| and < PRS = 50o. Calculate < PQR', ' 120o~ 110o~ 100o~ 80o~', ' 100o', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1365, '2007', 'The venn diagram shows the choice of food of a number of visitors to a canteen. If there were 35 visitors in all, find the value of x.', '5~4~3~2~', '5', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1366, '2007', 'The Venn diagram shows the choice of food of a number of visitors to a canteen. How many people took at least two kinds of food?', '10~12~15~17~', '17', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1367, '2007', 'If the ratio x:y = 3:5 and y:z = 4:7, find the ratio x:y:z', ' 15 : 28 : 84~ 12 : 20 : 35~ 3 : 5 : 4~ 5 : 4 : 7~', ' 12 : 20 : 35', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1368, '2007', 'In the diagram, |XR| = 4cm, |RZ| = 12cm, |SR| = n, |XZ| = m and SR||YZ. Find m in terms of n.', ' m = 2n~ m = 3n~ m = 4n~ m = 5n~', ' m = 4n', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1369, '2007', 'In the diagram, O is the centre of the circle and PQ is a diameter. Triangle RSO is an equilateral triangle of side 4cm. Find the area of the shaded region.', ' 43.36cm2~32.072~18.212~6.932~', '18.212', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1370, '2007', 'Find the product of 0.0409 and 0.0021 leaving your answer in the standard form', ' 8.6 x 10-6~ 8.6 x 105~ 8.6 x 10-4~ 8.6 x 105~', ' 8.6 x 105', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1371, '2007', 'Convert 425 to base three numeral', '2013~2103~2113~3433~', '2113', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1372, '2008', 'The volume of a cube is 512cm3. Find the length of its side', ' 6cm~ 7cm~ 8cm~ 9cm~', ' 8cm', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1373, '2008', 'Simplify x2−y22x2+xy−y2x2−y22x2+xy−y2\\frac{x^2 - y^2}{2x^2 + xy - y^2}', ' x+y2x+yx+y2x+y\\frac{x + y}{2x + y}~ x+y2x−yx+y2x−y\\frac{x + y}{2x - y}~ x−y2x−yx−y2x−y\\frac{x - y}{2x - y}~ x−y2x+yx−y2x+y\\frac{x - y}{2x + y}~', ' x−y2x−yx−y2x−y\\frac{x - y}{2x - y}', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1374, '2008', 'A train travels 60km in M minutes. If its average speed is 400km per hour, find the value of M.', '15~12~10~9~7', '9', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1375, '2008', 'If x% of 240 equals 12, find x', ' x = 1~ x = 3~ x = 5~ x = 7~', ' x = 5', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1376, '2008', 'Evaluate (3.2)2−(4.8)23.2+4.8(3.2)2−(4.8)23.2+4.8\\frac{(3.2)^2 - (4.8)^2}{3.2 + 4.8}', '-0.08~-1.6~-10.24~-12.8~', '-1.6', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1377, '2008', 'Simplify √50+10√2\\sqrt{50} + \\frac{10}{\\sqrt{2}}', '10~ 10√2\\sqrt{2}~20~ 20√2\\sqrt{2}~', ' 10√2\\sqrt{2}', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1378, '2008', 'P naira invested for 4 years at r% simple interest per annum yields 0.36 p naira interest. Find the value of r.', ' 119\\frac{1}{9}~ 149\\frac{4}{9}~9~11~', '9', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1379, '2008', 'A trader bought 100 tubers at 5 for N350.00. She sold them in sets of 4 for N290.00. Find her gain percent.', '0.036~0.035~0.034~0.025~', '0.036', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1380, '2008', 'If p-2g + 1 = g + 3p and p - 2 = 0, find g', '-2~-1~1~2~', '-1', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1381, '2008', 'Simplify 1x+1yx+y\\frac{\\frac{1}{x} + \\frac{1}{y}}{x + y}', ' 1x+y\\frac{1}{x + y}~ 1xy\\frac{1}{xy}~ x + y~ xy~', ' 1xy\\frac{1}{xy}', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1382, '2008', 'Simplify 3√27x3y93√27x3y9\\sqrt', ' 9xy3~ 3xy6~ 3xy3~ 9y3~', ' 3xy3', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1383, '2008', 'Given that x = 2 and y = -14\\frac{1}{4}, evaluate x2y−2xy5\\frac{x^2y - 2xy}{5}', ' zero~ 15\\frac{1}{5}~1~2~', ' zero', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1384, '2008', 'Factorize 5y2 + 2ay - 3a2', ' (a - y)(5y - 3a)~ (y - a)(5y - 3a)~ (y - a)(5y + 3a)~ (y + a)(5y - 3a)~', ' (y + a)(5y - 3a)', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1385, '2008', 'If 4y is 9 greater than the sum of y and 3x, by how much is y greater than x?', '3~6~9~12~', '3', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1386, '2008', 'If p = 35√qr\\frac{3}{5} \\sqrt{\\frac{q}{r}}, express q in terms of p and r', ' 925pr2\\frac{9}{25} pr^2~ 925p2r\\frac{9}{25} p^2r~ 259p2r\\frac{25}{9} p^2r~ 259pr2\\frac{25}{9} pr^2~', ' 259p2r\\frac{25}{9} p^2r', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1387, '2008', 'PQR is a sector of a circle centre O, radius 4cm. If PQR = 30o, find, correct to 3 significant figures, the area of sector PQR. ', ' 4.19cm2~ 8.38cm2~ 10.5cm2~ 20.9cm2~', ' 4.19cm2', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1388, '2008', 'The angles of a quadrilateral are (x + 10)o, 2yo, 90o and (100 - y)o, Find y in terms of x', ' y = 160 + x~ y = 100 + x~ y = 160 - x~ y = x - 100~', ' y = 160 - x', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1389, '2008', 'If xo is obtuse, which of the following is true?', ' x > 90~ 180 < x < 270~ x < 90~ 90 < x < 180~', ' 90 < x < 180', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1390, '2008', 'If tan x = 1, evaluate sin x + cos x, leaving your answer in the surd form', ' 2√2\\sqrt{2}~ 12√2\\frac{1}{2} \\sqrt{2}~ √2\\sqrt{2}~2~', ' √2\\sqrt{2}', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1391, '2008', 'If cos (x + 25)o = sin 45o, find the value of x', '20~30~45~60~', '20', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1392, '2008', 'If 2n = 128, find the value of (2n - 1)(5n - 2)', ' 5(106)~ 2(106)~ 5(105)~ 2(105)~', ' 2(105)', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1393, '2008', 'If x ', ' x varies directly as y~ x varies inversely as y~ x is partly constant and partly varies as y~ x vries jointly as 45 and directly as y~', ' x is partly constant and partly varies as y', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1394, '2008', 'Simplify log√8log4−log2\\frac{\\log \\sqrt{8}}{\\log 4 - \\log 2}', ' 23\\frac{2}{3}~ 12log2\\frac{1}{2} \\log 2~ 32\\frac{3}{2}~ log2\\log 2~', ' 32\\frac{3}{2}', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1395, '2008', 'An arc of a circle, radius 14cm, is 18.33cm long. Calculate to the nearest degree, the angle which the arc subtends at the centre of the circle. ', ' 11o~ 20o~ 22o~ 75o~', ' 75o', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1396, '2008', 'What is the length of an edge of a cube whose total surface area is X cm2 and whose volume is X2\\frac{X}{2}cm3?', '3~6~9~12~', '3', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1397, '2008', 'XY is a chord of circle centre O and radius 7cm. The chord XY which is 8cm long subtends an angle of 120o at the centre of the circle. Calculate the perimeter of the minor segment. ', ' 14.67cm~ 22.67cm~ 29.33cm~ 37.33cm~', ' 22.67cm', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1398, '2008', 'If p = 12\\frac{1}{2} and 1p−1=2p+x\\frac{1}{p - 1} = \\frac{2}{p + x}, find the value of x', ' -212\\frac{1}{2}~ -112\\frac{1}{2}~ 112\\frac{1}{2}~ 212\\frac{1}{2}~', ' -112\\frac{1}{2}', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1399, '2008', 'Find the quadratic equation whose roots are c and -c', ' x2 - c2 = 0~ x2 + 2cx = 0~ x2 + 2cx + c2 = 0~ x2 - 2cx + c2 = 0~', ' x2 - c2 = 0', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1400, '2008', 'Given the sets A = ', '1~ 34\\frac{3}{4}~ 12\\frac{1}{2}~ 14\\frac{1}{4}~', ' 34\\frac{3}{4}', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1401, '2008', 'Find the mean deviation of these numbers 10, 12, 14, 15, 17, 19.', '2.5~2.6~2.7~2.8~', '2.5', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1402, '2008', 'In the diagram, < QPR = 90o. If q2 = 25 - r2. Find the value of p.', '3~4~5~6~', '5', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1403, '2008', 'In the diagram, /PQ/ = /QR/ and /PR/ = /RS/ = /SP/, calculate the side of < QRS.', ' 150o~ 120o~ 90o~ 60o~', ' 90o', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1404, '2008', 'In the diagram, PQRS is a rhombus. /PR/ = 10cm and /QS/ = 24cm. Calculate the perimeter of the rhombus', ' 34cm~ 52cm~ 56cm~ 96cm~', ' 52cm', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1405, '2008', 'In the figure shown, PQS is a straight line. What is the value of ', ' 128o~ 108o~ 98o~ 78o~', ' 78o', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1406, '2008', 'In the diagram, PQ//RS, QU//PT and < PSR = 42o. Find angle x.', ' 84o~ 48o~ 42o~ 32o~', ' 42o', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1407, '2008', 'In the figure /PX/ = /XQ/, PQ//YZ and XV//QR. What is the ratio of the area of XYZQ to te area of ', ' 1:2~ 2:1~ 1:2~ 3:1~', ' 2:1', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1408, '2008', 'In the diagram, O is the centre of the circle and PQRS is a cyclic quadrilateral. Find the value of x.', ' 25o~ 65o~ 115o~ 130o~', ' 65o', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1409, '2008', 'If the perimeter of ', ' 19.5cm2~ 15.0cm2~ 13.0cm2~ 9.3cm2~', ' 15.0cm2', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1410, '2008', 'In the diagram, PX is a tangent to the circle and RST is an equilateral triangle. Calculate < PTS.', ' 60o~ 90o~ 120o~ 150o~', ' 120o', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1411, '2008', 'In the diagram, o, o, o, o, o and RS//TU, calculate y.', ' 102o~ 78o~ 70o~ 60o~', ' 102o', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1412, '2008', 'Every staff in an office owns either a Mercedes and/or a Toyota car. 20 own Mercedes, 15 own Toyota and 5 own both. How many staff are there in the office?', '25~30~35~45~', '30', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1413, '2009', 'Simplify: (312+413)', ' 11414\\frac{1}{4}~ 81212\\frac{1}{2}~35~47~', '47', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1414, '2009', 'If log 2 = x, log 3 = y and log 7 = z, find, in terms of x, y and z, the value of log (283283\\frac{28}{3})', ' 2x + y - z~ 2x + x - y~ x + y - 2x~ x + x - y~', ' 2x + x - y', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1415, '2009', 'Arrange the following numbers in descending orders of magnitude: 22three, 34five, 21six', ' 21six, 22three, 34five,~ 21six, 34five, 22three~ 22three, 34five,  21six~ 34five,  21six, 22three~', ' 34five,  21six, 22three', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1416, '2009', 'Find the value to which N3000.00 will amount in 5 years at 6% per annum simple interest', ' N3,900.00~ N3,750.00~ N3,600.00~ N3,300~', ' N3,900.00', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1417, '2009', 'Express the square root of 0.000144 in the standard form', ' 1.2 x 10-4~ 1.2 x 10-3~ 1.2 x 10-2~ 1.2 x 10-1~', ' 1.2 x 10-2', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1418, '2009', 'Two sets are disjoint if', ' they are both empty~ their union is an empty set~ their intersection is an empty set~ one of them is a subset of the other~', ' their intersection is an empty set', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1419, '2009', 'If 32x−23x=432x−23x=4\\frac{3}{2x} - \\frac{2}{3x} = 4, solve for x', ' 4545\\frac{4}{5}~ 413413\\frac{4}{13}~ 524524\\frac{5}{24}~ 13241324\\frac{13}{24}~', ' 524524\\frac{5}{24}', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1420, '2009', 'If (2x + 3)3 = 125, find the value of x', '1~2~3~4~', '1', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1421, '2009', 'Factorize x2 - 2x - 3xy completely', ' (x - 2)(3x - y)~ (x - 3y)(x - 2)~ (x - 3y)(x - 3y)~ (3x + y)(x - 2)~', ' (x - 3y)(x - 2)', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1422, '2009', 'The diameter of a bicycle wheel is 42cm. If the wheel makes 16 complete revolution, what will be the total distance covered by the wheel? ', ' 672cm~ 1056cm~ 2112cm~ 4224cm~', ' 2112cm', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1423, '2009', 'The interior angles of a pentagon are (2x + 5)o, (x + 20)o, xo, (3x - 20)o and (x + 15)o. Find the value of x', ' 80o~ 70o~ 65o~ 40o~', ' 65o', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1424, '2009', 'If M and N are two fixed points in a plane. Find the locus L = ', ' a line equal to MN~ line parallel to MN~ Perpendicular bisector of MN~ A circle centre P, radius MN~', ' Perpendicular bisector of MN', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1425, '2009', 'Given that sin 60o = √32√32\\frac{\\sqrt{3}}{2} and cos 60o = 1212\\frac{1}{2}, evaluate 1−sin60o1+cos60o1−sin60o1+cos60o\\frac{1 - sin 60^o}{1 + cos 60^o}', ' 2+√332+√33\\frac{2 + \\sqrt{3}}{3}~ 1−√331−√33\\frac{1 - \\sqrt{3}}{3}~ 1+√331+√33\\frac{1 + \\sqrt{3}}{3}~ 2−√332−√33\\frac{2 - \\sqrt{3}}{3}~', ' 2−√332−√33\\frac{2 - \\sqrt{3}}{3}', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1426, '2009', 'In an examination, Kofi scored x% in Physics, 50% in Chemistry and 70% in Biology. If his mean score for the three subjects was 55%, find x', '40~45~55~60~', '45', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1427, '2009', 'What is the median of the following scores: 22 35 41 63 74 82', '82~52~49~22~', '52', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1428, '2009', 'Express 2.7864 x 10-3 to 2 significant figures', '0.28~0.27~0.0028~0.0027~', '0.0028', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1429, '2009', 'If 92x = 1313\\frac{1}{3}(27x), find x', '2~1~-1~-2~', '-1', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1430, '2009', 'If 85% of x is N3230, what is the value of x?', ' N2745.50~ N3714.50~ N3800.00~ N4845.00~', ' N3800.00', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1431, '2009', 'The capacity of a water tank is 1,800 litres. If the tank is in form of a cuboid with base 600cm by 150 cm. Find the height of the tank', ' 2cm~ 20cm~ 200cm~ 2000cm~', ' 20cm', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1432, '2009', 'An arc of a circle subtends an angle of 60o at the centre. If the radius of the circle is 3cm, find , in terms of ', ' ~~~~', ' ', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1433, '2009', 'solve 2x+16−3x−142x+16−3x−14\\frac{2x + 1}{6} - \\frac{3x - 1}{4} = 0', '1~ 1515\\frac{1}{5}~ -1515\\frac{1}{5}~-1~', '1', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1434, '2009', 'If a positive integer, list the values of x which satisfy the equation 3x - 4 < 6 and x - 1 > 0', ' {1, 2, 3}~ {2, 3}~ {2, 3, 4}~ {2, 3, 4, 5}~', ' {2, 3}', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1435, '2009', 'If c and k are the roots of 6 - x - x2 = 0, find c + k', '2~1~-1~-3~', '-1', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1436, '2009', 'A bucket holds 10 litres of water. How many buckets of water will fill a reservoir of size 8m x 7m x 5m.(1 litre = 1000cm3)', '28~280~2800~28000~', '28000', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1437, '2009', 'A box contains black, white and red identical balls. The probability of picking a black ball at random from the box is 310310\\frac{3}{10} and the probability of picking a white ball at random is 2525\\frac{2}{5}. If there are 30 balls in the box, how many of them are red?', '3~7~9~12~', '9', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1438, '2009', 'Simplify; 12√32−√18√21232−182\\frac{1}{2}\\sqrt{32} - \\sqrt{18}  \\sqrt{2}', ' zero~ √22\\sqrt{2}~ 2√22\\sqrt{2}~ 4\\(\\sqrt{2}\\0~', ' zero', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1439, '2009', 'The angles of triangle are (x + 10)o, (2x - 40)o and (3x - 90)o. Which of the following accurately describes the triangle?', ' it is a scalene triangle~ it is right angled isosceles triangle~ t is an equilateral triangle~ It is an isosceles triangle but not right angled~', ' t is an equilateral triangle', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1440, '2009', 'Simplify (x - 3y)2 - (x + 3y)2', ' 2(x + 3y)~ (2x - 3y)~ -12xy~ 6xy~', ' -12xy', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1441, '2009', 'Find the quadratic equation whose roots are -1212\\frac{1}{2} and 3', ' 2x2 - 2x + 3 = 0~ 2x2 - 2x - 3 = 0~ 2x2 - 5x + 3 = 0~ 3x2 - 5x - 3 = 0~', ' 3x2 - 5x - 3 = 0', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1442, '2009', 'The following is the graph of a quadratic friction, find the co-ordinates of point P.', ' (0, 4)~ (4, 0)~ (0, -4)~ (-4, 0)~', ' (0, 4)', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1443, '2009', 'The following is the graph of a quadratic friction, find the value of x when y = 0.', ' 1, 2~ 1, 4~ 2, 3~ 1, 6~', ' 1, 4', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1444, '2009', 'PQRS is a trapezuim. QR//PS, /PQ/ = 5cm, /OR/ = 6cm, /PS/ = 10cm and angle QPS = 42o. Calculate the perpendicular distance between the parallel sides', ' 3.35cm~ 3.73cm~ 4.50cm~ 4.62cm~', ' 3.35cm', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1445, '2009', 'PQRS is a trapezium. QR//PS, /PQ/ = 5cm, /OR/ = 6cm, /PS/ = 10cm and angle QPS = 42o. Calculate, correct to the nearest cm2, the area of the trapezium (h = 3.35cm2 )', ' 27cm2~ 30cm2~ 36cm2~ 37cm2~', ' 27cm2', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1446, '2009', 'In the diagram, /TP/ = 12cm and it is 6cm from O, the centre of the circle, Calculate ', ' 120o~ 90o~ 60o~ 45o~', ' 90o', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1447, '2009', 'In the diagram, IG is parallel to JE, JEF = 120o and FHG = 130o, find the angle marked t.', ' 40o~ 70o~ 80o~ 100o~', ' 70o', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1448, '2009', 'In the diagram, \\SQ\\ = 4cm, \\PT\\ = 7cm. /TR/ = 5cm and ST//OR. If /SP/ = xcm, find the value of x.', '5.6~6.5~6.6~6.8~', '5.6', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1449, '2009', 'Using the diagram, find the bearing of X from Y.', ' 300o~ 240o~ 120o~ 60o~', ' 240o', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1450, '2009', 'If /XY/ = 50m, how far cast of X is Y?', ' 25.5m~ 40.6m~ 40.8m~ 43.3m~', ' 43.3m', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1451, '2009', 'In the diagram, angle 20o is subtended at the centre of the circle, find the value of x.', ' 30o~ 50o~ 80o~ 100o~', ' 50o', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1452, '2009', 'The pie chart shows the distribution of 4320 students who graduated from four departments in a university. How many students graduated from the science department?', '840~864~920~960~', '864', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1453, '2009', 'The pie chart shows the distribution of 4320 students who graduated from four departments in a university. If a student is picked at random from the four departments, what id the probability that he is not from the education department?', ' 310310\\frac{3}{10}~ 2929\\frac{2}{9}~ 518518\\frac{5}{18}~ 710710\\frac{7}{10}~', ' 710710\\frac{7}{10}', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1454, '2009', 'From the diagram which of the following statements are correct? i. XQ is a radius of a circle centre Q. ii. /XQ/ = /QY/. iii. /QX/ = /XY/', ' i only~ inand ii only~ ii and iii only~ i, ii and iii~', ' inand ii only', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1455, '2010', 'If 2 logx (33838\\frac{3}{8}) = 6, find the value of x', ' 3232\\frac{3}{2}~ 4343\\frac{4}{3}~ 2323\\frac{2}{3}~ 1212\\frac{1}{2}~', ' 3232\\frac{3}{2}', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1456, '2010', 'Simplify (1681)14', ' 2323\\frac{2}{3}~ 1212\\frac{1}{2}~ 8989\\frac{8}{9}~ 1313\\frac{1}{3}~', ' 1212\\frac{1}{2}', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1457, '2010', 'Solve the inequality -6(x + 3) ≤≤\\leq 4(x - 2)', ' x ≤≤\\leq 2~ x ≥≥\\geq -1~ x ≥≥\\geq -2~ x ≤≤\\leq -1~', ' x ≥≥\\geq -1', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1458, '2010', 'Simplify 0.000215 x 0.000028 and express your answer in standard form', ' 6.03 x 109~ 6.02 x 109~ 6.03 x 10-9~ 6.02 x 10-9~', ' 6.02 x 10-9', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1459, '2010', 'Factorize: x + y - ax = ay', ' (x - y)(1 - a)~ (x + y)(1 + a)~ (x + y)(1 - a)~ (x - y)(1 + a)~', ' (x + y)(1 - a)', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1460, '2010', 'A car uses one litre of petrol for every 14km. If one of petrol cost N63.00, how far can the car go with N900.00 worth of petrol?', ' 420km~ 405km~ 210km~ 200km~', ' 200km', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1461, '2010', 'Correct 0.002473 to 3 significant figure', '0.002~0.0024~0.00247~0.0025~', '0.00247', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1462, '2010', 'The sum of 2 consecutive whole numbers is 5656\\frac{5}{6} of their product, find the numbers', ' 3, 4~ 1, 2~ 2, 3~ 0, 1~', ' 2, 3', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1463, '2010', 'A casting is made up of Copper and Zinc. If 65% of the casting is Zinc and there are 147g of Copper. What is the mass of the casting?', ' 320g~ 420g~ 520g~ 620g~', ' 420g', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1464, '2010', 'Given that P = {x : 1 ≤x≤6≤x≤6\\leq x \\leq 6}, and Q = {x : 2 ≤x≤10\\leq x \\leq 10} where x is an integer. Find n(P ∩\\cap)', '4~6~8~10~', '4', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1465, '2010', 'The sum of 6 and one-third of x is one more than twice x, find x', ' x = 7~ x = 5~ x = 3~ x = 2~', ' x = 3', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1466, '2010', 'An empty rectangular tank is 250cm long and 120cm wide. If 180 litres of water is poured into the tank. Calculate the height of the water', ' 6.0cm~ 5.5cm~ 5.0cm~ 4.5cm~', ' 6.0cm', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1467, '2010', 'Given that 5n+3252n−2\\frac{5^{n +3}}{25^{2n -2}} = 5o, find n', ' n = 1~ n = 2~ n = 3~ n = 5~', ' n = 3', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1468, '2010', 'No. of pets01234No. of students845103\\begin{array}{c|c} \\text{No. of pets} ', ' 15\\frac{1}{5}~ 25\\frac{2}{5}~ 35\\frac{3}{5}~ 45\\frac{4}{5}~', ' 35\\frac{3}{5}', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1469, '2010', 'Simplify 2√3\\sqrt{3} - 6√3+3√27\\frac{6}{\\sqrt{3}} + \\frac{3}{\\sqrt{27}}', '1~ 13√3\\frac{1}{3}\\sqrt{3}~ 2√3−523\\sqrt{3} - 5\\frac{2}{3}~ 6√3\\sqrt{3} - 17~', ' 13√3\\frac{1}{3}\\sqrt{3}', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1470, '2010', 'If sin 3y = cos 2y and 0o ≤\\leq 90o, find the value of y', ' 18o~ 36o~ 54o~ 90o~', ' 18o', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1471, '2010', 'A rectangle has length xcm and width (x - 1)cm. If the perimeter is 16cm. Find the value of x', ' 312\\frac{1}{2}cm~ 4cm~ 412\\frac{1}{2}cm~ 5cm~', ' 412\\frac{1}{2}cm', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1472, '2010', 'Given that tan x = 1, where 0o ≥\\geq x 90o, evaluate 1−sin2xcosx\\frac{1 - \\sin^2 x}{\\cos x}', ' 2√2\\sqrt{2}~ √2\\sqrt{2}~ √22\\frac{\\sqrt{2}}{2}~ 12\\frac{1}{2}~', ' √22\\frac{\\sqrt{2}}{2}', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1473, '2010', 'What is the length of a rectangular garden whose perimeter is 32cm and area 39cm2?', ' 25cm~ 18cm~ 13cm~ 9cm~', ' 13cm', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1474, '2010', 'Simplify 22+x+22−x\\frac{2}{2 + x} + \\frac{2}{2 - x}', ' 44−x3\\frac{4}{4 - x^3}~ 84−x2\\frac{8}{4 - x^2}~ 4x4−x2\\frac{4x}{4 - x^2}~ 8−4x4−x2\\frac{8 - 4x}{4 - x^2}~', ' 84−x2\\frac{8}{4 - x^2}', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1475, '2010', 'The sum of the exterior of an n-sided convex polygon is half the sum of its interior angle. find n', '6~8~9~12~', '6', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1476, '2010', 'If y = y(2√x2+m)3N\\frac{y(2\\sqrt{x^2 + m})}{3N}, make x the subject of the formular', ' √9y2N2−2m3\\frac{\\sqrt{9y^2 N^2 - 2m}}{3}~ √9y2N2−4m2\\frac{\\sqrt{9y^2 N^2 - 4m}}{2}~ √9y2N2−3m2\\frac{\\sqrt{9y^2 N^2 - 3m}}{2}~ √9y2N−3m2\\frac{\\sqrt{9y^2 N - 3m}}{2}~', ' √9y2N2−4m2\\frac{\\sqrt{9y^2 N^2 - 4m}}{2}', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1477, '2010', 'The nth term of the sequence -2, 4, -8, 16.... is given by', ' Tn = 2n~ Tn = (-2)n~ Tn = (-2n)~ Tn = n~', ' Tn = (-2)n', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1478, '2010', 'How many times, correct to the nearest whole number, will a man run round  circular track of diameter 100m to cover a distance of 1000m?', '3~4~5~6~', '3', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1479, '2010', 'Bola sold an article for N6,900.00 and made a profit of 15%. If he sold it for N6,600.00 he would make a', ' profit of 13%~ loss of 12%~ loss of 10%~ loss of 5%~', ' loss of 10%', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1480, '2010', 'The mean age of R men in a club is 50 years, Two men aged 55 and 63, left the club and the mean age reduced by 1 year. Find the value of R', '18~20~22~28~', '20', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1481, '2010', 'x0246y1234\\begin{array}{c|c} x ', ' y = 2x~ y = x + 1~ y = x~ y = 12x+1\\frac{1}{2}x + 1~', ' y = 12x+1\\frac{1}{2}x + 1', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1482, '2010', 'What is the value of x when y = 5?', '8~9~10~11~', '8', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1483, '2010', 'The subtraction below is in base seven. Find the missing number.5 1 6 2seven-2 6 4 4seven--------2 * 1 5--------', '2~3~4~5~', '2', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1484, '2010', 'If the sum of the roots of the equation (x - p)(2x - 1) - 0 is 1, find the value of x', ' 112\\frac{1}{2}~ 12\\frac{1}{2}~ -32\\frac{3}{2}~ -112\\frac{1}{2}~', ' 112\\frac{1}{2}', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1485, '2010', 'If x km/h = y m/s, then y =', ' 718718\\frac{7}{18}x~ 11201120\\frac{11}{20}x~ 415\\frac{4}{15}x~ 518\\frac{5}{18}x~', ' 518\\frac{5}{18}x', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1486, '2010', 'The mean of the numbers 2, 5, 2x and 7 is less than or equal to 5. Find the range of the values of x', ' x ≤\\leq 3~ x ≥\\geq 3~~~', ' x ≥\\geq 3', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1487, '2010', 'In an athletic composition, the probability that an athlete wins a 100m race is 18\\frac{1}{8} and the probability that he wins in high jump is 14\\frac{1}{4}. What is the probability that he wins only one of the events?', ' 332\\frac{3}{32}~ 73\\frac{7}{3}~ 53\\frac{5}{3}~ 516\\frac{5}{16}~', ' 516\\frac{5}{16}', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1488, '2010', 'In the diagram, < PSR = 220o, < SPQ = 58o and < PQR = 41o. Calculate the obtuse angle QRS.', ' 90o~ 100o~ 121o~ 60o~', ' 121o', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1489, '2010', 'The bar chart shows the marks distribution in am English test. If 50% is the pass mark, how many students passed the test?', '100~85~80~70~', '85', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1490, '2010', 'The bar chart shows the marks distribution in am English test. What percentage of the students had marks ranging from 35 to 50?', ' 5513\\frac{1}{3}%~0.6~0.65~ 6623\\frac{2}{3}%~', ' 6623\\frac{2}{3}%', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1491, '2010', 'What is the value of m in the diagram?', ' 20o~ 30o~ 40o~ 50o~', ' 30o', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1492, '2010', 'In the diagram, QR//ST, /PQ/ = /PR/ and < PST = 75o. Find the value of y.', ' 105o~ 110o~ 130o~ 150o~', ' 105o', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1493, '2010', 'In the diagram, triangles HKL and HIJ are similar. Which of the following ratios is equal to LHJH\\frac{LH}{JH}?', ' KLJI\\frac{KL}{JI}~ HKJK\\frac{HK}{JK}~ JIKL\\frac{JI}{KL}~ HKLK\\frac{HK}{LK}~', ' KLJI\\frac{KL}{JI}', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1494, '2010', 'In the diagram, < ROS = 66o and < POQ = 3x. some construction lines are shown. Calculate the value of x.', ' 10o~ 11o~ 22o~ 35o~', ' 11o', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1495, '2010', 'In the diagram, O is the centre of the circle, < SQR = 60o, < SPR = y and < SOR = 3x. Find the value of (x + y).', ' 110o~ 100o~ 80o~ 70o~', ' 100o', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1496, '2010', 'The shaded portion in the diagram is the solution of', ' x + y ≤≤\\leq 3~ x + y < 3~ x + y > 3~ x + y ≥≥\\geq 3~', ' x + y < 3', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1497, '2010', 'In the diagram, GI is a tangent to the circle at H. If EF//GI, calculate the size of ', ' 126o~ 72o~ 64cmo~ 32cmo~', ' 72o', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1498, '2010', 'The diagram is a net right rectangular pyramid. Calculate the total surface area.', ' 208cm2~ 112cm2~ 92cm2~ 76cm2~', ' 92cm2', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1499, '2010', 'The diagram shows a rectangular cardboard from which a semi-circle is cut off. Calculate the area of the remaining part.', ' 44cm2~ 99cm2~ 154cm2~ 165cm2~', ' 99cm2', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1500, '2010', 'In the diagram, 0 is the centre of the circle. Find the value x.', '34~29~17~14~', '14', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1501, '2010', 'In the diagram, < WOX = 60o, < YOE = 50o and < OXY = 30o. What is the bearing of x from y?', ' 300o~ 240o~ 190o~ 150o~', ' 300o', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1502, '2011', 'Simplify 1025−623+31025−623+310\\frac{2}{5} - 6 \\frac{2}{3} + 3', ' 6415415\\frac{4}{15}~ 611151115\\frac{11}{15}~ 7415415\\frac{4}{15}~ 711151115\\frac{11}{15}~', ' 611151115\\frac{11}{15}', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1503, '2011', 'Solve the inequality -6(x + 3) ≤\\leq 4(x - 2)', ' x ≤\\leq 2~ x ≥\\geq -1~ x ≥\\geq -2~ x ≤\\leq -1~', ' x ≥\\geq -1', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1504, '2011', 'If N112.00 exchanges for D14.95, calculate the value of D1.00 in naira', '0.13~7.49~8~13~', '7.49', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1505, '2011', 'Solve for x in the equation; 35\\frac{3}{5}(2x - 1) = 14\\frac{1}{4}(5x - 3)', ' zero~1~2~3~', '3', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1506, '2011', 'Given that cos xo = 1r\\frac{1}{r}, express tan x in terms of r', ' 1√r\\frac{1}{\\sqrt{r}}~ √r\\sqrt{r}~ √r2+1\\sqrt{r^2 + 1}~ √r2−1\\sqrt{r^2 - 1}~', ' √r2−1\\sqrt{r^2 - 1}', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1507, '2011', 'Solve the equation; 3x - 2y = 7, x + 2y = -3', ' x = 1, y = -2~ x = 1, y = 3~ x = -2, y = -1~ x = 4, y = -3~', ' x = 1, y = -2', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1508, '2011', 'If a number is chosen at random from the set {x: 4 ≤x≤45\\leq x \\leq 45}. Find the probability that it is a multiple of 3 or a multiple of 4', ' 112\\frac{1}{12}~ 512\\frac{5}{12}~ 72\\frac{7}{2}~ 1112\\frac{11}{12}~', ' 72\\frac{7}{2}', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1509, '2011', 'One of the factors of (mn - nq - n2 + mq) is (m - n). The other factor is', ' (n - q)~ (q - n)~ (n + q)~ (q - m)~', ' (n - q)', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1510, '2011', 'A cylindrical container has a base radius of 14cm and height 18cm. How many litres of liquid can it hold?  correct to the nearest litre ', '11~14~16~18~', '11', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1511, '2011', 'A regular polygon of n sides has each exterior angle to 45o. Find the value of n', '6~8~12~15~', '8', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1512, '2011', 'Esther was facing S 20o W. She turned 90o in the clock wise direction. What direction is she facing?', ' N 70o W~ S 70o W~ N 20o W~ S 20o E~', ' S 70o W', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1513, '2011', 'The cross section section of a uniform prism is a right-angled triangle with sides 3cm. 4cm and 5cm. If its length is 10cm. Calculate the total surface area', ' 142cm2~ 132cm2~ 122cm2~ 112cm2~', ' 132cm2', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1514, '2011', 'From the equation whose roots are x = 12\\frac{1}{2} and -23\\frac{2}{3}', ' 6x2 - x + 2 = 0~ 6x2 - x - 2 = 0~ 6x2 + x + 2 = 0~ 6x2 + x - 2 = 0~', ' 6x2 + x - 2 = 0', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1515, '2011', 'Simplify log√27log√81\\frac{\\log \\sqrt{27}}{\\log \\sqrt{81}}', '3~2~ 32\\frac{3}{2}~ 34\\frac{3}{4}~', ' 34\\frac{3}{4}', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1516, '2011', 'Which of these angles can be constructed using ruler and a pair of compasses only?', ' 115o~ 125o~ 135o~ 145o~', ' 135o', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1517, '2011', 'The perimeter of a sector of a circle of radius 4cm is (', ' 45o~ 60o~ 75o~ 90o~', ' 45o', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1518, '2011', 'The length of a piece of stick is 1.75m. A girl measured it as 1.80m. Find the percentage error', ' 287\\frac{28}{7}%~ 297\\frac{29}{7}%~0.05~ 207\\frac{20}{7}%~', ' 207\\frac{20}{7}%', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1519, '2011', 'What is the value of 3 in the number 42.7531?', ' 310000\\frac{3}{10000}~ 31000\\frac{3}{1000}~ 3100\\frac{3}{100}~ 110\\frac{1}{10}~', ' 3100\\frac{3}{100}', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1520, '2011', 'The height of a cylinder is equal to its radius. If the volume is 0.216 ', ' 0.46m~ 0.60m~ 0.87m~ 1.80m~', ' 0.46m', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1521, '2011', 'Factorize the expression: am + bn - an - bm', ' (a - b)(m + n)~ (a - b)(m - n)~ (a + b)(m - n)~ (a + b)(m + n)~', ' (a - b)(m - n)', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1522, '2011', 'The graph represents the relation y = x2 - 3x - 3. Find the value of x for which x2 - 3x = 7.', ' -1.55, 4.44~ 1.55, -4.55~ -1.55, -4.55~ 1.55, 4.55~', ' -1.55, 4.44', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0);
INSERT INTO `exam_simulation` (`id`, `examYear`, `questions`, `options`, `answer`, `explaination`, `imageDir`, `topic`, `package_id`, `subject_id`, `status`) VALUES
(1523, '2011', 'Simplify mn+(m−1)5n=(m−2)10nmn+(m−1)5n=(m−2)10n\\frac{m}{n} + \\frac{(m - 1)}{5n} = \\frac{(m - 2)}{10n} where n ', ' m−310nm−310n\\frac{m - 3}{10n}~ 11m10n11m10n\\frac{11m}{10n}~ m+110nm+110n\\frac{m + 1}{10n}~ 11m+410n11m+410n\\frac{11m + 4}{10n}~', ' 11m10n11m10n\\frac{11m}{10n}', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1524, '2011', 'If √72+√32−3√18=x√8\\sqrt{72} + \\sqrt{32} - 3 \\sqrt{18} = x \\sqrt{8}, Find the value of x', '1~ 34\\frac{3}{4}~ 12\\frac{1}{2}~ 14\\frac{1}{4}~', ' 12\\frac{1}{2}', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1525, '2011', 'G varies directly as the square of H, If G is 4 when H is 3, find H when G = 100', '15~25~75~225~', '15', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1526, '2011', 'Given that n(p) = 19, m(P ', '26~31~36~50~', '26', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1527, '2011', 'What must be added to (2x - 3y) to get (x 0 2y)?', ' 5y - x~ y - x~ x - 5x~ x - y~', ' y - x', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1528, '2011', 'Find the smaller value of x that satisfies the equation x2 + 7x + 10 = 0', '-5~-2~2~5~', '-5', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1529, '2011', 'The perpendicular bisectors of the sides of an acute-angled triangle are drawn. Which of these statements is correct? They intersect', ' on one of the vertices~ at a midpoint of a side~ inside the triangle~ outside the triangle~', ' inside the triangle', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1530, '2011', 'A rectangular garden measures 18.6m by 12.5m. Calculate, correct to three significant figures, the area of the garden', ' 230m2~ 231m2~ 232m2~ 233m2~', ' 233m2', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1531, '2011', 'John pours 96 litres of red oil into a rectangular container with length 220cm and breadth 40cm. Calculate, correct to the nearest cm, the height of the oil in the container', ' 11cm~ 18cm~ 21cm~ 34cm~', ' 11cm', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1532, '2011', 'in a quiz competition, a student answers n questions correctly and was given D(n + 50) for each question correctly answered. If he gets D600.00 altogether, how many questions did he answer correctly?', '18~15~12~10~', '10', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1533, '2011', 'If 27x = 9y. Find the value of xy\\frac{x}{y}', ' 13\\frac{1}{3}~ 23\\frac{2}{3}~ 112\\frac{1}{2}~3~', ' 23\\frac{2}{3}', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1534, '2011', 'a boy looks through a window of a building and sees a mango fruit on the ground 50m away from the foot of the building. If the window is 9m from the ground, calculate, correct to the nearest degree, the angle of depression of the mango from the window', ' 9o~ 10o~ 11o~ 12o~', ' 10o', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1535, '2011', 'The diagram shows a cyclic quadrilateral PQRS with its diagonals intersecting at K. Which of the following triangles is similar to triangle QKR?', ' ~~~~', '', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1536, '2011', 'In the diagram, PO and OR are radii, |PQ| = |QR| and reflex o. Calculate the value x.', ' 60o~ 55o~ 50o~ 45o~', ' 60o', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1537, '2011', 'Find the size of the angle marked x in the diagram.', ' 108o~ 112o~ 128o~ 142o~', ' 128o', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1538, '2011', 'The histogram shows the age distribution of members of a club. How many members are in the club?', '52~50~48~40~', '52', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1539, '2011', 'The histogram shows the age distribution of members of a club. What is their modal age?', '44.5~42.5~41.5~40.5~', '44.5', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1540, '2011', 'In the diagram, /MN/, /OP/, /QOP/ = 125o. What is the size of ', ' 11o~ 120o~ 130o~ 160o~', ' 120o', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1541, '2011', 'The graph represents the relation y = x2 - 3x - 3. What is the equation of the line of symmetry of the graph?', ' y = 0.5~ x = 1.0~ x = 1.5~ y = 4.6~', ' y = 4.6', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1542, '2011', 'From the diagram, Which of the following statements are true? i. m = q ii. n = q iii. n + p = 180o iv. p + m = 180o', ' i and iii~ i and iv~ ii and iii~ ii and ic~', ' i and iii', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1543, '2011', 'Using the cumulative frequency curve, estimate the median of the data represented on the graph.', '35.5~36.5~37.5~38.5~', '36.5', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1544, '2011', 'Using the cumulative frequency curve. What is the 80th percentile?', '45.5~46.5~47.5~48.5~', '47.5', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1545, '2011', 'The Venn diagram shows the number of students in a class who like reading(R), dancing(D) and swimming(S). How many students like dancing and swimming?', '7~9~11~13~', '7', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1546, '2012', 'Find the coefficient of m in the expression of (m2−112)(m+23)(\\frac{m}{2} - 1\\frac{1}{2})(m + \\frac{2}{3})', ' −16-\\frac{1}{6}~ −12-\\frac{1}{2}~ −1-1~ −116-1\\frac{1}{6}~', ' −116-1\\frac{1}{6}', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1547, '2012', 'The volume of a cuboid is 54 cm3. If the length, width and height of the cuboid are in the ratio 2:1:2 respectively, find its total surface area', ' 108 cm2~ 90 cm2~ 80 cm2~ 72 cm2~', ' 90 cm2', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1548, '2012', 'A side and a diagonal of a rhombus are 10cm and 12cm respectively, Find its area', ' 20 cm2~ 24 cm2~ 48 cm2~ 96 cm2~', ' 96 cm2', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1549, '2012', 'Factorise completely: 32x2y - 3y3', ' 16x2y(2 - 3xy2)~ 8xy(4x - 6x2y2)~ 8x2y(4 - 6xy2)~ 16xy(2x - 3x2y2)~', ' 16x2y(2 - 3xy2)', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1550, '2012', 'In the diagram, MN, PQ and RS are three intersecting straight lines. Which of the following statements is/are true?  i. t = y  ii. x + y + z + m = 180°  iii. x + m + n = 180°  iv. x + n = m + z', ' i and iv~ ii~ iii~ iv~', ' iii', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1551, '2012', 'The bar chart shows the frequency distribution of marks scored by students in a class test. How many students are in the class?', '10~24~25~30~', '25', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1552, '2012', 'The bar chart shows the frequency distribution of marks scored by students in a class test. What is the median of the distribution?', '2~4~6~8~', '2', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1553, '2012', '', '11.1~10.5~10.3~9.9~', '11.1', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1554, '2012', 'The diagram is a circle of radius |QR| = 4cm. ', ' 2.32cm~ 1.84cm~ 0.62cm~ 0.26cm~', ' 0.62cm', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1555, '2012', 'Simplify 1p\\frac{1}{p} - 1q\\frac{1}{q} + pq\\frac{p}{q} - qp\\frac{q}{p}', ' 1p−q\\frac{1}{p - q}~ −1p+q\\frac{-1}{p + q}~ 1pq\\frac{1}{pq}~ 1pq(p−q)\\frac{1}{pq(p - q)}~', ' −1p+q\\frac{-1}{p + q}', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1556, '2012', 'Express 302.10495 correct to five significant figures', '302.1~302.11~302.105~302.1049~', '302.1', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1557, '2012', 'Simplify; 3√5×4√62√2×3√2\\frac{3\\sqrt{5} \\times 4\\sqrt{6}}{2 \\sqrt{2} \\times 3\\sqrt{2}}', ' √2\\sqrt{2}~ √5\\sqrt{5}~ 2√2\\sqrt{2}~ 2√5\\sqrt{5}~', ' 2√5\\sqrt{5}', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1558, '2012', 'In 1995, the enrollment of two schools X and Y were 1,050 and 1,190 respectively. Find the ration of the enrollments of X and Y', ' 50:11~ 15:17~ 13:55~ 12:11~', ' 15:17', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1559, '2012', 'Convert 3510 to number in base 2', '1011~10011~100011~11001~', '100011', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1560, '2012', 'The nth term of a sequence is Tn = 5 + (n - 1)2. Evaluate T4 - T6', '30~16~-16~-30~', '-16', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1561, '2012', 'Mr. Manu travelled from Accra to Pamfokromb a distance of 720km in 8 hours. What will be his speed in m/s?', ' 25m/s~ 150m/s~ 250m/s~ 500m/s~', ' 25m/s', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1562, '2012', 'If N2,500.00 amounted to N3,50.00 in 4 years at simple interest, find the rate at which the interest was charged', '0.05~ 712\\frac{1}{2}%~0.08~0.1~', '0.1', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1563, '2012', 'Solve for x in the equation; 1x+23x=13\\frac{1}{x} + \\frac{2}{3x} = \\frac{1}{3}', '5~4~3~1~', '5', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1564, '2012', 'Simplify: 54k2−63k+1\\frac{54k^2 - 6}{3k + 1}', ' 6(1 - 3k2)~ 6(3k2 - 1)~ 6(3k - 1)~ 6(1 - 3k)~', ' 6(3k - 1)', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1565, '2012', 'Make p the subject of the relation: q = 3pr+s23pr+s2\\frac{3p}{r} + \\frac{s}{2}', ' p = 2q−rs62q−rs6\\frac{2q - rs}{6}~ p = 2qr - sr - 3~ p = 2qr−s62qr−s6\\frac{2qr - s}{6}~ p = 2qr−rs62qr−rs6\\frac{2qr - rs}{6}~', ' p = 2qr−rs62qr−rs6\\frac{2qr - rs}{6}', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1566, '2012', 'If x + y = 2y - x + 1 = 5, find the value of x', '3~2~1~-1~', '2', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1567, '2012', 'The sum of 12 and one third of n is 1 more than twice n. Express the statement in the form of an equation', ' 12n - 6 = 0~ 3n - 12 = 0~ 2n - 35 = 0~ 5n - 33 = 0~', ' 5n - 33 = 0', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1568, '2012', 'Solve the inequality: −m2−54≥5m12−76−m2−54≥5m12−76\\frac{-m}{2} - \\frac{5}{4} \\geq \\frac{5m}{12} - \\frac{7}{6}', ' m ≤54≤54\\leq \\frac{5}{4}~ m ≥54≥54\\geq \\frac{5}{4}~ m ≤−111≤−111\\leq - \\frac{1}{11}~ m ≥−111≥−111\\geq - \\frac{1}{11}~', ' m ≤−111≤−111\\leq - \\frac{1}{11}', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1569, '2012', 'The curved surface area of a cylindrical tin is 704cm2. If the radius of its base is 8cm, find the height. ', ' 14cm~ 9cm~ 8cm~ 7cm~', ' 14cm', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1570, '2012', 'The lengths of the minor and major arcs 54cm and 126cm respectively. Calculate the angle of the major sector', ' 360o~ 252o~ 246o~ 234o~', ' 252o', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1571, '2012', 'If cos(x + 40)o = 0.0872, what is the value of x?', ' 85o~ 75o~ 65o~ 45o~', ' 45o', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1572, '2012', 'A kite flies on a taut string of length 50m inclined at tan angle 54o to the horizontal ground. The height of the kite above the ground is', ' 50 tan 30o~ 50 sin 54o~ 50 tan 54o~ 50 sin 36o~', ' 50 sin 54o', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1573, '2012', 'Given that the mean of the scores 15, 21, 17, 26, 18 and 29 is 21, calculate the standard deviation of the scores', ' √10√10\\sqrt{10}~4~5~ √30\\sqrt{30}~', '5', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1574, '2012', 'A bag contains 4 red and 6 black balls of the same size. If the balls are shuffled briskly and two balls are drawn one after the other without replacement, find the probability of picking balls of different colours', ' 815815\\frac{8}{15}~ 13251325\\frac{13}{25}~ 11151115\\frac{11}{15}~ 13151315\\frac{13}{15}~', ' 815815\\frac{8}{15}', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1575, '2012', 'Which of these statements about y = 8√m\\sqrt{m} is correct?', ' log y = log 8 x log √m\\sqrt{m}~ log y = 3 log 2 x 12\\frac{1}{2} log m~ log y = 3 log 2 - 12\\frac{1}{2} log m~ log y = 3 log 2 + 12\\frac{1}{2} log m~', ' log y = 3 log 2 + 12\\frac{1}{2} log m', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1576, '2012', 'If x + 0.4y = 3 and y = 12\\frac{1}{2}x, find the value of (x + y)', ' 114\\frac{1}{4}~ 212\\frac{1}{2}~ 334\\frac{3}{4}~5~', ' 334\\frac{3}{4}', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1577, '2012', 'Express 3 - ', ' 3xyy\\frac{3xy}{y}~ x−4yy\\frac{x - 4y}{y}~ 4y−xy\\frac{4y - x}{y}~ 3 - x−yy\\frac{x - y}{y}~', ' 4y−xy\\frac{4y - x}{y}', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1578, '2012', 'If p = {prime factors of 210} and Q = {prime less than 10}, find p ∩\\cap Q', ' {1,2, 3}~ {2, 3, 5}~ {1, 3, 5,7}~ {2,3,5,7}~', ' {2,3,5,7}', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1579, '2012', 'Alfred spent 14\\frac{1}{4} of his money on food, 13\\frac{1}{3} on clothing and save the rest. If he saved N72,20.00, how much did he spend on food?', ' N43,200.00~ N43,000.00~ N42,200.00~ N40,000.00~', ' N43,200.00', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1580, '2012', 'If x and y are variables and k is a constant, which of the following describes an inverse relationship between x and y?', ' y = kx~ y = kx\\frac{k}{x}~ y = k√x\\sqrt{x}~ y = x + k~', ' y = kx\\frac{k}{x}', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1581, '2012', 'In the diagram, |QR| = 10m, |SR| = 8m, < QPS = 30o, < QRP = 90o and |PS| = x, Find x.', ' 1.32m~ 6.32m~ 9.32m~17.32~', ' 9.32m', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1582, '2012', 'In the diagram, O is a circle centre of the circle PQRS and < PSR = 86o. If < PQR = xo, find x.', ' 108o~ 172o~ 130o~ 50o~', ' 172o', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1583, '2012', 'The diagram is a circle centre O. If < SPR = 2m and < SQR = n, express m in terms of n.', ' m = n2n2\\frac{n}{2}~ m = 2n~ m = n - 2~ m = n + 2~', ' m = n2n2\\frac{n}{2}', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1584, '2012', 'In the diagram, MQ//RS, < TUV = 70o and < RLV = 30o. Find the value of x.', ' 150o~ 110o~ 100o~ 95o~', ' 100o', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1585, '2012', 'The position of three ships P,Q and R at sea are illustrated in the diagram. The arrows indicated the North direction. The bearing of Q from P is 050o and < PQR = 72o. Calculate the bearing of R and Q.', ' 155o~ 80o~ 158o~ 91o~', ' 158o', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1586, '2012', 'In the diagram, MN//PO, < PMN = 112o, < PNO = 129o', ' 51o~ 54o~ 56o~ 68o~', ' 54o', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1587, '2012', 'The diagram is a circle with centre P. PRST are points on the circle. Find the value of ', ' 144o~ 72o~ 40o~ 36o~', ' 144o', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1588, '2012', 'In the diagram, |SR| = |QR|. o and o, find ', ' 65o~ 45o~ 25o~ 19o~', ' 19o', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1589, '2012', 'The graph is that of y = 2x2 - 5x - 3. For what value of x will y be negative? For what value of x will y be negative?', ' -12≤x\\frac{1}{2} \\leq x < 3~ -12<x≤3\\frac{1}{2} < x \\leq 3~ -12<x<3\\frac{1}{2} < x < 3~ -12≤x≤3\\frac{1}{2} \\leq x \\leq 3~', ' -12<x<3\\frac{1}{2} < x < 3', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1590, '2012', 'The diagram is a polygon. Find the largest of its interior angles.', ' 300o~ 100o~ 120o~ 150o~', ' 150o', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1591, '2012', 'Simplify 3x+152y+6xy\\frac{3}{x} + \\frac{15}{2y} + \\frac{6}{xy}', ' 2y−5x4\\frac{2y - 5x}{4}~ 9(2x−5x)x2y2\\frac{9(2x - 5x)}{x^2y^2}~ 5x−2y2\\frac{5x - 2y}{2}~ c2y218y−45x\\frac{c^2y^2}{18y - 45x}~ 42y−5x\\frac{4}{2y - 5x}', ' 2y−5x4\\frac{2y - 5x}{4}', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1592, '2013', 'In what number base is the addition 465 + 24 + 225 = 1050?', ' ten~ nine~ eight~ seven~', ' seven', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1593, '2013', 'Simplify 178×225634', '999~ 4124124\\frac{1}{2}~222~ 1212\\frac{1}{2}~', ' 1212\\frac{1}{2}', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1594, '2013', 'If √50−K√8=2√250−K8=22\\sqrt{50} - K\\sqrt{8} = \\frac{2}{\\sqrt{2}}, find KKK', '-2~-1~1~2~', '2', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1595, '2013', 'Four oranges sell for ', ' ~~~~', '', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1596, '2013', 'Simplify: x2−y2(x+y)2+(x−y)2(3x+3y)x2−y2(x+y)2+(x−y)2(3x+3y)\\frac{x^2 - y^2}{(x + y)^2} + \\frac{(x - y)^2}{(3x + 3y)}', ' x−y3x−y3\\frac{x - y}{3}~ x+yx+yx + y~ 3x−y3x−y\\frac{3}{x - y}~ x−yx−yx - y~', ' 3x−y3x−y\\frac{3}{x - y}', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1597, '2013', 'If x=64x=64x = 64 and y=27,y=27,y = 27, evaluate: x12−y13y−x23x12−y13y−x23\\frac{x^{\\frac{1}{2}} - y^{\\frac{1}{3}}}{y - x^{\\frac{2}{3}}}', ' 2152152\\frac{1}{5}~111~ 511511\\frac{5}{11}~ 11431143\\frac{11}{43}~', ' 511511\\frac{5}{11}', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1598, '2013', 'A chord is 2cm from the centre of a circle. If the radius of the circle is 5cm, find the length of the chord', ' 2√2121\\sqrt{21}cm~ √4242\\sqrt{42}cm~ 2√1919\\sqrt{19}cm~ √2121\\sqrt{21}cm~', ' 2√2121\\sqrt{21}cm', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1599, '2013', 'A cube and cuboid have the same base area. The volume of the cube is 64 cm3 while that of the cuboid is 80 cm3. Find the height of the cuboid', ' 1 cm~ 3 cm~ 5 cm~ 6 cm~', ' 1 cm', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1600, '2013', 'If sin x = 513513\\frac{5}{13} and 0° ≤≤\\leq x ≤≤\\leq 90°, find the value of (cos x - tan x)', ' 713713\\frac{7}{13}~ 12131213\\frac{12}{13}~ 7915679156\\frac{79}{156}~ 209156209156\\frac{209}{156}~', ' 7915679156\\frac{79}{156}', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1601, '2013', 'An object is 6m away from the base of a mast. The angle of depression of the object from the top of the mast is 50°, Find, correct to 2 decimal places, the height of the mast', ' 8.60m~ 7.51m~ 7.15m~ 1.19m~', ' 7.15m', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1602, '2013', 'Which of the following is not a probability of Mary scoring 85% in a mathematics test?', '0.15~0.57~0.94~1.01~', '1.01', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1603, '2013', 'If p = (y : 2y ≥≥\\geq 6) and Q = (y : y -3 ≥≥\\geq 4), where y is an integer, find p∩∩\\capQ', ' {3, 4}~ {3, 7}~ {3, 4, 5, 6, 7}~ {4, 5, 6}~', ' {3, 4, 5, 6, 7}', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1604, '2013', 'Find the values of k in the equation 6k2 = 5k + 6', ' {−23,−32−23,−32\\frac{-2}{3}, \\frac{-3}{2}}~ {−23,32−23,32\\frac{-2}{3}, \\frac{3}{2}}~ {23,−3223,−32\\frac{2}{3}, \\frac{-3}{2}}~ {23,3223,32\\frac{2}{3}, \\frac{3}{2}}~', ' {−23,32−23,32\\frac{-2}{3}, \\frac{3}{2}}', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1605, '2013', 'If y varies directly as the square root of (x + 1) and y = 6 when x = 3, find x when y = 9', '8~7~6~5~', '8', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1606, '2013', 'The graph of the relation y = x2 + 2x + k passes through the point (2, 0). Find the values of k', ' zero~-2~-4~-8~', '-8', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1607, '2013', 'What is the locus of the point X which moves relative to two fixed points P and M on a plane such that', ' the bisector of the straight line joining P and M~ an arc of a circle with PM as a chord~ the bisector of angle PXM~ a circle centre X and radius PM~', ' an arc of a circle with PM as a chord', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1608, '2013', 'When a number is subtracted from 2, the result equals 2 less than one-fifth of the number. Find the number', '111111~ 152152\\frac{15}{2}~555~ 5252\\frac{5}{2}~', '555', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1609, '2013', 'Express 2x+3−1x−22x+3−1x−2\\frac{2}{x + 3} - \\frac{1}{x - 2} as a simple fraction', ' x−7x2+x−6x−7x2+x−6\\frac{x - 7}{x^2 + x - 6}~ x−1x2+x−6x−1x2+x−6\\frac{x - 1}{x^2 + x - 6}~ x−2x2+x−6\\frac{x - 2}{x^2 + x - 6}~ x−27x2+x−6\\frac{x - 27}{x^2 + x - 6}~', ' x−7x2+x−6x−7x2+x−6\\frac{x - 7}{x^2 + x - 6}', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1610, '2013', 'Given that P = x2 + 4x - 2, Q = 2x - 1 and Q - P = 2, find x.', '-2~-1~1~2~', '-1', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1611, '2013', 'A pyramid has a rectangular base with dimensions 12m by 8m. If its height is 14m, calculate the volume', ' 322 m3~ 448 m3~ 632 m3~ 840 m3~', ' 448 m3', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1612, '2013', 'The slant height of a cone is 5cm and the radius of its base is 3cm. Find, correct to the nearest whole number, the volume of the cone. (Take ', ' 48cm3~ 47cm3~ 38cm3~ 12cm3~', ' 38cm3', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1613, '2013', 'The distance between two towns is 50km. It is represented on a map by 5cm. Find the scale used', ' 1: 1,000,000~ 1: 500,000~ 1: 100,000~ 1: 10,000~', ' 1: 1,000,000', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1614, '2013', 'Given that (x + 2)(x2 - 3x + 2) + 2(x + 2)(x - 1) = (x + 2) M, find M', ' (x + 2)2~ x(x + 2)~ xv + 2~ x2 - x~', ' x2 - x', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1615, '2013', 'An open cone with base radius 28cm and perpendicular height 96cm was stretched to form sector of a circle. Calculate the arc of the sector (Take ', ' 8800cm2~ 8448cm2~ 4400cm2~ 4224cm2~', ' 8800cm2', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1616, '2013', 'In the diagram, the height of a flagpole |TF| and the length of its shadow |FL| re in the ratio 6:8. Using k as a constant of proportionality, find the shortest distance between T and L', ' 7K units~ 10K units~ 12K units~ 14K units~', ' 10K units', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1617, '2013', 'the diagrams, |XZ| = |MN|, |ZY| = |MO| and |XY| = |NO|. Which of the following statements is true?', ' ~~~~', '', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1618, '2013', 'The pie chart shows the distribution of 600 mathematics textbooks for Arts, Business, Science and Technical Classes. How many textbooks are for the technical class?', '100~150~200~250~', '250', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1619, '2013', 'The pie chart shows the distribution of 600 mathematics textbooks for Arts, Business, Science and Technical Classes. What percentage of the total number of textbooks belongs to science?', ' 1212121212\\frac{1}{2}%~ 20562056\\frac{205}{6}%~2525.25~ 41234123\\frac{412}{3}%~', ' 1212121212\\frac{1}{2}%', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1620, '2013', 'Multiply 2.7 x 10-4 by 6.3 x 106 and leave your answers in standard form', ' 1.7 x 103~ 1.70 x 103~ 1.701 x 103~ 17.01 x 103~', ' 1.701 x 103', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1621, '2013', 'If 9(2 - x) = 3, find x', '1~ 32\\frac{3}{2}~2~ 52\\frac{5}{2}~', ' 32\\frac{3}{2}', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1622, '2013', 'If Un = n(n2 + 1), evaluate U5 - U4', '18~56~62~80~', '62', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1623, '2013', 'A sales boy gave a change of N68 instead of N72. Calculate his percentage error', '0.04~ 559\\frac{5}{9}%~ 51517\\frac{15}{17}%~0.07~', ' 559\\frac{5}{9}%', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1624, '2013', 'If 12\\frac{1}{2}x + 2y = 3 and 32\\frac{3}{2}x and 32\\frac{3}{2}x - 2y = 1, find (x + y)', '3~2~1~5~', '3', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1625, '2013', 'Given that p13\\frac{1}{3} = 3√qr\\frac{3\\sqrt{q}}{r}, make q the subject of the equation', ' q = p√r\\sqrt{r}~ q = p3r~ q = pr3~ q = pr13\\frac{1}{3}~', ' q = pr13\\frac{1}{3}', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1626, '2013', 'The bearing of Y from X is 060o and the bearing of Z from Y = 060o. Find the bearing of X from Z', ' 300o~ 240o~ 180o~ 120o~', ' 240o', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1627, '2013', 'In the diagram, PRST is a square. If |PQ| = 24cm. |QR| = 10cm and < PQR = 90o, find the perimeter of the polygon PQRST.', ' 112cm~ 98cm~ 86cm~ 84cm~', ' 112cm', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1628, '2013', 'In the diagram, PQRS is a rhombus and < PSQ = 35o. Calculate the size of < PRO.', ' 65o~ 55o~ 45o~ 35o~', ' 35o', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1629, '2013', 'Find the value of m in the diagram.', ' 34o~ 27o~ 23o~ 17o~', ' 17o', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1630, '2013', 'In the diagram, O is the centre of the circle. OM||XZ and < ZOM = 25o.', ' 50o~ 55o~ 60o~ 65o~', ' 60o', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1631, '2013', 'Using the histogram, estimate the mode of distribution.', '51.5~52.5~53.5~54.5~', '53.5', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1632, '2013', 'Using the histogram, what is the median class?', ' 60.5 - 70.5~ 50.5 - 60.5~ 40.5 - 50.5~ 30.5 - 40.5~', ' 40.5 - 50.5', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1633, '2014', 'If 23x = 325, find the value of x', '7~6~5~4~', '7', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1634, '2014', 'Simplify: √12(√48−√3)12(48−3)\\sqrt{12} (\\sqrt{48} - \\sqrt{3})', '18~16~14~12~', '18', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1635, '2014', 'Given that x > y and 3 < y, which of the following is/are true? i. y > 3 ii. x < 3 iii. x > y > 3', ' i~ i and ii~ i and iii~ i, ii and iii~', ' i and iii', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1636, '2014', 'Three quarters of a number added to two and a half of the number gives 13. Find the number', '4~5~6~7~', '4', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1637, '2014', 'Find the truth set of the equation x2 = 3(2x + 9)', ' {x : x = 3, x = 9}~ {x : x = -3, x = -9}~ {x : x = 3, x = -9}~ {x : x = -3, x = 9}~', ' {x : x = -3, x = 9}', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1638, '2014', 'The coordinates of points P and Q are (4, 3) and (2, -1) respectively. Find the shortest distance between P and Q.', ' 10√22\\sqrt{2}~ 4√55\\sqrt{5}~ 5√22\\sqrt{2}~ 2√55\\sqrt{5}~', ' 2√55\\sqrt{5}', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1639, '2014', 'Make u the subject of formula, E = m2gm2g\\frac{m}{2g}(v2 - u2)', ' u = √v2−2Egmv2−2Egm\\sqrt{v^2 - \\frac{2Eg}{m}}~ u = √v2m−2Eg4v2m−2Eg4\\sqrt{\\frac{v^2}{m} - \\frac{2Eg}{4}}~ u = √v−2Egmv−2Egm\\sqrt{v- \\frac{2Eg}{m}}~ u = √2v2Egm2v2Egm\\sqrt{\\frac{2v^2Eg}{m}}~', ' u = √v2−2Egmv2−2Egm\\sqrt{v^2 - \\frac{2Eg}{m}}', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1640, '2014', 'If x varies inversely as y and y varies directly as z, what is the relationship between x and z?', ' x ~~~~', '', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1641, '2014', 'Find the gradient of the line joining the points (2, -3) and 2, 5)', '9~1~2~ undefined~', ' undefined', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1642, '2014', 'The area of a sector a circle with diameter 12cm is 66cm2. If the sector is folded to form a cone, calculate the radius of the base of the cone ', ' 3.0cm~ 3.5cm~ 7.0cm~ 7.5cm~', ' 3.5cm', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1643, '2014', 'A chord, 7cm long, is drawn in a circle with radius 3.7cm. Calculate the distance of the chord from the centre of the circle', ' 0.7cm~ 1.2cm~ 2.0cm~ 2.5cm~', ' 1.2cm', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1644, '2014', 'Which of the following is a measure of dispersion?', ' range~ percentile~ median~ quartile~', ' range', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1645, '2014', 'A ship sails x km due east to a point E and continues x km due north to F. Find the bearing the bearing of f from the starting point.', ' 045°~ 090°~ 135°~ 225°~', ' 045°', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1646, '2014', 'A box contains 13 currency notes, all of which are either ', '4~6~8~9~', '9', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1647, '2014', 'If x : y = 3 : 2 and y : z = 5 : 4, find the value of x in the ratio x : y : z', '8~10~15~20~', '15', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1648, '2014', 'Given that cos x = 12131213\\frac{12}{13}, evaluate 1−tanxtanx1−tanxtanx\\frac{1 - \\tan x}{\\tan x}', ' 513513\\frac{5}{13}~ 5757\\frac{5}{7}~ 7575\\frac{7}{5}~ 135135\\frac{13}{5}~', ' 7575\\frac{7}{5}', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1649, '2014', 'Approximate 0.0033780 to 3 significant figures', '338~0.338~0.00338~0.003~', '0.00338', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1650, '2014', 'Simplify 82×4n+122n×1682×4n+122n×16\\frac{8^2 \\times 4^n + 1}{2^{2n} \\times 16}', '16~8~4~1~', '4', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1651, '2014', 'If 2x−3−3x−2\\frac{2}{x - 3} - \\frac{3}{x - 2} is equal to p(x−3)(x−2)\\frac{p}{(x - 3)(x - 2)}, find p', ' -x - 5~ -(x + 3)~ 5x - 13~ 5 - x~', ' 5 - x', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1652, '2014', 'Subtract 12\\frac{1}{2}(a - b - c) from the sum of 12\\frac{1}{2}(a - b + c) and 12\\frac{1}{2}(a + b - c)', ' 12\\frac{1}{2} (a + b + c)~ 12\\frac{1}{2} (a - b - c)~ 12\\frac{1}{2} (a - b + c)~ 12\\frac{1}{2} (a + b - c)~', ' 12\\frac{1}{2} (a + b + c)', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1653, '2014', 'A man\'s eye level is 1.7m above the horizontal ground and 13m from a vertical pole. If the pole is 8.3m high, calculate, correct to the nearest degree, the angle of elevation of the top of the pole from his eyes.', ' 33°~ 32°~ 27°~ 26°~', ' 27°', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1654, '2014', 'A chord subtends an angle of 120° at the centre of a circle of radius 3.5cm. Find the perimeter of the minor sector containing the chord, ', ' 1413\\frac{1}{3}cm~ 1256\\frac{5}{6}cm~ 817\\frac{1}{7}cm~ 713\\frac{1}{3}cm~', ' 1413\\frac{1}{3}cm', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1655, '2014', 'In parallelogram PQRS, QR is produced to M such that |QR| = |RM|. What fraction of the area of PQMS is the area of PRMS?', ' 14\\frac{1}{4}~ 13\\frac{1}{3}~ 23\\frac{2}{3}~ 34\\frac{3}{4}~', ' 23\\frac{2}{3}', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1656, '2014', 'In a cumulative frequency graph, the lower quartile is 18 years while the 60th percentile is 48 years. What percentage of the distribution is at most 18 years or greater than 48 years?', '0.15~0.35~0.65~0.85~', '0.65', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1657, '2014', 'If a number is selected at random from each of the sets p = {1, 2, 3} and Q = {2, 3, 5}, find the probability that the sum of the numbers is prime', ' 59\\frac{5}{9}~ 149\\frac{4}{9}~ 13\\frac{1}{3}~ 29\\frac{2}{9}~', ' 13\\frac{1}{3}', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1658, '2014', 'If log 5.957 = 0.7750, find log 3√0.0005957\\sqrt', '4.1986~2.925~1.5917~1.2853~', '2.925', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1659, '2014', 'The probability of an event P happening is 15\\frac{1}{5} and that of events are independent, what is the probability that neither of them happens?', ' 45\\frac{4}{5}~ 34\\frac{3}{4}~ 35\\frac{3}{5}~ 120\\frac{1}{20}~', ' 35\\frac{3}{5}', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1660, '2014', 'Find the number of terms in the Arithmetic Progression (A.P): 2, -9, -20,... -141.', '11~12~13~14~', '14', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1661, '2014', 'In what modulus is it true that 9 + 8 = 5?', ' mod 10~ mod 11~ mod 12~ mod 13~', ' mod 12', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1662, '2014', 'The radii of the base of two cylindrical tins, P and Q are r and 2r respectively. If the water level in p is 10cm high, would be the height of the same quantity of water in Q?', ' 2.5cm~ 5.0cm~ 7.5cm~ 20.0cm~', ' 2.5cm', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1663, '2014', 'height23456frequency24531height23456frequency24531\\begin{array}{c|c}\\text{height} ', '3.8~3~2.8~2.3~', '3.8', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1664, '2014', 'Each exterior angle of a polygon is 30o. Calculate the sum of the interior angles', ' 540o~ 720o~ 1080o~ 1800o~', ' 1800o', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1665, '2014', 'In the diagram, the value of x + y = 220o. Find the value of n.', ' 20o~ 40o~ 60o~ 80o~', ' 40o', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1666, '2014', 'In the diagram < QPT = < PTS = 90o, < PQR = 110 and < TSR = 20o. Find the size of the obtuse angle QRS.', ' 140o~ 130o~ 120o~ 110o~', ' 130o', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1667, '2014', 'In the diagram, PQR is a straight line, (m + n) = 120o and (n + r) = 100o. Find (m + r).', ' 110o~ 120o~ 140o~ 160o~', ' 140o', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1668, '2014', 'In the diagram, ST is parallel to UW, < WVT = xo, < VUT = yo, < RSV = 45o and < VTU = 20o. Find the value of x.', '20~45~65~135~', '45', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1669, '2014', 'In the diagram, ST is parallel to UW, < WVT = xo, < VUT = yo, < RSV = 45o and < VTU = 20o. Calculate the value of y.', '20~25~45~65~', '25', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1670, '2014', 'The graph given is for the relation y = 2x2 + x - 1. What are the coordinates of the point S?', ' (1, 0.2)~ (1, 0.4)~ (1, 2.0)~ (1, 40)~', ' (1, 2.0)', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1671, '2014', 'The graph given is for the relation y = 2x2 + x - 1. Find the minimum value of y.', '0~-0.65~-1.25~-2.1~', '-1.25', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1672, '2014', 'In the figures, PQ is a tangent to the circle at R and UT is parallel to PQ. if < TRQ = xo, find < URT in terms of x.', ' 2xo~ (90 - x)o~ (90 + x)o~ (180 - 2x)o~', ' (180 - 2x)o', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1673, '2014', 'Determine the value of m in the diagram.', ' 80o~ 90o~ 110o~ 150o~', ' 90o', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1674, '2014', 'In the diagram, O is the centre of the circle of the circle, PR is a tangent to the circle at Q, < SOQ = 86o. Calculate the value of < SQR.', ' 43o~ 47o~ 54o~ 86o~', ' 43o', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1675, '2015', 'If {X: 2 d- x d- 19; X integer} and 7 + x = 4 (mod 9), find the highest value of x', '2~5~15~18~', '15', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1676, '2015', 'The sum 110112, 11112 and 10m10n02. Find the value of m and n.', ' m = 0, n = 0~ m = 1, n = 0~ m = 0, n = 1~ m = 1, n = 1~', ' m = 0, n = 1', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1677, '2015', 'If 27x×31−x92x=127x×31−x92x=1\\frac{27^x \\times 3^{1 - x}}{9^{2x}} = 1, find the value of x.', '1~ 1212\\frac{1}{2}~ -1212\\frac{1}{2}~-1~', ' 1212\\frac{1}{2}', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1678, '2015', 'Find the 7th term of the sequence: 2, 5, 10, 17, 6,...', '37~48~50~63~', '50', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1679, '2015', 'Given that logx 64 = 3, evaluate x log8', '6~9~12~24~', '12', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1680, '2015', 'If 2n = y, Find 2(2+n3)(2+n3)^{(2 + \\frac{n}{3})}', ' 4y1313^\\frac{1}{3}~ 4y−3−3^-3~ 2y1313^\\frac{1}{3}~ 2y−3−3^-3~', ' 4y1313^\\frac{1}{3}', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1681, '2015', 'Factorize completely: 6ax - 12by - 9ay + 8bx', ' (2a - 3b)(4x + 3y)~ (3a + 4b)(2x - 3y)~ (3a - 4b)(2x + 3y)~ (2a + 3b)(4x -3y)~', ' (3a + 4b)(2x - 3y)', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1682, '2015', 'Find the equation whose roots are 3434\\frac{3}{4} and -4', ' 4x2 - 13x + 12 = 0~ 4x2 - 13x - 12 = 0~ 4x2 + 13x - 12 = 0~ 4x2 + 13x + 12 = 0~', ' 4x2 + 13x - 12 = 0', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1683, '2015', 'If m = 4, n = 9 and r = 16., evaluate mnmn\\frac{m}{n} - 17979\\frac{7}{9} + nrnr\\frac{n}{r}', ' 1516516\\frac{5}{16}~ 1116116\\frac{1}{16}~ 516516\\frac{5}{16}~ - 137483748\\frac{37}{48}~', ' - 137483748\\frac{37}{48}', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1684, '2015', 'Adding 42 to a given positive number gives the same result as squaring the number. Find the number', '14~13~7~6~', '7', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1685, '2015', 'Ada draws the graph of y = x2 - x - 2 and y = 2x - 1 on the same axes. Which of these equations is she solving?', ' x2 - x - 3 = 0~ x2 - 3x - 1 = 0~ x2 - 3x - 3 = 0~ x2 +  3x - 1 = 0~', ' x2 - 3x - 1 = 0', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1686, '2015', 'The volume of a cone of height 3cm is 3812\\frac{1}{2}cm3. Find the radius of its base. ', ' 3.0cm~ 3.5cm~ 4.0cm~ 4.5cm~', ' 3.5cm', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1687, '2015', 'The dimension of a rectangular tank are 2m by 7m by 11m. If its volume is equal to that of a cylindrical tank of height 4cm, calculate the base radius of the cylindrical tank. ', ' 14cm~ 7m~ 312\\frac{1}{2}m~ 134\\frac{3}{4}m~', ' 312\\frac{1}{2}m', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1688, '2015', 'In the diagram, O is the centre. If PQ//RS, what is ', ' 40o~ 50o~ 60o~ 80o~', ' 40o', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1689, '2015', 'In the diagram, PTR is a tangent to the centre O. If angles TON = 180o, Calculate the size of angle PTN.', ' 132o~ 126o~ 108o~ 102o~', ' 126o', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1690, '2015', 'In the diagram, PQ//RT, QR//SU, what is x?', ' 134o~ 132o~ 96o~ 48o~', ' 132o', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1691, '2015', 'In the diagram, O is the centre, ¯RT\\bar{RT} is a diameter, < PQT = 33o^o and ', ' 73o~ 67o~ 57o~ 37o~', ' 57o', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1692, '2015', 'In the diagram, O is the centre, ¯RT\\bar{RT} is a diameter, < PQT = 33o^o', ' 76o~ 71o~ 38o~ 33o~', ' 71o', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1693, '2015', 'x011224y0512\\begin{array}{c|c}', '1~2~3~4~', '2', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1694, '2015', 'PQRT is square. If x is the midpoint of PQ, Calculate correct to the nearest degree, LPXS', ' 53o~ 55o~ 63o~ 65o~', ' 63o', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1695, '2015', 'The angle of elevation of an aircraft from a point K on the horizontal ground 30', ' 400.00m~ 692.82m~ 923.76m~ 1,600.99m~', ' 1,600.99m', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1696, '2015', 'The population of students in a school is 810. If this is represented on a pie chart, calculate the sectoral angle for a class of 7 students', ' 32o~ 45o~ 60o~ 75o~', ' 32o', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1697, '2015', 'The scores of twenty students in a test are as follows: 44, 47, 48, 49, 50, 51, 52, 53, 53, 54, 58, 59, 60, 61, 63, 65, 67, 70, 73, 75. Find the third quartile.', '62~63~64~65~', '63', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1698, '2015', 'Scores0−45−910−14Frequency212\\begin{array}{c|c} Scores ', '5.6~6.2~6.6~7~', '7', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1699, '2015', 'The probability that kebba, Ebou and Omar will hit a target are 23\\frac{2}{3}, 34\\frac{3}{4} and 45\\frac{4}{5} respectively. Find the probability that only Kebba will hit the target.', ' 25\\frac{2}{5}~ 760\\frac{7}{60}~ 130\\frac{1}{30}~ 160\\frac{1}{60}~', ' 130\\frac{1}{30}', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1700, '2015', 'In the diagram, VW//YZ, |WX| = 6cm, |XY| = 16cm, |YZ| = 20cm and |ZX| = 12cm. Calculate |VX|.', ' 3cm~ 4cm~ 6cm~ 8cm~', ' 8cm', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1701, '2015', 'Tom will be 25 years old in n years\' time. If he is 5 years younger than Bade\'s present age.', ' (30 - n)years~ (20 - n)years~ (25 - n)years~ (30 + n)years~', ' (30 - n)years', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1702, '2015', 'If √2+√33\\frac{\\sqrt{2} + \\sqrt{3}}{3} is simplified as m + n√6\\sqrt{6}, find the value of (m + n)', ' 13\\frac{1}{3}~ 23\\frac{2}{3}~ 113\\frac{1}{3}~ 123\\frac{2}{3}~', ' 113\\frac{1}{3}', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1703, '2015', 'In the given diagram, ¯QT\\bar{QT} and ¯PR\\bar{PR} are straight lines. Find n.', ' 35o~ 40o~ 55o~ 60o~', ' 35o', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1704, '2015', 'Make K the subject of the relation T = √TK−HK−HTK−HK−H\\sqrt{\\frac{TK - H}{K - H}}', ' K = H(T2−1)T2−TH(T2−1)T2−T\\frac{H(T^2 - 1)}{T^2 - T}~ K = HT(T−1)2HT(T−1)2\\frac{HT}{(T - 1)^2}~ K = H(T2+1)TH(T2+1)T\\frac{H(T^2 + 1)}{T}~ K = H(T−1)TH(T−1)T\\frac{H(T - 1)}{T}~', ' K = H(T2−1)T2−TH(T2−1)T2−T\\frac{H(T^2 - 1)}{T^2 - T}', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1705, '2015', 'Which of the following is used to determine the mode of a grouped data?', ' Bar chart~ Frequency  polygon~ Ogive~ Histogram~', ' Histogram', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1706, '2015', 'The area of a rhombus is 110cm', '5~4~3~2.5~', '5', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1707, '2015', 'Simplify: 3x−yxy+2x+3y2xy+123x−yxy+2x+3y2xy+12\\frac{3x - y}{xy} +  \\frac{2x + 3y}{2xy} + \\frac{1}{2}', ' 4x+5y−xy2xy4x+5y−xy2xy\\frac{4x + 5y - xy}{2xy}~ 5y−4x+xy2xy5y−4x+xy2xy\\frac{5y - 4x + xy}{2xy}~ 5x+4y−xy2xy5x+4y−xy2xy\\frac{5x + 4y - xy}{2xy}~ 4x−5y+xy2xy4x−5y+xy2xy\\frac{4x - 5y + xy}{2xy}~', ' 4x−5y+xy2xy4x−5y+xy2xy\\frac{4x - 5y + xy}{2xy}', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1708, '2015', 'A farmer uses 2525\\frac{2}{5} of his land to grow cassava, 1313\\frac{1}{3} of the remaining for yam and the rest for maize. Find the part of the land used for maize', ' 215215\\frac{2}{15}~ 2525\\frac{2}{5}~ 2323\\frac{2}{3}~ 4545\\frac{4}{5}~', ' 2525\\frac{2}{5}', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1709, '2015', 'The rate of consumption of petrol by a vehicle varies directly as the square of the distance covered. If 4 litres of petrol is consumed on a distance of 15km. how far would the vehicle go on 9 litres of petrol?', ' 221212\\frac{1}{2}km~ 30km~ 331212\\frac{1}{2}km~ 45km~', ' 221212\\frac{1}{2}km', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1710, '2015', 'A trader bought 100 oranges at 5 for N40.00 and 20 for N120.00. Find the profit or loss percent', ' 20% profit~ 20% loss~ 25% profit~ 25% loss~', ' 25% loss', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1711, '2015', 'Find the value of p if 1414\\frac{1}{4}p + 3q = 10 and 2p - q = 7', '4~3~-3~-4~', '4', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1712, '2015', 'Calculate the mean deviation of 5, 3, 0, 7, 2, 1', '0~2~2.5~3~', '2', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1713, '2015', 'In the diagram, the shaded part is carpet laid in a room with dimensions 3.5m by 2.2m leaving a margin of 0.5m round it. Find area of the margin.', ' 4.7m2~ 4.9m2~ 5.7m2~ 5.9m2~', ' 4.7m2', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1714, '2015', 'Two angles of a pentagon are in the ratio 2:3. The others are 60o each. Calculate the smaller of the two angles', ' 72o~ 100o~ 120o~ 144o~', ' 144o', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1715, '2015', 'A letter is selected from the letters of the English alphabet. What is the probability that the letter selected is from the word MATHEMATICS?', ' 913913\\frac{9}{13}~ 11261126\\frac{11}{26}~ 413413\\frac{4}{13}~ 126126\\frac{1}{26}~', ' 413413\\frac{4}{13}', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1716, '2015', 'In a circle radius rcm, a chord 16√3cm√3cm\\sqrt{3}cm long is 10cmfrom the centre of the circle. Find, correct to the nearest cm, the value of r', ' 22cm~ 17cm~ 16cm~ 15cm~', ' 17cm', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1717, '2015', 'In the diagram, ¯OX¯OX\\bar{OX} bisects < YXZ and ¯OZ¯OZ\\bar{OZ} bisects < YZX. If < XYZ = 68o, calculate the value of < XOZ.', ' 68o~ 72o~ 112o~ 124o~', ' 124o', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1718, '2016', 'The curved surface area of a cylinder 5cm high is 110cm2. Find the radius of its base ', ' 2.6cm~ 3.5cm~ 3.6cm~ 7.0cm~', ' 3.5cm', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1719, '2016', 'Make S the subject of the relation p = s + sm2nrsm2nr\\frac{sm^2}{nr}', ' s = nrpnr+m2nrpnr+m2\\frac{nrp}{nr + m^2}~ s = nr + m2mrpm2mrp\\frac{m^2}{mrp}~ s = nrpmrnrpmr\\frac{nrp}{mr} + m2~ s = nrpnrnrpnr\\frac{nrp}{nr} + m2~', ' s = nrpnrnrpnr\\frac{nrp}{nr} + m2', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1720, '2016', 'Simplify:(10√3√5−√1510√3√5−√15\\frac{10\\sqrt{3}}{\\sqrt{5}} - \\sqrt{15})2', '75~15~8.66~3.87~', '15', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1721, '2016', 'The distance, d, through which a stone falls from rest varies directly as the square of the time, t, taken. If the stone falls 45cm in 3 seconds, how far will it fall in 6 seconds?', ' 90cm~ 135cm~ 180cm~ 225cm~', ' 180cm', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1722, '2016', 'Which of following is a valid conclusion from the premise. \"Nigeria footballers are good footballers\"?', ' Joseph plays football in Nigeria therefore he is a good footballer~ Joseph is a good footballer therefore he is a Nigerian footballer~ Joseph is a Nigerian footballer therefore he is a good footballer~ Joseph plays good football therefore he is a Nigerian footballer~', ' Joseph is a Nigerian footballer therefore he is a good footballer', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1723, '2016', 'On a map, 1cm represent 5km. Find the area on the map that represents 100km2.', ' 2cm2~ 4cm2~ 8cm2~ 8cm2~', ' 4cm2', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1724, '2016', 'Simplify; 3n−1×27n+181n3n−1×27n+181n\\frac{3^{n - 1} \\times 27^{n + 1}}{81^{n}}', ' 32n~9~ 3n~ 3n + 1~', '9', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0);
INSERT INTO `exam_simulation` (`id`, `examYear`, `questions`, `options`, `answer`, `explaination`, `imageDir`, `topic`, `package_id`, `subject_id`, `status`) VALUES
(1725, '2016', 'What sum of money will amount to D10,400 in 5 years at 6% simple interest?', ' D8,000.00~ D10,000.00~ D12,000.00~ D16,000.00~', ' D8,000.00', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1726, '2016', 'The roots of a quadratic equation are 4343\\frac{4}{3} and -3737\\frac{3}{7}. Find the equation', ' 21x2 - 19x - 12 = 0~ 21x2 + 37x - 12 = 0~ 21x2 - x + 12 = 0~ 21x2 + 7x - 4 = 0~', ' 21x2 - 19x - 12 = 0', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1727, '2016', 'Find the values of y for which the expression y2−9y+18y2+4y−21y2−9y+18y2+4y−21\\frac{y^2 - 9y + 18}{y^2 + 4y - 21} is undefined', ' 6, -7~ 3, -6~ 3, -7~ -3, -7~', ' 3, -7', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1728, '2016', 'Given that 2x + y = 7 and 3x - 2y = 3, by how much is 7x greater than 10?', '1~3~7~17~', '7', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1729, '2016', 'Simplify; 21−x−1x\\frac{2}{1 - x} - \\frac{1}{x}', ' x+1x(1−x)\\frac{x + 1}{x(1 - x)}~ 3x−1x(1−x)\\frac{3x - 1}{ x(1 - x)}~ 3x+1x(1−x)\\frac{3x + 1}{ x(1 - x)}~ x+1x(1−x)\\frac{x  + 1}{ x(1 - x)}~', ' 3x−1x(1−x)\\frac{3x - 1}{ x(1 - x)}', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1730, '2016', 'Factorize; (2x + 3y)2 - (x - 4y)2', ' (3x - y)(x + 7y)~ (3x + y)(2x - 7y)~ (3x + y)(x - 7y)~ (3x - y)(2x + 7y)~', ' (3x - y)(x + 7y)', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1731, '2016', 'The volume of a pyramid with height 15cm is 90cm3. If its base is a rectangle with dimension xcm by 6cm, find the value of x', '3~5~6~8~', '3', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1732, '2016', 'Calculate the gradient of the line PQ', ' 35\\frac{3}{5}~ 23\\frac{2}{3}~ 32\\frac{3}{2}~ 53\\frac{5}{3}~', ' 32\\frac{3}{2}', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1733, '2016', 'A straight line passes through the point P(1,2) and Q(5,8). Calculate the length PQ.', ' 4√114\\sqrt{11}~ 4√104\\sqrt{10}~ 2√172\\sqrt{17}~ 2√132\\sqrt{13}~', ' 2√132\\sqrt{13}', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1734, '2016', 'If cos θ\\theta = x and sin 60o = x + 0.50o < θ\\theta < 90o, find, correct to the nearest degree, the value of θ\\theta', ' 32o~ 40o~ 60o~ 69o~', ' 69o', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1735, '2016', 'Age(years)1314151617Frequency1024853Age(years)1314151617Frequency1024853\\begin{array}{c|c} Age(years) ', '50~55~60~65~', '50', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1736, '2016', 'In the diagram, ¯YW¯YW\\bar{YW} is a tangent to the circle at X, |UV| = |VX| and < VXW = 50o. Find the value of < VXY.', ' 70o~ 80o~ 105o~ 110o~', ' 80o', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1737, '2016', 'In the diagram, ¯PF¯PF\\bar{PF}, ¯QT¯QT\\bar{QT}, ¯RG¯RG\\bar{RG} intersect at S and PG||RG. If < SPQ = 113o and < RST = 220o, find < PSQ.', ' 22o~ 45o~ 67o~ 89o~', ' 45o', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1738, '2016', 'In the diagram, O is the centre of the circle, < XOZ = (10cm)o and < XWZ = mo. Calculate the value of m.', '30~36~40~72~', '30', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1739, '2016', 'Kweku walked 8m up to slope and was 3m above the ground. If he walks 12m further up the slope, how far above the ground will he be?', ' 4.5m~ 6.0m~ 7.5m~ 9.0m~', ' 7.5m', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1740, '2016', 'In the diagram, TS is a tangent to the circle at S. |PR| and < PQR = 177o. Calculate < PST.', ' 54o~ 44o~ 34o~ 27o~', ' 54o', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1741, '2016', 'In the diagram, PR||SV||WY|, TX||QY|, < PQT = 48o and < TXW = 60o. Find < TQU.', ' 120o~ 108o~ 72o~ 60o~', ' 72o', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1742, '2016', 'In the diagram, TX is perpendicular to UW, |UX| = 1cm and |TX| = |WX| = √3√3\\sqrt{3}cm. Find UTW.', ' 135o~ 105o~ 75o~ 60o~', ' 75o', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1743, '2016', 'The figure is a pie chart which represents the expenditure of a family in a year. If the total income of the family was Le 10,800,000.00, how much was spent on food?', ' Le 2,250,000.00~ Le 22,700,000.00~ Le 3,600,000.00~ Le 4,500,000.00~', ' Le 3,600,000.00', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1744, '2016', 'A fair die is thrown two times. What is the probability that the sum of the scores is at least 10?', ' 536\\frac{5}{36}~ 16\\frac{1}{6}~ 518\\frac{5}{18}~ 23\\frac{2}{3}~', ' 16\\frac{1}{6}', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1745, '2016', 'The marks of eight students in a test are: 3, 10, 4, 5, 14, 13, 16 and 7. Find the range', '16~14~13~11~', '13', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1746, '2016', 'If log2(3x - 1) = 5, find x.', '2~3.67~8.67~11~', '11', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1747, '2016', 'A sphere of radius rcm has the same volume as cylinder of radius 3cm and height 4cm. Find the value of r', ' 23\\frac{2}{3}~2~3~6~', '3', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1748, '2016', 'Express 1975 correct to 2 significant figures', '20~1900~1980~2000~', '2000', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1749, '2016', 'The diagram, MOPQ is a trapezium with QP||MO, MQ||NP, NQ||OP, |QP| = 9cm and the height of Δ\\Delta QNP = 6cm, calculate the area of the trapezium.', ' 96cm2~ 90cm2~ 81cm2~ 27cm2~', ' 81cm2', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1750, '2016', 'The perimeter of a sector of a circle of radius 21cm is 64cm. Find the angle of the sector ', ' 70o~ 60o~ 55o~ 42o~', ' 60o', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1751, '2016', 'Examine M\' ∩\\cap N from the venn diagram.', ' {f, g}~ {e}~ {e, f, g}~ {e, f, g}~', ' {f, g}', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1752, '2016', 'If 20(mod 9) is equivalent to y(mod 6), find y.', '1~2~3~4~', '2', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1753, '2016', 'Simplify (p−r)2−r22p2−4pr\\frac{(p - r)^2 - r^2}{2p^2 - 4pr}', ' 12\\frac{1}{2}~ p - 2r~ 1p−2r\\frac{1}{p - 2r}~ 2pp−2r\\frac{2p}{p - 2r}~', ' 12\\frac{1}{2}', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1754, '2016', 'In the diagram, O is the centre of the circle, < QPS = 100o, < PSQ = 60o and < QSR. Calculate < SQR.', ' 20o~ 40o~ 60o~ 80o~', ' 20o', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1755, '2016', 'A bag contains 5 red and 4 blue identical balls. Id two balls are selected at random from the bag, one after the other, with replacement, find the probability that the first is red and the second is blue', ' 2929\\frac{2}{9}~ 518518\\frac{5}{18}~ 20812081\\frac{20}{81}~ 5959\\frac{5}{9}~', ' 20812081\\frac{20}{81}', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1756, '2016', 'Find the next three terms of the sequence; 0, 1, 1, 2, 3, 5, 8...', ' 13, 19, 23~ 9, 11, 13~ 11, 15, 19~ 13, 21, 34~', ' 13, 21, 34', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1757, '2016', 'Find the lower quartile of the distribution illustrated by the cumulative frequency curve.', '17.5~19~27.5~28~', '19', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1758, '2016', 'The ratio of the exterior angle to the interior angle of a regular polygon is 1:11. How many sides has the polygon?', '30~24~18~12~', '24', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1759, '2016', 'Halima is n years old. Her brother\'s age is 5 years more than half of her age. How old is her brother?', ' n2+52n2+52\\frac{n}{2} + \\frac{5}{2}~ n2n2\\frac{n}{2} - 5~ 5 - n2n2\\frac{n}{2}~ n2n2\\frac{n}{2} + 5~', ' n2n2\\frac{n}{2} + 5', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1760, '2016', 'In the diagram MN is a chord of a circle KMN centre O and radius 10cm. If < MON = 140o, find, correct to the nearest cm, the length of the chord MN.', ' 19cm~ 18cm~ 17cm~ 12cm~', ' 19cm', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0),
(1761, '2016', 'From the diagram, which of the following is true?', ' m + n + p = 180o~ m + n = 180o~ m = p + n~ n = m + p~', ' n = m + p', 'No Official Explaination!', 'None', '', 'WAEC-PK-1', 'WAEC-MATH', 0);

-- --------------------------------------------------------

--
-- Table structure for table `exam_simulation2`
--

CREATE TABLE `exam_simulation2` (
  `id` int(11) NOT NULL,
  `examYear` varchar(200) NOT NULL,
  `questions` longtext NOT NULL,
  `parent_que` longtext NOT NULL,
  `child_que` longtext NOT NULL,
  `gen_child_que_desc` longtext NOT NULL,
  `imageDir` varchar(500) DEFAULT NULL,
  `explanation` longtext NOT NULL,
  `parent_que_expl` longtext NOT NULL,
  `child_que_exp` longtext NOT NULL,
  `gen_child_que_expl` longtext NOT NULL,
  `package_id` varchar(200) NOT NULL,
  `subject_id` varchar(200) NOT NULL,
  `status` int(1) NOT NULL DEFAULT 0
) ENGINE=InnoDB DEFAULT CHARSET=utf8mb4;

--
-- Dumping data for table `exam_simulation2`
--

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
(1, '2015', 'Question\n\n(a) Define strain.\n(b) A rubber band is stretched to twice its original length. Calculate the strain on the rubber band.\n\n', '', '(a) Define strain.\n(b) A rubber band is stretched to twice its original length. Calculate the strain on the rubber band.', '(a) Define strain.~(b) A rubber band is stretched to twice its original length. Calculate the strain on the rubber band.', NULL, '\n\nExplanation\n(a) Strain is the ratio of the original length of an object.\n(b) Strain = \\(\\frac{e}{l_o} = \\frac{2l_o - l_o}{l_o}\\) = 1\n\n', '', 'Explanation\n(a) Strain is the ratio of the original length of an object.\n(b) Strain = \\(\\frac{e}{l_o} = \\frac{2l_o - l_o}{l_o}\\) = 1\n\n', '(a) Strain is the ratio of the original length of an object.~(b) Strain = \\(\\frac{e}{l_o} = \\frac{2l_o - l_o}{l_o}\\) = 1\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(2, '2015', '\n\nQuestion\n\nState three materials used for making optical fibres.\n\n', 'State three materials used for making optical fibres.', '', '', NULL, '\n\nExplanation\nMaterials for making optical fibre are Glass, Polycarbonate, Teflon sheath, Silica, quartz, Silicon, Sapphire. Feromagnetic\n\n', 'Materials for making optical fibre are Glass, Polycarbonate, Teflon sheath, Silica, quartz, Silicon, Sapphire. Feromagnetic\n\n', '', '', 'WAEC-PK-1', 'WAEC-PHY', 1),
(3, '2015', '\n\nQuestion\n\nName three classes of magnetic materials.\n\n', 'Name three classes of magnetic materials.', '', '', NULL, '\n\nExplanation\nThree classes of magnetic materials are\n(i) Diamagnetic\n(ii) Paramagnetic\n(iii) Feromagnetic\n\n', 'Three classes of magnetic materials are\n1. Diamagnetic\n2. Paramagnetic\n3. Feromagnetic\n\n', '', '', 'WAEC-PK-1', 'WAEC-PHY', 1),
(4, '2015', '\n\nQuestion\n\n(a) What is an intrinsic semiconductor?\n(b) Distinguish between the p-type and n-type semi-conductors.\n\n', '', '(a) What is an intrinsic semiconductor?\n(b) Distinguish between the p-type and n-type semi-conductors.', '(a) What is an intrinsic semiconductor?~(b) Distinguish between the p-type and n-type semi-conductors.', NULL, '\n\nExplanation\n(a) An intrinsic semiconductor contains equal number of free electrons and holes.\n(b)\n\n\n\n\np— TYPE\n\nn - TYPE\n\n\n\ni\nThe majority carrier of electricity is holes.\n\nThe majority carrier of electricity is electron.\n\n\n\nii\nCarries a net positive charge\n\nCarries a net negative charge\n\n\n\niii\nThe impurity is trivalent\n\nThe impurity is pentavalent.\n\n\n\n\n\n', '', 'Explanation\n(a) An intrinsic semiconductor contains equal number of free electrons and holes.\n(b)\n\n\n\n\np— TYPE\n\nn - TYPE\n\n\n\ni\nThe majority carrier of electricity is holes.\n\nThe majority carrier of electricity is electron.\n\n\n\nii\nCarries a net positive charge\n\nCarries a net negative charge\n\n\n\niii\nThe impurity is trivalent\n\nThe impurity is pentavalent.\n\n\n\n\n\n', '(a) An intrinsic semiconductor contains equal number of free electrons and holes.~(b)\n\n\n\n\np— TYPE\n\nn - TYPE\n\n\n\ni\nThe majority carrier of electricity is holes.\n\nThe majority carrier of electricity is electron.\n\n\n\nii\nCarries a net positive charge\n\nCarries a net negative charge\n\n\n\niii\nThe impurity is trivalent\n\nThe impurity is pentavalent.\n\n\n\n\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(5, '2015', '\n\nQuestion\n\nA missile is projected so as to attain its maximum range. Calculate the maximum height attained if the initial velocity of projection is 200 ms\\(^{-1}\\). [g = 10ms\\(^{-2}\\)]\n\n', 'A missile is projected so as to attain its maximum range. Calculate the maximum height attained if the initial velocity of projection is 200 ms\\(^{-1}\\). [g = 10ms\\(^{-2}\\)]', '', '', NULL, '\n\nExplanation\nH = \\(\\frac{U^2sin^2 \\theta}{2g}\\)\nH = \\(\\frac{200^2(sin 45^o)^2}{20}\\)\n= 999.8m or 100m\n\n', 'H = \\(\\frac{U^2sin^2 \\theta}{2g}\\)\nH = \\(\\frac{200^2(sin 45^o)^2}{20}\\)\n= 999.8m or 100m\n\n', '', '', 'WAEC-PK-1', 'WAEC-PHY', 1),
(6, '2015', '\n\nQuestion\n\nA black body radiates maximum energy when its surface temperature T and the corresponding wavelength \\(\\lambda\\)max are related by the equation \\(\\lambda\\)max T = constant. Given the values of the constant and surface temperature as 2.9 x 10\\(^{-3}\\) mK and 57°C respectively; Calculate the frequency of the energy radiated.\n\n', 'A black body radiates maximum energy when its surface temperature T and the corresponding wavelength \\(\\lambda\\)max are related by the equation \\(\\lambda\\)max T = constant. Given the values of the constant and surface temperature as 2.9 x 10\\(^{-3}\\) mK and 57°C respectively; Calculate the frequency of the energy radiated.', '', '', NULL, '\n\nExplanation\n\\(\\lambda\\) max = \\(\\frac{2.9 \\times 10^{-3}}{330}\\) = 8.8 x 10\\(^{-6}\\)m\nv = f\\(\\lambda\\) or f = \\(\\frac{v}{\\lambda}\\)\nwhere v = velocity of light = 3.0 x10\\(^{-8}\\) m/s\n\\(\\lambda\\) = 8.8 x 10\\(^{-6}\\)\nf = \\(\\frac{v}{\\lambda}\\)\n= \\(\\frac{3.0 \\times 10^8}{8.8 \\times 10^{-6}}\\)\n= 3.4 x 10\\(^{13}\\) Hz\n\n', '\\(\\lambda\\) max = \\(\\frac{2.9 \\times 10^{-3}}{330}\\) = 8.8 x 10\\(^{-6}\\)m\nv = f\\(\\lambda\\) or f = \\(\\frac{v}{\\lambda}\\)\nwhere v = velocity of light = 3.0 x10\\(^{-8}\\) m/s\n\\(\\lambda\\) = 8.8 x 10\\(^{-6}\\)\nf = \\(\\frac{v}{\\lambda}\\)\n= \\(\\frac{3.0 \\times 10^8}{8.8 \\times 10^{-6}}\\)\n= 3.4 x 10\\(^{13}\\) Hz\n\n', '', '', 'WAEC-PK-1', 'WAEC-PHY', 1),
(7, '2015', '\n\nQuestion\n\na) What does the acronym LASER stand for?\nb) What is a laser?\n\n', 'a) What does the acronym LASER stand for?\nb) What is a laser?', '', '', NULL, '\n\nExplanation\n(a) LASER - Light Amplification by Stimulated Emission of Radiation.\n(b) A laser is a device that generates an intense beam of coherent monochromatic light.\n\n', '', 'Explanation\n(a) LASER - Light Amplification by Stimulated Emission of Radiation.\n(b) A laser is a device that generates an intense beam of coherent monochromatic light.\n\n', '(a) LASER - Light Amplification by Stimulated Emission of Radiation.~(b) A laser is a device that generates an intense beam of coherent monochromatic light.\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(8, '2015', '\n\nQuestion\n\n(a) Define uniform acceleration.\n(b) Forces act on a car in motion. List the\n(i) horizontal forces and their directions;\n(ii) vertical forces and their directions\n\n(c) A car starts from rest and accelerate uniformly for 20s to attain a speed of 25 ms\\(^{-1}\\). It maintains this speed for 30s before decelerating uniformly to rest. The total time for the journey is 60s.\n(i) Sketch a velocity-tune graph for the motion.\n(ii) Use the graph to determine the (\\(\\alpha\\)) total distance travelled by the car (\\(\\beta\\)) deceleration of the car.\n\nThe figure here illustrates force-extension graph for a stretched spiral spring. Determine the work done on the spring.\n\n', '', '(a) Define uniform acceleration.\n(b) Forces act on a car in motion. List the\n(i) horizontal forces and their directions;\n(ii) vertical forces and their directions\n\n(c) A car starts from rest and accelerate uniformly for 20s to attain a speed of 25 ms\\(^{-1}\\). It maintains this speed for 30s before decelerating uniformly to rest. The total time for the journey is 60s.\n(i) Sketch a velocity-tune graph for the motion.\n(ii) Use the graph to determine the (\\(\\alpha\\)) total distance travelled by the car (\\(\\beta\\)) deceleration of the car.\n\nThe figure here illustrates force-extension graph for a stretched spiral spring. Determine the work done on the spring.', '(a) Define uniform acceleration.~(b) Forces act on a car in motion. List the\n(i) horizontal forces and their directions;\n(ii) vertical forces and their directions\n\n(c) A car starts from rest and accelerate uniformly for 20s to attain a speed of 25 ms\\(^{-1}\\). It maintains this speed for 30s before decelerating uniformly to rest. The total time for the journey is 60s.\n(i) Sketch a velocity-tune graph for the motion.\n(ii) Use the graph to determine the (\\(\\alpha\\)) total distance travelled by the car (\\(\\beta\\)) deceleration of the car.\n\nThe figure here illustrates force-extension graph for a stretched spiral spring. Determine the work done on the spring.~(c) A car starts from rest and accelerate uniformly for 20s to attain a speed of 25 ms\\(^{-1}\\). It maintains this speed for 30s before decelerating uniformly to rest. The total time for the journey is 60s.\n(i) Sketch a velocity-tune graph for the motion.\n(ii) Use the graph to determine the (\\(\\alpha\\)) total distance travelled by the car (\\(\\beta\\)) deceleration of the car.\n\nThe figure here illustrates force-extension graph for a stretched spiral spring. Determine the work done on the spring.', NULL, '\n\nExplanation\n(a) Uniform acceleration is the constant time rate of increase in velocity.\n(b) (i) Horizontal Force;\nForward force - Thrust\nbackward force - Frictional force\n(ii) Vertical Force\nUpward force - Normal reaction.\nDownward force - Weight of the car.\n\n(c)\n\n\n\n\n(ii) (\\(\\lambda\\)) Total distance\n= Area of trapezium OABC\n= \\(\\frac{1}{2}\\)[AB + OC] AD\n= \\(\\frac{1}{2}\\)[60+30] x 25\n=1125m\n(\\(\\beta\\)) Deceleration = slope of BC\n= \\(\\frac{BE}{CE} = \\frac{25}{10}\\)\n= 25m/s\\(^2\\)\n\n(d) Work done = \\(\\frac{1}{2}\\)Fe\n= \\(\\frac{1}{2}\\) x 12 x \\(\\frac{0.5}{100}\\)J\n= 0.03J\n\n', '', 'Explanation\n(a) Uniform acceleration is the constant time rate of increase in velocity.\n(b) (i) Horizontal Force;\nForward force - Thrust\nbackward force - Frictional force\n(ii) Vertical Force\nUpward force - Normal reaction.\nDownward force - Weight of the car.\n\n(c)\n\n\n\n\n(ii) (\\(\\lambda\\)) Total distance\n= Area of trapezium OABC\n= \\(\\frac{1}{2}\\)[AB + OC] AD\n= \\(\\frac{1}{2}\\)[60+30] x 25\n=1125m\n(\\(\\beta\\)) Deceleration = slope of BC\n= \\(\\frac{BE}{CE} = \\frac{25}{10}\\)\n= 25m/s\\(^2\\)\n\n(d) Work done = \\(\\frac{1}{2}\\)Fe\n= \\(\\frac{1}{2}\\) x 12 x \\(\\frac{0.5}{100}\\)J\n= 0.03J\n\n', '(a) Uniform acceleration is the constant time rate of increase in velocity.~(b) (i) Horizontal Force;\nForward force - Thrust\nbackward force - Frictional force\n(ii) Vertical Force\nUpward force - Normal reaction.\nDownward force - Weight of the car.\n\n(c)\n\n\n\n\n(ii) (\\(\\lambda\\)) Total distance\n= Area of trapezium OABC\n= \\(\\frac{1}{2}\\)[AB + OC] AD\n= \\(\\frac{1}{2}\\)[60+30] x 25\n=1125m\n(\\(\\beta\\)) Deceleration = slope of BC\n= \\(\\frac{BE}{CE} = \\frac{25}{10}\\)\n= 25m/s\\(^2\\)\n\n(d) Work done = \\(\\frac{1}{2}\\)Fe\n= \\(\\frac{1}{2}\\) x 12 x \\(\\frac{0.5}{100}\\)J\n= 0.03J\n~(c)\n\n\n\n\n(ii) (\\(\\lambda\\)) Total distance\n= Area of trapezium OABC\n= \\(\\frac{1}{2}\\)[AB + OC] AD\n= \\(\\frac{1}{2}\\)[60+30] x 25\n=1125m\n(\\(\\beta\\)) Deceleration = slope of BC\n= \\(\\frac{BE}{CE} = \\frac{25}{10}\\)\n= 25m/s\\(^2\\)\n\n(d) Work done = \\(\\frac{1}{2}\\)Fe\n= \\(\\frac{1}{2}\\) x 12 x \\(\\frac{0.5}{100}\\)J\n= 0.03J\n~(d) Work done = \\(\\frac{1}{2}\\)Fe\n= \\(\\frac{1}{2}\\) x 12 x \\(\\frac{0.5}{100}\\)J\n= 0.03J\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(9, '2015', '\n\nQuestion\n\n(a) List two factors each that affect heat loss by:\n(i) radiation;\n(ii) convection.\n\n(b) State two factors that determine the quantity of heat in a body.\n(c) Explain the statement: The vecilic latent heat of vaporization of mercury is 2.72 x 10\\(^5\\) Jkg\\(^{-1}\\).\n(d)A jug of heat capacity 250 Jkg\\(^{-1}\\) contains water at 28°C. An electric heater of resistance 35\\(\\Omega\\) connected to a 220 V source is used to raise the temperature of the water until it boils at 100°C in 4 minutes. After. another 5 minutes, 300 g of water has evaporated. Assuming no heat is lost to the surroundings, calculate the:\n(i) mass of water in the jug before heating;\n(ii) specific latent heat of vaporization of steam. [Specific heat capacity of water = 4200 kg\\(^{-1}\\)K\\(^{-1}\\)]\n\n', '', '(a) List two factors each that affect heat loss by:\n(i) radiation;\n(ii) convection.\n\n(b) State two factors that determine the quantity of heat in a body.\n(c) Explain the statement: The vecilic latent heat of vaporization of mercury is 2.72 x 10\\(^5\\) Jkg\\(^{-1}\\).\n(d)A jug of heat capacity 250 Jkg\\(^{-1}\\) contains water at 28°C. An electric heater of resistance 35\\(\\Omega\\) connected to a 220 V source is used to raise the temperature of the water until it boils at 100°C in 4 minutes. After. another 5 minutes, 300 g of water has evaporated. Assuming no heat is lost to the surroundings, calculate the:\n(i) mass of water in the jug before heating;\n(ii) specific latent heat of vaporization of steam. [Specific heat capacity of water = 4200 kg\\(^{-1}\\)K\\(^{-1}\\)]', '(a) List two factors each that affect heat loss by:\n(i) radiation;\n(ii) convection.\n~(b) State two factors that determine the quantity of heat in a body.\n(c) Explain the statement: The vecilic latent heat of vaporization of mercury is 2.72 x 10\\(^5\\) Jkg\\(^{-1}\\).\n(d)A jug of heat capacity 250 Jkg\\(^{-1}\\) contains water at 28°C. An electric heater of resistance 35\\(\\Omega\\) connected to a 220 V source is used to raise the temperature of the water until it boils at 100°C in 4 minutes. After. another 5 minutes, 300 g of water has evaporated. Assuming no heat is lost to the surroundings, calculate the:\n(i) mass of water in the jug before heating;\n(ii) specific latent heat of vaporization of steam. [Specific heat capacity of water = 4200 kg\\(^{-1}\\)K\\(^{-1}\\)]~(c) Explain the statement: The vecilic latent heat of vaporization of mercury is 2.72 x 10\\(^5\\) Jkg\\(^{-1}\\).\n(d)A jug of heat capacity 250 Jkg\\(^{-1}\\) contains water at 28°C. An electric heater of resistance 35\\(\\Omega\\) connected to a 220 V source is used to raise the temperature of the water until it boils at 100°C in 4 minutes. After. another 5 minutes, 300 g of water has evaporated. Assuming no heat is lost to the surroundings, calculate the:\n(i) mass of water in the jug before heating;\n(ii) specific latent heat of vaporization of steam. [Specific heat capacity of water = 4200 kg\\(^{-1}\\)K\\(^{-1}\\)]~(d)A jug of heat capacity 250 Jkg\\(^{-1}\\) contains water at 28°C. An electric heater of resistance 35\\(\\Omega\\) connected to a 220 V source is used to raise the temperature of the water until it boils at 100°C in 4 minutes. After. another 5 minutes, 300 g of water has evaporated. Assuming no heat is lost to the surroundings, calculate the:\n(i) mass of water in the jug before heating;\n(ii) specific latent heat of vaporization of steam. [Specific heat capacity of water = 4200 kg\\(^{-1}\\)K\\(^{-1}\\)]', NULL, '\n\nExplanation\n(a) (i) Radiation:\nSurface area, Temperature.\n(ii) Convection:\nNature, Density/Viscosity of the fluid, Thermal conductivity of fluid, Specific heat capacity of fluid and Exposed surface area.\n(b) Factors that determine the quantity of heat in a body: Heat/thermal capacity and. Temperature\n(c) It means that 2.72 x 105J of heat energy is required/ needed to change I kg of mercury at its boiling point to vapour (without temperature change)\n(d) (i) heat supplied by heater = ivt = \\(\\frac{v^2}{R}\\)t\nheat gained by water = M\\(_w\\)C\\(_w\\)(\\(\\theta_2\\) - \\(\\theta_1\\))\nwhere M\\(_w\\) = mass of water\nC\\(_w\\) = s.h.c of water\n\\(\\theta_2\\) = final temp. of water\n\\(\\theta_1\\) = initial temp. of water\nHeat gained by jug = C\\(_j\\) (\\(\\theta_2\\) - \\(\\theta_1\\))\n\\(\\frac{v^2t}{R}\\) = M\\(_w\\)C\\(_w\\) (\\(\\theta_2\\) - \\(\\theta_1\\))\n= \\(\\frac{220^2 \\times 4 \\times 60}{35}\\)\n= M\\(_w\\) x 4200[100 - 28] + 250[100 - 28]\nM\\(_w\\) = 1.038kg\n\n', '', 'Explanation\n(a) (i) Radiation:\nSurface area, Temperature.\n(ii) Convection:\nNature, Density/Viscosity of the fluid, Thermal conductivity of fluid, Specific heat capacity of fluid and Exposed surface area.\n(b) Factors that determine the quantity of heat in a body: Heat/thermal capacity and. Temperature\n(c) It means that 2.72 x 105J of heat energy is required/ needed to change I kg of mercury at its boiling point to vapour (without temperature change)\n(d) (i) heat supplied by heater = ivt = \\(\\frac{v^2}{R}\\)t\nheat gained by water = M\\(_w\\)C\\(_w\\)(\\(\\theta_2\\) - \\(\\theta_1\\))\nwhere M\\(_w\\) = mass of water\nC\\(_w\\) = s.h.c of water\n\\(\\theta_2\\) = final temp. of water\n\\(\\theta_1\\) = initial temp. of water\nHeat gained by jug = C\\(_j\\) (\\(\\theta_2\\) - \\(\\theta_1\\))\n\\(\\frac{v^2t}{R}\\) = M\\(_w\\)C\\(_w\\) (\\(\\theta_2\\) - \\(\\theta_1\\))\n= \\(\\frac{220^2 \\times 4 \\times 60}{35}\\)\n= M\\(_w\\) x 4200[100 - 28] + 250[100 - 28]\nM\\(_w\\) = 1.038kg\n\n', '(a) (i) Radiation:\nSurface area, Temperature.\n(ii) Convection:\nNature, Density/Viscosity of the fluid, Thermal conductivity of fluid, Specific heat capacity of fluid and Exposed surface area.~(b) Factors that determine the quantity of heat in a body: Heat/thermal capacity and. Temperature\n(c) It means that 2.72 x 105J of heat energy is required/ needed to change I kg of mercury at its boiling point to vapour (without temperature change)\n(d) (i) heat supplied by heater = ivt = \\(\\frac{v^2}{R}\\)t\nheat gained by water = M\\(_w\\)C\\(_w\\)(\\(\\theta_2\\) - \\(\\theta_1\\))\nwhere M\\(_w\\) = mass of water\nC\\(_w\\) = s.h.c of water\n\\(\\theta_2\\) = final temp. of water\n\\(\\theta_1\\) = initial temp. of water\nHeat gained by jug = C\\(_j\\) (\\(\\theta_2\\) - \\(\\theta_1\\))\n\\(\\frac{v^2t}{R}\\) = M\\(_w\\)C\\(_w\\) (\\(\\theta_2\\) - \\(\\theta_1\\))\n= \\(\\frac{220^2 \\times 4 \\times 60}{35}\\)\n= M\\(_w\\) x 4200[100 - 28] + 250[100 - 28]\nM\\(_w\\) = 1.038kg\n~(c) It means that 2.72 x 105J of heat energy is required/ needed to change I kg of mercury at its boiling point to vapour (without temperature change)\n(d) (i) heat supplied by heater = ivt = \\(\\frac{v^2}{R}\\)t\nheat gained by water = M\\(_w\\)C\\(_w\\)(\\(\\theta_2\\) - \\(\\theta_1\\))\nwhere M\\(_w\\) = mass of water\nC\\(_w\\) = s.h.c of water\n\\(\\theta_2\\) = final temp. of water\n\\(\\theta_1\\) = initial temp. of water\nHeat gained by jug = C\\(_j\\) (\\(\\theta_2\\) - \\(\\theta_1\\))\n\\(\\frac{v^2t}{R}\\) = M\\(_w\\)C\\(_w\\) (\\(\\theta_2\\) - \\(\\theta_1\\))\n= \\(\\frac{220^2 \\times 4 \\times 60}{35}\\)\n= M\\(_w\\) x 4200[100 - 28] + 250[100 - 28]\nM\\(_w\\) = 1.038kg\n~(d) (i) heat supplied by heater = ivt = \\(\\frac{v^2}{R}\\)t\nheat gained by water = M\\(_w\\)C\\(_w\\)(\\(\\theta_2\\) - \\(\\theta_1\\))\nwhere M\\(_w\\) = mass of water\nC\\(_w\\) = s.h.c of water\n\\(\\theta_2\\) = final temp. of water\n\\(\\theta_1\\) = initial temp. of water\nHeat gained by jug = C\\(_j\\) (\\(\\theta_2\\) - \\(\\theta_1\\))\n\\(\\frac{v^2t}{R}\\) = M\\(_w\\)C\\(_w\\) (\\(\\theta_2\\) - \\(\\theta_1\\))\n= \\(\\frac{220^2 \\times 4 \\times 60}{35}\\)\n= M\\(_w\\) x 4200[100 - 28] + 250[100 - 28]\nM\\(_w\\) = 1.038kg\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(10, '2015', '\n\nQuestion\n\n(a) Define dffraction.\n\n(b)(i) Explain critical angle. The diagram here illustrates a ray of light passing through a rectangular transparent plastic block \\(\\alpha\\) Determine the value of the critical angle. \\(\\beta\\) Calculate the refractive index of the block.\n\n(c) A pipe closed at one end has fundamental frequency of 200Hz. The frequency of the first overtone of the closed pipe is equal to the frequency of the first overtone of an open pipe. Calculate the:\n(i) fundamental frequency of the open pipe;\n(ii) length of the closed pipe;\n(iii) length of the open pipe. [Speed of sound in air = 330 ms\\(^{-1}\\)]\n\n', '', '(a) Define dffraction.\n\n(b)(i) Explain critical angle. The diagram here illustrates a ray of light passing through a rectangular transparent plastic block \\(\\alpha\\) Determine the value of the critical angle. \\(\\beta\\) Calculate the refractive index of the block.\n\n(c) A pipe closed at one end has fundamental frequency of 200Hz. The frequency of the first overtone of the closed pipe is equal to the frequency of the first overtone of an open pipe. Calculate the:\n(i) fundamental frequency of the open pipe;\n(ii) length of the closed pipe;\n(iii) length of the open pipe. [Speed of sound in air = 330 ms\\(^{-1}\\)]', '(a) Define dffraction.\n~(b)(i) Explain critical angle. The diagram here illustrates a ray of light passing through a rectangular transparent plastic block \\(\\alpha\\) Determine the value of the critical angle. \\(\\beta\\) Calculate the refractive index of the block.\n\n(c) A pipe closed at one end has fundamental frequency of 200Hz. The frequency of the first overtone of the closed pipe is equal to the frequency of the first overtone of an open pipe. Calculate the:\n(i) fundamental frequency of the open pipe;\n(ii) length of the closed pipe;\n(iii) length of the open pipe. [Speed of sound in air = 330 ms\\(^{-1}\\)]~(c) A pipe closed at one end has fundamental frequency of 200Hz. The frequency of the first overtone of the closed pipe is equal to the frequency of the first overtone of an open pipe. Calculate the:\n(i) fundamental frequency of the open pipe;\n(ii) length of the closed pipe;\n(iii) length of the open pipe. [Speed of sound in air = 330 ms\\(^{-1}\\)]', NULL, '\n\nExplanation\n(a) Diffraction is the bending or spreading of a wave round an obstacle or as it passes through an opening.\n(b) (i) Critical angle is the angle of incidence in the denser medium when an angle of refraction in the lens dense medium is 90°.\n(ii) (\\(\\alpha\\))Critical angle = 90° -.44° = 46°\n(\\(\\beta\\))n = \\(\\frac{1}{sin c} = \\frac{1}{\\sin 46^o}\\) = 1.39\n\n(c) (i) Closed Pipe\n1st Overtone = 3f\\(_o\\) = 3 x 200 = 600Hz\nOpen Pipe\nAll harmonics are possible overtone = 2f\\(_o\\)\n2f\\(_o\\) = 600\nf\\(_o\\) = 300Hz\n\n(ii) Closed Pipe\nf\\(_o\\) = \\(\\frac{v}{4l}\\)\nl = \\(\\frac{v}{4f_o}\\)\n\\(\\frac{330}{4 \\times 200}\\)\n= 0.412m\nOpen Pipe\n\\(f_1\\) = \\(\\frac{v}{l}\\)\n600 = \\(\\frac{300}{l}\\)\nl = 0.55m\n\nor\n\\(f_1\\) = \\(\\frac{v}{2l}\\)\n300 = \\(\\frac{300}{2 \\times l}\\)\nl = \\(\\frac{330}{600}\\)\n= 0.55m\n\n\n', '', 'Explanation\n(a) Diffraction is the bending or spreading of a wave round an obstacle or as it passes through an opening.\n(b) (i) Critical angle is the angle of incidence in the denser medium when an angle of refraction in the lens dense medium is 90°.\n(ii) (\\(\\alpha\\))Critical angle = 90° -.44° = 46°\n(\\(\\beta\\))n = \\(\\frac{1}{sin c} = \\frac{1}{\\sin 46^o}\\) = 1.39\n\n(c) (i) Closed Pipe\n1st Overtone = 3f\\(_o\\) = 3 x 200 = 600Hz\nOpen Pipe\nAll harmonics are possible overtone = 2f\\(_o\\)\n2f\\(_o\\) = 600\nf\\(_o\\) = 300Hz\n\n(ii) Closed Pipe\nf\\(_o\\) = \\(\\frac{v}{4l}\\)\nl = \\(\\frac{v}{4f_o}\\)\n\\(\\frac{330}{4 \\times 200}\\)\n= 0.412m\nOpen Pipe\n\\(f_1\\) = \\(\\frac{v}{l}\\)\n600 = \\(\\frac{300}{l}\\)\nl = 0.55m\n\nor\n\\(f_1\\) = \\(\\frac{v}{2l}\\)\n300 = \\(\\frac{300}{2 \\times l}\\)\nl = \\(\\frac{330}{600}\\)\n= 0.55m\n\n\n', '(a) Diffraction is the bending or spreading of a wave round an obstacle or as it passes through an opening.~(b) (i) Critical angle is the angle of incidence in the denser medium when an angle of refraction in the lens dense medium is 90°.\n(ii) (\\(\\alpha\\))Critical angle = 90° -.44° = 46°\n(\\(\\beta\\))n = \\(\\frac{1}{sin c} = \\frac{1}{\\sin 46^o}\\) = 1.39\n\n(c) (i) Closed Pipe\n1st Overtone = 3f\\(_o\\) = 3 x 200 = 600Hz\nOpen Pipe\nAll harmonics are possible overtone = 2f\\(_o\\)\n2f\\(_o\\) = 600\nf\\(_o\\) = 300Hz\n\n(ii) Closed Pipe\nf\\(_o\\) = \\(\\frac{v}{4l}\\)\nl = \\(\\frac{v}{4f_o}\\)\n\\(\\frac{330}{4 \\times 200}\\)\n= 0.412m\nOpen Pipe\n\\(f_1\\) = \\(\\frac{v}{l}\\)\n600 = \\(\\frac{300}{l}\\)\nl = 0.55m\n\nor\n\\(f_1\\) = \\(\\frac{v}{2l}\\)\n300 = \\(\\frac{300}{2 \\times l}\\)\nl = \\(\\frac{330}{600}\\)\n= 0.55m\n\n~(c) (i) Closed Pipe\n1st Overtone = 3f\\(_o\\) = 3 x 200 = 600Hz\nOpen Pipe\nAll harmonics are possible overtone = 2f\\(_o\\)\n2f\\(_o\\) = 600\nf\\(_o\\) = 300Hz\n\n(ii) Closed Pipe\nf\\(_o\\) = \\(\\frac{v}{4l}\\)\nl = \\(\\frac{v}{4f_o}\\)\n\\(\\frac{330}{4 \\times 200}\\)\n= 0.412m\nOpen Pipe\n\\(f_1\\) = \\(\\frac{v}{l}\\)\n600 = \\(\\frac{300}{l}\\)\nl = 0.55m\n\nor\n\\(f_1\\) = \\(\\frac{v}{2l}\\)\n300 = \\(\\frac{300}{2 \\times l}\\)\nl = \\(\\frac{330}{600}\\)\n= 0.55m\n\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(11, '2015', '\n\nQuestion\n\n(a) Define:\n(i) reactance;\n(ii) impedance in an a.c.\nThe diagram here illustrates an a.c. generator. When the coil is rotated, an e.m.f is induced in the coil.\n\n(i) Explain why an e.m.f. is induced.\n(ii) State the purpose of the slip-rings.\n(iii) Name and state the law used to determine the direction of the induced current.\n(iv) State two ways to increase the induced e.m.f.\n\n(c) A lamp is rated 12 V. 6 W. Calculate the amount of energy transformed by the lamp in 5 minutes.\n\n', '', '(a) Define:\n(i) reactance;\n(ii) impedance in an a.c.\nThe diagram here illustrates an a.c. generator. When the coil is rotated, an e.m.f is induced in the coil.\n\n(i) Explain why an e.m.f. is induced.\n(ii) State the purpose of the slip-rings.\n(iii) Name and state the law used to determine the direction of the induced current.\n(iv) State two ways to increase the induced e.m.f.\n\n(c) A lamp is rated 12 V. 6 W. Calculate the amount of energy transformed by the lamp in 5 minutes.', '~(c) A lamp is rated 12 V. 6 W. Calculate the amount of energy transformed by the lamp in 5 minutes.', NULL, '\n\nExplanation\n(a)(i) Reactance is the opposition to the flow of a.c offered by a capacitor or an inductor or both.\n(ii) Impedance is the overall opposition of a mixed circuit containing a resistor, an inductor and/or a capacitor.\n\n(b) (i) Emf is induced because the coil rotates at a steady speed about a fixed axis in the magnetic field of the elctromagnet, it cuts through the magnetic lines of force.\n(ii) The direction of current is reversed in each half of the rotation of the coil.\n(iii) The law used to determine the direction of the induced current is Lenz\'s law. It states that the direction of the induced current is such as to oppose the change producing it.\n(iv) The induced e.m.f can be increased by any two of the following: increased speed of rotation, efficient core design, increased number of turns in the coil and increased area of coil.\n\n(c) E = Power x Time\n= 6 x 5 x 60\n=1800J\n\n', '', 'Explanation\n(a)(i) Reactance is the opposition to the flow of a.c offered by a capacitor or an inductor or both.\n(ii) Impedance is the overall opposition of a mixed circuit containing a resistor, an inductor and/or a capacitor.\n\n(b) (i) Emf is induced because the coil rotates at a steady speed about a fixed axis in the magnetic field of the elctromagnet, it cuts through the magnetic lines of force.\n(ii) The direction of current is reversed in each half of the rotation of the coil.\n(iii) The law used to determine the direction of the induced current is Lenz\'s law. It states that the direction of the induced current is such as to oppose the change producing it.\n(iv) The induced e.m.f can be increased by any two of the following: increased speed of rotation, efficient core design, increased number of turns in the coil and increased area of coil.\n\n(c) E = Power x Time\n= 6 x 5 x 60\n=1800J\n\n', '(a)(i) Reactance is the opposition to the flow of a.c offered by a capacitor or an inductor or both.\n(ii) Impedance is the overall opposition of a mixed circuit containing a resistor, an inductor and/or a capacitor.\n~(b) (i) Emf is induced because the coil rotates at a steady speed about a fixed axis in the magnetic field of the elctromagnet, it cuts through the magnetic lines of force.\n(ii) The direction of current is reversed in each half of the rotation of the coil.\n(iii) The law used to determine the direction of the induced current is Lenz\'s law. It states that the direction of the induced current is such as to oppose the change producing it.\n(iv) The induced e.m.f can be increased by any two of the following: increased speed of rotation, efficient core design, increased number of turns in the coil and increased area of coil.\n\n(c) E = Power x Time\n= 6 x 5 x 60\n=1800J\n~(c) E = Power x Time\n= 6 x 5 x 60\n=1800J\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(12, '2015', '\n\nQuestion\n\n(a) Define binding energy in an atom.\n(b) List three evidence to support the claim that X-rays are electromagnetic waves.\n(c) List three peaceful uses of nuclear energy.\n(d) Light of wavelength 4.5 x 10\\(^{-7}\\) in is incident on a metal resulting in the emission of photo electrons. If the work function of the metal is 3.0 x 10\\(^{-9}\\) J, calculate the:\n(i) frequency of the incident light;\n(ii) energy of the incident light;\n(iii) energy of the photoelectrons. [Speed of light = 3.0 x 10\\(^8\\) ms\\(^{-1}\\), h = 6.6 x 10\\(^{-34}\\) Js]\n\n', '', '(a) Define binding energy in an atom.\n(b) List three evidence to support the claim that X-rays are electromagnetic waves.\n(c) List three peaceful uses of nuclear energy.\n(d) Light of wavelength 4.5 x 10\\(^{-7}\\) in is incident on a metal resulting in the emission of photo electrons. If the work function of the metal is 3.0 x 10\\(^{-9}\\) J, calculate the:\n(i) frequency of the incident light;\n(ii) energy of the incident light;\n(iii) energy of the photoelectrons. [Speed of light = 3.0 x 10\\(^8\\) ms\\(^{-1}\\), h = 6.6 x 10\\(^{-34}\\) Js]', '(a) Define binding energy in an atom.~(b) List three evidence to support the claim that X-rays are electromagnetic waves.\n(c) List three peaceful uses of nuclear energy.\n(d) Light of wavelength 4.5 x 10\\(^{-7}\\) in is incident on a metal resulting in the emission of photo electrons. If the work function of the metal is 3.0 x 10\\(^{-9}\\) J, calculate the:\n(i) frequency of the incident light;\n(ii) energy of the incident light;\n(iii) energy of the photoelectrons. [Speed of light = 3.0 x 10\\(^8\\) ms\\(^{-1}\\), h = 6.6 x 10\\(^{-34}\\) Js]~(c) List three peaceful uses of nuclear energy.\n(d) Light of wavelength 4.5 x 10\\(^{-7}\\) in is incident on a metal resulting in the emission of photo electrons. If the work function of the metal is 3.0 x 10\\(^{-9}\\) J, calculate the:\n(i) frequency of the incident light;\n(ii) energy of the incident light;\n(iii) energy of the photoelectrons. [Speed of light = 3.0 x 10\\(^8\\) ms\\(^{-1}\\), h = 6.6 x 10\\(^{-34}\\) Js]~(d) Light of wavelength 4.5 x 10\\(^{-7}\\) in is incident on a metal resulting in the emission of photo electrons. If the work function of the metal is 3.0 x 10\\(^{-9}\\) J, calculate the:\n(i) frequency of the incident light;\n(ii) energy of the incident light;\n(iii) energy of the photoelectrons. [Speed of light = 3.0 x 10\\(^8\\) ms\\(^{-1}\\), h = 6.6 x 10\\(^{-34}\\) Js]', NULL, '\n\nExplanation\n(a) Binding energy is the minimum energy required to seperate the nucleus of an atom.\n\n(b) Three evidences in support of the claim that x -rays are electromagnetic waves are:\n(i) They can be polarized.\n(ii) They can travel through a vacuum.\n(iii) They travel with the speed of light.\n(iv) They are not affected by electric field.\n(v) They are not affected by magnetic field.\n\n(c) Three peaceful uses of nuclear energy are:\n(i) It is used in the treatment of tumour.\n(ii) It is used to power submarines.\n(iii) It is used to generate electricity.\n\n(d) (1) c = f\\(\\lambda\\) or f = \\(\\frac{c}{\\lambda}\\)\nf = \\(\\frac{3.0 \\times 10^8}{4.5 \\times 10^{-7}}\\)\nf = 6.67 x10\\(^{14}\\)Hz\n\n(ii) E = hf\n= 6.67 x10\\(^{-34}\\) x 6.67 x10\\(^{14}\\)J\n= 4.4 x10\\(^{-19}\\)J\n\n(iii) K.E = E- W\n= 4.4 x 10\\(^{-19}\\)\n=1.4 x 10\\(^{-19}\\) - 3.0 x 10\\(^{-19}\\)\n=1.4 x 10\\(^{-19}\\)J\n\n', '', 'Explanation\n(a) Binding energy is the minimum energy required to seperate the nucleus of an atom.\n\n(b) Three evidences in support of the claim that x -rays are electromagnetic waves are:\n(i) They can be polarized.\n(ii) They can travel through a vacuum.\n(iii) They travel with the speed of light.\n(iv) They are not affected by electric field.\n(v) They are not affected by magnetic field.\n\n(c) Three peaceful uses of nuclear energy are:\n(i) It is used in the treatment of tumour.\n(ii) It is used to power submarines.\n(iii) It is used to generate electricity.\n\n(d) (1) c = f\\(\\lambda\\) or f = \\(\\frac{c}{\\lambda}\\)\nf = \\(\\frac{3.0 \\times 10^8}{4.5 \\times 10^{-7}}\\)\nf = 6.67 x10\\(^{14}\\)Hz\n\n(ii) E = hf\n= 6.67 x10\\(^{-34}\\) x 6.67 x10\\(^{14}\\)J\n= 4.4 x10\\(^{-19}\\)J\n\n(iii) K.E = E- W\n= 4.4 x 10\\(^{-19}\\)\n=1.4 x 10\\(^{-19}\\) - 3.0 x 10\\(^{-19}\\)\n=1.4 x 10\\(^{-19}\\)J\n\n', '(a) Binding energy is the minimum energy required to seperate the nucleus of an atom.\n~(b) Three evidences in support of the claim that x -rays are electromagnetic waves are:\n(i) They can be polarized.\n(ii) They can travel through a vacuum.\n(iii) They travel with the speed of light.\n(iv) They are not affected by electric field.\n(v) They are not affected by magnetic field.\n\n(c) Three peaceful uses of nuclear energy are:\n(i) It is used in the treatment of tumour.\n(ii) It is used to power submarines.\n(iii) It is used to generate electricity.\n\n(d) (1) c = f\\(\\lambda\\) or f = \\(\\frac{c}{\\lambda}\\)\nf = \\(\\frac{3.0 \\times 10^8}{4.5 \\times 10^{-7}}\\)\nf = 6.67 x10\\(^{14}\\)Hz\n\n(ii) E = hf\n= 6.67 x10\\(^{-34}\\) x 6.67 x10\\(^{14}\\)J\n= 4.4 x10\\(^{-19}\\)J\n\n(iii) K.E = E- W\n= 4.4 x 10\\(^{-19}\\)\n=1.4 x 10\\(^{-19}\\) - 3.0 x 10\\(^{-19}\\)\n=1.4 x 10\\(^{-19}\\)J\n~(c) Three peaceful uses of nuclear energy are:\n(i) It is used in the treatment of tumour.\n(ii) It is used to power submarines.\n(iii) It is used to generate electricity.\n\n(d) (1) c = f\\(\\lambda\\) or f = \\(\\frac{c}{\\lambda}\\)\nf = \\(\\frac{3.0 \\times 10^8}{4.5 \\times 10^{-7}}\\)\nf = 6.67 x10\\(^{14}\\)Hz\n\n(ii) E = hf\n= 6.67 x10\\(^{-34}\\) x 6.67 x10\\(^{14}\\)J\n= 4.4 x10\\(^{-19}\\)J\n\n(iii) K.E = E- W\n= 4.4 x 10\\(^{-19}\\)\n=1.4 x 10\\(^{-19}\\) - 3.0 x 10\\(^{-19}\\)\n=1.4 x 10\\(^{-19}\\)J\n~(d) (1) c = f\\(\\lambda\\) or f = \\(\\frac{c}{\\lambda}\\)\nf = \\(\\frac{3.0 \\times 10^8}{4.5 \\times 10^{-7}}\\)\nf = 6.67 x10\\(^{14}\\)Hz\n\n(ii) E = hf\n= 6.67 x10\\(^{-34}\\) x 6.67 x10\\(^{14}\\)J\n= 4.4 x10\\(^{-19}\\)J\n\n(iii) K.E = E- W\n= 4.4 x 10\\(^{-19}\\)\n=1.4 x 10\\(^{-19}\\) - 3.0 x 10\\(^{-19}\\)\n=1.4 x 10\\(^{-19}\\)J\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(13, '2015', '\n\nQuestion\n\nA particle is dropped from a vertical height h and falls freely for a time t. With the aid of a sketch, explain how h varies with \\(t^2\\)\n\n', 'A particle is dropped from a vertical height h and falls freely for a time t. With the aid of a sketch, explain how h varies with \\(t^2\\)', '', '', NULL, '\n\nExplanation\nThe sketch shows that the height decreases heavily with time (t\\(^2\\))\n\n\n\n', 'The sketch shows that the height decreases heavily with time (t\\(^2\\))\n\n\n\n', '', '', 'WAEC-PK-1', 'WAEC-PHY', 1),
(14, '2015', '\n\nQuestion\n\nA particle is projected horizontally at 15ms\\(^{-1}\\) from a height of 20m. Calculate the horizontal distance covered by the particle just before hitting the ground. [g= 10ms\\(^{-2}\\)]\n\n', 'A particle is projected horizontally at 15ms\\(^{-1}\\) from a height of 20m. Calculate the horizontal distance covered by the particle just before hitting the ground. [g= 10ms\\(^{-2}\\)]', '', '', NULL, '\n\nExplanation\nLet R represent the horizontal distance covered at time t.\nUsing h = ut + \\(\\frac{1}{2}gt^2\\)\nbut U\\(_v\\) = 0\nh = \\(\\frac{1}{2} gt^{2}\\)\n20 = \\(\\frac{1}{2} \\times 10 \\times t^2\\)\n20 = 5t\\(^2\\)\nt\\(^2\\) = \\(\\frac{20}{5}\\) = 4\nt = 2s\nRange(R) = U\\(_x\\) x t\n= 15 x 2\n= 30m\nor\nR = u x \\(\\sqrt{\\frac{2h}{g}}\\)\n= 15 x \\(\\sqrt{\\frac{2 \\times 20}{10}}\\)\n= 15 x \\(\\sqrt{4}\\)\n= 15 x 2\n= 30m\n\n', 'Let R represent the horizontal distance covered at time t.\nUsing h = ut + \\(\\frac{1}{2}gt^2\\)\nbut U\\(_v\\) = 0\nh = \\(\\frac{1}{2} gt^{2}\\)\n20 = \\(\\frac{1}{2} \\times 10 \\times t^2\\)\n20 = 5t\\(^2\\)\nt\\(^2\\) = \\(\\frac{20}{5}\\) = 4\nt = 2s\nRange(R) = U\\(_x\\) x t\n= 15 x 2\n= 30m\nor\nR = u x \\(\\sqrt{\\frac{2h}{g}}\\)\n= 15 x \\(\\sqrt{\\frac{2 \\times 20}{10}}\\)\n= 15 x \\(\\sqrt{4}\\)\n= 15 x 2\n= 30m\n\n', '', '', 'WAEC-PK-1', 'WAEC-PHY', 1),
(15, '2015', '\n\nQuestion\n\nList three phenomena which can be explained by the molecular theory of matter.\n\n', 'List three phenomena which can be explained by the molecular theory of matter.', '', '', NULL, '\n\nExplanation\n(i) Osmosis\n(ii) Brownian motion\n(iii) Diffusion\n(iv) Capilarity\n(v) Viscosity\n\n', '', 'Explanation\n(i) Osmosis\n(ii) Brownian motion\n(iii) Diffusion\n(iv) Capilarity\n(v) Viscosity\n\n', '', 'WAEC-PK-1', 'WAEC-PHY', 1),
(16, '2015', '\n\nQuestion\n\nA spiral spring has a length of 14cm when a force of 4N is hung on it. A force of 6N extends the spring by 4cm. Calculate the unstretched length of the spring\n\n', 'A spiral spring has a length of 14cm when a force of 4N is hung on it. A force of 6N extends the spring by 4cm. Calculate the unstretched length of the spring', '', '', NULL, '\n\nExplanation\nUsing hookes law\n\\(\\frac{F_1}{e_1} = \\frac{F_2}{e_2}\\)\nbut e\\(_1\\) = l\\(_1\\) = l\\(_o\\)\n\\(\\frac{F_1}{l_1 - l_o}\\) = \\(\\frac{F_2}{e_2}\\)\n\\(\\frac{4}{14 - \\_o}\\) = \\(\\frac{6}{4}\\);  16 = 6(14 - \\(_o\\))\n14 = 84 - 6\\(l_o\\); 16 - 84 = -6\\(l_o\\)\n\\(l_o = \\frac{-68}{-6}\\) = 11.33cm\n\n\n', 'Using hookes law\n\\(\\frac{F_1}{e_1} = \\frac{F_2}{e_2}\\)\nbut e\\(_1\\) = l\\(_1\\) = l\\(_o\\)\n\\(\\frac{F_1}{l_1 - l_o}\\) = \\(\\frac{F_2}{e_2}\\)\n\\(\\frac{4}{14 - \\_o}\\) = \\(\\frac{6}{4}\\);  16 = 6(14 - \\(_o\\))\n14 = 84 - 6\\(l_o\\); 16 - 84 = -6\\(l_o\\)\n\\(l_o = \\frac{-68}{-6}\\) = 11.33cm\n\n\n', '', '', 'WAEC-PK-1', 'WAEC-PHY', 1),
(17, '2015', '\n\nQuestion\n\n(a) State two factors on which surface tension depends.\n(b) How can mosquito larvae be made to sink in stagnant water?\n\n', '', '(a) State two factors on which surface tension depends.\n(b) How can mosquito larvae be made to sink in stagnant water?', '(a) State two factors on which surface tension depends.~(b) How can mosquito larvae be made to sink in stagnant water?', NULL, '\n\nExplanation\n(a)(i) Purity of liquid\n(ii) Temperature of liquid\n(iii) Nature / type of liquid.\n\n(b) Mosquito larvae can be made to sink by the addition of any of;\n(i) Detergent / Soap\n(ii) Alcohol\n(iii) Heat\n(iv) Camphor .\n\n', '', 'Explanation\n(a)(i) Purity of liquid\n(ii) Temperature of liquid\n(iii) Nature / type of liquid.\n\n(b) Mosquito larvae can be made to sink by the addition of any of;\n(i) Detergent / Soap\n(ii) Alcohol\n(iii) Heat\n(iv) Camphor .\n\n', '(a)(i) Purity of liquid\n(ii) Temperature of liquid\n(iii) Nature / type of liquid.\n~(b) Mosquito larvae can be made to sink by the addition of any of;\n(i) Detergent / Soap\n(ii) Alcohol\n(iii) Heat\n(iv) Camphor .\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(18, '2015', '\n\nQuestion\n\nList three advantages of fluorescent tubes over filament bulbs.\n\n', 'List three advantages of fluorescent tubes over filament bulbs.', '', '', NULL, '\n\nExplanation\nAdvantages of flourescent tubes over filament bulbs are;\n(i) They produce less heat\n(ii) They last longer\n(iii) They are brighter\n(iv) They consume less electrical energy\n\n', '', 'Explanation\nAdvantages of flourescent tubes over filament bulbs are;\n(i) They produce less heat\n(ii) They last longer\n(iii) They are brighter\n(iv) They consume less electrical energy\n\n', '', 'WAEC-PK-1', 'WAEC-PHY', 1),
(19, '2015', '\n\nQuestion\n\nList three advantages of p - n junetion diode over diode valve.\n\n', 'List three advantages of p - n junetion diode over diode valve.', '', '', NULL, '\n\nExplanation\nThe advantages of P – n; junction diode over diode valves are:\n(i) It is not bulky\n(ii) It does not need time to warm up\n(iii) It is cheaper to manufacture in large quantities.\n(iv) It needs low voltage to operate.\n\n', '', 'Explanation\nThe advantages of P – n; junction diode over diode valves are:\n(i) It is not bulky\n(ii) It does not need time to warm up\n(iii) It is cheaper to manufacture in large quantities.\n(iv) It needs low voltage to operate.\n\n', '', 'WAEC-PK-1', 'WAEC-PHY', 1),
(20, '2015', '\n\nQuestion\n\n(a) State two deductions that can be made from a displacement-time graph.\n(b) If the distance beween two equal masses is doubled and their individual masses are also doubled, what would happen to the force between them? Support your answer quantitatively.\n(c) State two factors that affect the maximum height attained by a bullet fired from a gun.\n(d) State two practical examples of mechanical resonance. A body is released from rest at the top of a plane inclined at 30° to the horizontal and 4.0 m high. If the coefficient of friction between the body and the plane is 0.3, calculate the time the body takes to reach the bottom of the plane.\n\n', '', '(a) State two deductions that can be made from a displacement-time graph.\n(b) If the distance beween two equal masses is doubled and their individual masses are also doubled, what would happen to the force between them? Support your answer quantitatively.\n(c) State two factors that affect the maximum height attained by a bullet fired from a gun.\n(d) State two practical examples of mechanical resonance. A body is released from rest at the top of a plane inclined at 30° to the horizontal and 4.0 m high. If the coefficient of friction between the body and the plane is 0.3, calculate the time the body takes to reach the bottom of the plane.', '(a) State two deductions that can be made from a displacement-time graph.~(b) If the distance beween two equal masses is doubled and their individual masses are also doubled, what would happen to the force between them? Support your answer quantitatively.\n(c) State two factors that affect the maximum height attained by a bullet fired from a gun.\n(d) State two practical examples of mechanical resonance. A body is released from rest at the top of a plane inclined at 30° to the horizontal and 4.0 m high. If the coefficient of friction between the body and the plane is 0.3, calculate the time the body takes to reach the bottom of the plane.~(c) State two factors that affect the maximum height attained by a bullet fired from a gun.\n(d) State two practical examples of mechanical resonance. A body is released from rest at the top of a plane inclined at 30° to the horizontal and 4.0 m high. If the coefficient of friction between the body and the plane is 0.3, calculate the time the body takes to reach the bottom of the plane.~(d) State two practical examples of mechanical resonance. A body is released from rest at the top of a plane inclined at 30° to the horizontal and 4.0 m high. If the coefficient of friction between the body and the plane is 0.3, calculate the time the body takes to reach the bottom of the plane.', NULL, '\n\nExplanation\n(a) From a displacement - time graph, the following can be deduced\n(i) Uniform velocity\n(ii) Instantaneous velocity\n(iii) Initial displacement\n\n(b) The force would remain the same Gram =\n\\(F_2 = \\frac{Gmm}{r^2}\\)\nF\\(_2\\) = \\(\\frac{G2m \\times 2m}{(2r)^2}\\)\n=\\(\\frac{G4m^2}{4r^2}\\)\n= \\(\\frac{Gm^2}{r^2}\\)\n= \\(\\frac{Gmm}{r^2}\\)\n\\(F_1 = F_2\\)\n\n(c)Factors that affect maximum height of a bullet include\n(i) Initial velocity / speed\n(ii) Acceleration due to gravity\n(iii) Angle of projection\n(iv) Air resistance\n\n(d)Practical examples of mechanical resonance include:\n(i) Shattering of glass due to high pitch note.\n(ii) Throwing of 2 paper rider from a vibrating stretched string\n(iii) Collapse of a bridge by matching soldiers\n(iv) Car bodies rattle (at very high speed)\n\n(e)\n\nacceleration = g sin \\(\\theta\\) – \\(\\mu\\) cos \\(\\theta\\)\na =10 x sin 30 – 0.3 x 10 x cos30\n=10 x 0.5 –0.3 x 10 x 0.8660\n= 5 – 2.598\n= 2.4m/s\\(^2\\)\nDistance covered, s = \\(\\frac{4}{\\sin30}\\)\n= \\(\\frac{4}{0.5}\\)\n= 8m\nUsing S = Ut + \\(\\frac{1}{2}\\)at\\(^2\\)\n8 =\\(\\frac{1}{2}\\) x 2.4 x r\\(^2\\)\n8 = 1.2\\(t^2\\)\n\\(r^2\\) = \\(\\frac{8}{1.2}\\)\n= 2.58s\n\n', '', 'Explanation\n(a) From a displacement - time graph, the following can be deduced\n(i) Uniform velocity\n(ii) Instantaneous velocity\n(iii) Initial displacement\n\n(b) The force would remain the same Gram =\n\\(F_2 = \\frac{Gmm}{r^2}\\)\nF\\(_2\\) = \\(\\frac{G2m \\times 2m}{(2r)^2}\\)\n=\\(\\frac{G4m^2}{4r^2}\\)\n= \\(\\frac{Gm^2}{r^2}\\)\n= \\(\\frac{Gmm}{r^2}\\)\n\\(F_1 = F_2\\)\n\n(c)Factors that affect maximum height of a bullet include\n(i) Initial velocity / speed\n(ii) Acceleration due to gravity\n(iii) Angle of projection\n(iv) Air resistance\n\n(d)Practical examples of mechanical resonance include:\n(i) Shattering of glass due to high pitch note.\n(ii) Throwing of 2 paper rider from a vibrating stretched string\n(iii) Collapse of a bridge by matching soldiers\n(iv) Car bodies rattle (at very high speed)\n\n(e)\n\nacceleration = g sin \\(\\theta\\) – \\(\\mu\\) cos \\(\\theta\\)\na =10 x sin 30 – 0.3 x 10 x cos30\n=10 x 0.5 –0.3 x 10 x 0.8660\n= 5 – 2.598\n= 2.4m/s\\(^2\\)\nDistance covered, s = \\(\\frac{4}{\\sin30}\\)\n= \\(\\frac{4}{0.5}\\)\n= 8m\nUsing S = Ut + \\(\\frac{1}{2}\\)at\\(^2\\)\n8 =\\(\\frac{1}{2}\\) x 2.4 x r\\(^2\\)\n8 = 1.2\\(t^2\\)\n\\(r^2\\) = \\(\\frac{8}{1.2}\\)\n= 2.58s\n\n', '(a) From a displacement - time graph, the following can be deduced\n(i) Uniform velocity\n(ii) Instantaneous velocity\n(iii) Initial displacement\n~(b) The force would remain the same Gram =\n\\(F_2 = \\frac{Gmm}{r^2}\\)\nF\\(_2\\) = \\(\\frac{G2m \\times 2m}{(2r)^2}\\)\n=\\(\\frac{G4m^2}{4r^2}\\)\n= \\(\\frac{Gm^2}{r^2}\\)\n= \\(\\frac{Gmm}{r^2}\\)\n\\(F_1 = F_2\\)\n\n(c)Factors that affect maximum height of a bullet include\n(i) Initial velocity / speed\n(ii) Acceleration due to gravity\n(iii) Angle of projection\n(iv) Air resistance\n\n(d)Practical examples of mechanical resonance include:\n(i) Shattering of glass due to high pitch note.\n(ii) Throwing of 2 paper rider from a vibrating stretched string\n(iii) Collapse of a bridge by matching soldiers\n(iv) Car bodies rattle (at very high speed)\n\n(e)\n\nacceleration = g sin \\(\\theta\\) – \\(\\mu\\) cos \\(\\theta\\)\na =10 x sin 30 – 0.3 x 10 x cos30\n=10 x 0.5 –0.3 x 10 x 0.8660\n= 5 – 2.598\n= 2.4m/s\\(^2\\)\nDistance covered, s = \\(\\frac{4}{\\sin30}\\)\n= \\(\\frac{4}{0.5}\\)\n= 8m\nUsing S = Ut + \\(\\frac{1}{2}\\)at\\(^2\\)\n8 =\\(\\frac{1}{2}\\) x 2.4 x r\\(^2\\)\n8 = 1.2\\(t^2\\)\n\\(r^2\\) = \\(\\frac{8}{1.2}\\)\n= 2.58s\n~(c)Factors that affect maximum height of a bullet include\n(i) Initial velocity / speed\n(ii) Acceleration due to gravity\n(iii) Angle of projection\n(iv) Air resistance\n\n(d)Practical examples of mechanical resonance include:\n(i) Shattering of glass due to high pitch note.\n(ii) Throwing of 2 paper rider from a vibrating stretched string\n(iii) Collapse of a bridge by matching soldiers\n(iv) Car bodies rattle (at very high speed)\n\n(e)\n\nacceleration = g sin \\(\\theta\\) – \\(\\mu\\) cos \\(\\theta\\)\na =10 x sin 30 – 0.3 x 10 x cos30\n=10 x 0.5 –0.3 x 10 x 0.8660\n= 5 – 2.598\n= 2.4m/s\\(^2\\)\nDistance covered, s = \\(\\frac{4}{\\sin30}\\)\n= \\(\\frac{4}{0.5}\\)\n= 8m\nUsing S = Ut + \\(\\frac{1}{2}\\)at\\(^2\\)\n8 =\\(\\frac{1}{2}\\) x 2.4 x r\\(^2\\)\n8 = 1.2\\(t^2\\)\n\\(r^2\\) = \\(\\frac{8}{1.2}\\)\n= 2.58s\n~(d)Practical examples of mechanical resonance include:\n(i) Shattering of glass due to high pitch note.\n(ii) Throwing of 2 paper rider from a vibrating stretched string\n(iii) Collapse of a bridge by matching soldiers\n(iv) Car bodies rattle (at very high speed)\n\n(e)\n\nacceleration = g sin \\(\\theta\\) – \\(\\mu\\) cos \\(\\theta\\)\na =10 x sin 30 – 0.3 x 10 x cos30\n=10 x 0.5 –0.3 x 10 x 0.8660\n= 5 – 2.598\n= 2.4m/s\\(^2\\)\nDistance covered, s = \\(\\frac{4}{\\sin30}\\)\n= \\(\\frac{4}{0.5}\\)\n= 8m\nUsing S = Ut + \\(\\frac{1}{2}\\)at\\(^2\\)\n8 =\\(\\frac{1}{2}\\) x 2.4 x r\\(^2\\)\n8 = 1.2\\(t^2\\)\n\\(r^2\\) = \\(\\frac{8}{1.2}\\)\n= 2.58s\n~(e)\n\nacceleration = g sin \\(\\theta\\) – \\(\\mu\\) cos \\(\\theta\\)\na =10 x sin 30 – 0.3 x 10 x cos30\n=10 x 0.5 –0.3 x 10 x 0.8660\n= 5 – 2.598\n= 2.4m/s\\(^2\\)\nDistance covered, s = \\(\\frac{4}{\\sin30}\\)\n= \\(\\frac{4}{0.5}\\)\n= 8m\nUsing S = Ut + \\(\\frac{1}{2}\\)at\\(^2\\)\n8 =\\(\\frac{1}{2}\\) x 2.4 x r\\(^2\\)\n8 = 1.2\\(t^2\\)\n\\(r^2\\) = \\(\\frac{8}{1.2}\\)\n= 2.58s\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
(21, '2015', '\n\nQuestion\n\n(a) Define stable equilibrium as applied to a rigid body.\n(b) Sketch a block and tackle system of pulleys with a velocity ratio of 3.\n(c) At the beginning of a race, a tyre of volume 8.0 x 10\\(^{-4}\\) at 20°C-has a gas pressure of 4.5 x 10\\(^5\\) Pa. Calculate the temperature of the gas in the tyre at the end of the race if the pressure has risen to 4.6 x 1 0\\(^5\\) Pa.\n(d)(i)\n\n\n\n\n\nIce point\n273k\n\n\n\n373k\nSteam point\n\n\n\n\nResistance/\\(\\Omega\\)\n\n5.67\n7.75\n\n\n\nPressure/Pa\n\n7.13 x 10\\(^4\\)\n9.74 x 10\\(^4\\)\n\n\n\n\nThe table above shows readings of the resistance and pressure of a platinum resistance thermometer and a constant-volume gas thermometer respectively, when immersed in the same liquid bath. Use this data to determine the temperature of the bath on the: (\\(\\alpha\\)) resistance thermometer; (\\(\\beta\\)) gas thermometer\n(ii) By what percentage is the temperature measured on the platinum resistance thermometer in error?\n\n\n(a) Define stable equilibrium as applied to a rigid body.\n(b) Sketch a block and tackle system of pulleys with a velocity ratio of 3.\n(c) At the beginning of a race, a tyre of volume 8.0 x 10\\(^{-4}\\) at 20°C-has a gas pressure of 4.5 x 10\\(^5\\) Pa. Calculate the temperature of the gas in the tyre at the end of the race if the pressure has risen to 4.6 x 1 0\\(^5\\) Pa.\n(d)(i)\n\n\n\n\n\nIce point\n273k\n\n\n\n373k\nSteam point\n\n\n\n\nResistance/\\(\\Omega\\)\n\n5.67\n7.75\n\n\n\nPressure/Pa\n\n7.13 x 10\\(^4\\)\n9.74 x 10\\(^4\\)\n\n\n\n', '', '(a) Define stable equilibrium as applied to a rigid body.\n(b) Sketch a block and tackle system of pulleys with a velocity ratio of 3.\n(c) At the beginning of a race, a tyre of volume 8.0 x 10\\(^{-4}\\) at 20°C-has a gas pressure of 4.5 x 10\\(^5\\) Pa. Calculate the temperature of the gas in the tyre at the end of the race if the pressure has risen to 4.6 x 1 0\\(^5\\) Pa.\n(d)(i)\n\n\n\n\n\nIce point\n273k\n\n\n\n373k\nSteam point\n\n\n\n\nResistance/\\(\\Omega\\)\n\n5.67\n7.75\n\n\n\nPressure/Pa\n\n7.13 x 10\\(^4\\)\n9.74 x 10\\(^4\\)\n\n\n\n\nThe table above shows readings of the resistance and pressure of a platinum resistance thermometer and a constant-volume gas thermometer respectively, when immersed in the same liquid bath. Use this data to determine the temperature of the bath on the: (\\(\\alpha\\)) resistance thermometer; (\\(\\beta\\)) gas thermometer\n(ii) By what percentage is the temperature measured on the platinum resistance thermometer in error?\n\n\n(a) Define stable equilibrium as applied to a rigid body.\n(b) Sketch a block and tackle system of pulleys with a velocity ratio of 3.\n(c) At the beginning of a race, a tyre of volume 8.0 x 10\\(^{-4}\\) at 20°C-has a gas pressure of 4.5 x 10\\(^5\\) Pa. Calculate the temperature of the gas in the tyre at the end of the race if the pressure has risen to 4.6 x 1 0\\(^5\\) Pa.\n(d)(i)\n\n\n\n\n\nIce point\n273k\n\n\n\n373k\nSteam point\n\n\n\n\nResistance/\\(\\Omega\\)\n\n5.67\n7.75\n\n\n\nPressure/Pa\n\n7.13 x 10\\(^4\\)\n9.74 x 10\\(^4\\)\n\n', '(a) Define stable equilibrium as applied to a rigid body.\n(b) Sketch a block and tackle system of pulleys with a velocity ratio of 3.\n(c) At the beginning of a race, a tyre of volume 8.0 x 10\\(^{-4}\\) at 20°C-has a gas pressure of 4.5 x 10\\(^5\\) Pa. Calculate the temperature of the gas in the tyre at the end of the race if the pressure has risen to 4.6 x 1 0\\(^5\\) Pa.\n(d)(i)\n\n\n\n\n\nIce point\n273k\n\n\n\n373k\nSteam point\n\n\n\n\nResistance/\\(\\Omega\\)\n\n5.67\n7.75\n\n\n\nPressure/Pa\n\n7.13 x 10\\(^4\\)\n9.74 x 10\\(^4\\)\n\n\n\n\nThe table above shows readings of the resistance and pressure of a platinum resistance thermometer and a constant-volume gas thermometer respectively, when immersed in the same liquid bath. Use this data to determine the temperature of the bath on the: (\\(\\alpha\\)) resistance thermometer; (\\(\\beta\\)) gas thermometer\n(ii) By what percentage is the temperature measured on the platinum resistance thermometer in error?\n\n\n(a) Define stable equilibrium as applied to a rigid body.~(b) Sketch a block and tackle system of pulleys with a velocity ratio of 3.\n(c) At the beginning of a race, a tyre of volume 8.0 x 10\\(^{-4}\\) at 20°C-has a gas pressure of 4.5 x 10\\(^5\\) Pa. Calculate the temperature of the gas in the tyre at the end of the race if the pressure has risen to 4.6 x 1 0\\(^5\\) Pa.\n(d)(i)\n\n\n\n\n\nIce point\n273k\n\n\n\n373k\nSteam point\n\n\n\n\nResistance/\\(\\Omega\\)\n\n5.67\n7.75\n\n\n\nPressure/Pa\n\n7.13 x 10\\(^4\\)\n9.74 x 10\\(^4\\)\n\n\n\n\nThe table above shows readings of the resistance and pressure of a platinum resistance thermometer and a constant-volume gas thermometer respectively, when immersed in the same liquid bath. Use this data to determine the temperature of the bath on the: (\\(\\alpha\\)) resistance thermometer; (\\(\\beta\\)) gas thermometer\n(ii) By what percentage is the temperature measured on the platinum resistance thermometer in error?\n\n\n(a) Define stable equilibrium as applied to a rigid body.\n(b) Sketch a block and tackle system of pulleys with a velocity ratio of 3.\n(c) At the beginning of a race, a tyre of volume 8.0 x 10\\(^{-4}\\) at 20°C-has a gas pressure of 4.5 x 10\\(^5\\) Pa. Calculate the temperature of the gas in the tyre at the end of the race if the pressure has risen to 4.6 x 1 0\\(^5\\) Pa.\n(d)(i)\n\n\n\n\n\nIce point\n273k\n\n\n\n373k\nSteam point\n\n\n\n\nResistance/\\(\\Omega\\)\n\n5.67\n7.75\n\n\n\nPressure/Pa\n\n7.13 x 10\\(^4\\)\n9.74 x 10\\(^4\\)\n\n~(c) At the beginning of a race, a tyre of volume 8.0 x 10\\(^{-4}\\) at 20°C-has a gas pressure of 4.5 x 10\\(^5\\) Pa. Calculate the temperature of the gas in the tyre at the end of the race if the pressure has risen to 4.6 x 1 0\\(^5\\) Pa.\n(d)(i)\n\n\n\n\n\nIce point\n273k\n\n\n\n373k\nSteam point\n\n\n\n\nResistance/\\(\\Omega\\)\n\n5.67\n7.75\n\n\n\nPressure/Pa\n\n7.13 x 10\\(^4\\)\n9.74 x 10\\(^4\\)\n\n\n\n\nThe table above shows readings of the resistance and pressure of a platinum resistance thermometer and a constant-volume gas thermometer respectively, when immersed in the same liquid bath. Use this data to determine the temperature of the bath on the: (\\(\\alpha\\)) resistance thermometer; (\\(\\beta\\)) gas thermometer\n(ii) By what percentage is the temperature measured on the platinum resistance thermometer in error?\n\n\n(a) Define stable equilibrium as applied to a rigid body.\n(b) Sketch a block and tackle system of pulleys with a velocity ratio of 3.\n(c) At the beginning of a race, a tyre of volume 8.0 x 10\\(^{-4}\\) at 20°C-has a gas pressure of 4.5 x 10\\(^5\\) Pa. Calculate the temperature of the gas in the tyre at the end of the race if the pressure has risen to 4.6 x 1 0\\(^5\\) Pa.\n(d)(i)\n\n\n\n\n\nIce point\n273k\n\n\n\n373k\nSteam point\n\n\n\n\nResistance/\\(\\Omega\\)\n\n5.67\n7.75\n\n\n\nPressure/Pa\n\n7.13 x 10\\(^4\\)\n9.74 x 10\\(^4\\)\n\n~(d)(i)\n\n\n\n\n\nIce point\n273k\n\n\n\n373k\nSteam point\n\n\n\n\nResistance/\\(\\Omega\\)\n\n5.67\n7.75\n\n\n\nPressure/Pa\n\n7.13 x 10\\(^4\\)\n9.74 x 10\\(^4\\)\n\n\n\n\nThe table above shows readings of the resistance and pressure of a platinum resistance thermometer and a constant-volume gas thermometer respectively, when immersed in the same liquid bath. Use this data to determine the temperature of the bath on the: (\\(\\alpha\\)) resistance thermometer; (\\(\\beta\\)) gas thermometer\n(ii) By what percentage is the temperature measured on the platinum resistance thermometer in error?\n\n\n(a) Define stable equilibrium as applied to a rigid body.\n(b) Sketch a block and tackle system of pulleys with a velocity ratio of 3.\n(c) At the beginning of a race, a tyre of volume 8.0 x 10\\(^{-4}\\) at 20°C-has a gas pressure of 4.5 x 10\\(^5\\) Pa. Calculate the temperature of the gas in the tyre at the end of the race if the pressure has risen to 4.6 x 1 0\\(^5\\) Pa.\n(d)(i)\n\n\n\n\n\nIce point\n273k\n\n\n\n373k\nSteam point\n\n\n\n\nResistance/\\(\\Omega\\)\n\n5.67\n7.75\n\n\n\nPressure/Pa\n\n7.13 x 10\\(^4\\)\n9.74 x 10\\(^4\\)\n\n', NULL, '\n\nExplanation\n(a) A body is said to be in stable equillibrium if it tends to return to its original position after being slightly displaced.\n(b)\n\ni.e. V.R = Number of pulleys in the system\nV.R = 3\n\n(c) \\(P_1 = 4.5 \\times 10^5\\)Pa; \\(_2 = 4.6 \\times 10^5\\)Pa\nT\\(_1\\) = 20\\(^{o}\\) + 273 = 293k\nT\\(_2\\) = ?\n\\(\\frac{P_1}{T_1} = \\frac{P_2}{T_2}; \\frac{4.5 \\times 10^5}{293} = \\frac{4.6 \\times 10^5}{T_2}\\)\n\\(T_2 = \\frac{4.6 \\times 10^5 \\times 293}{4.5 \\times 10^5}\\)\n\\(T_2\\) = 299.5k or 26.5\\(^o\\)C\n\n(d) \\(\\alpha\\) Resistance thermometer\n\\(\\frac{a}{b} = \\frac{c}{d}; \\frac{R \\theta - 5.67}{7.75 - 5.67} = \\frac{R - 273}{373 - 273}\\)\n\\(\\frac{R \\theta - 5.67}{1,88} = \\frac{R - 273}{100}\\)\n\\(\\beta\\) Gas thermometer\n\\(\\frac{a}{b} = \\frac{c}{d}; \\frac{G\\theta - (7.13) x 10^4}{(9.74 - 7.13) \\times 10^4} = \\frac{G - 273}{373 - 273}\\)\n\\(\\frac{G\\theta - 7.13}{2.16} = \\frac{G - 273}{100}\\)\n\n\n', '', 'Explanation\n(a) A body is said to be in stable equillibrium if it tends to return to its original position after being slightly displaced.\n(b)\n\ni.e. V.R = Number of pulleys in the system\nV.R = 3\n\n(c) \\(P_1 = 4.5 \\times 10^5\\)Pa; \\(_2 = 4.6 \\times 10^5\\)Pa\nT\\(_1\\) = 20\\(^{o}\\) + 273 = 293k\nT\\(_2\\) = ?\n\\(\\frac{P_1}{T_1} = \\frac{P_2}{T_2}; \\frac{4.5 \\times 10^5}{293} = \\frac{4.6 \\times 10^5}{T_2}\\)\n\\(T_2 = \\frac{4.6 \\times 10^5 \\times 293}{4.5 \\times 10^5}\\)\n\\(T_2\\) = 299.5k or 26.5\\(^o\\)C\n\n(d) \\(\\alpha\\) Resistance thermometer\n\\(\\frac{a}{b} = \\frac{c}{d}; \\frac{R \\theta - 5.67}{7.75 - 5.67} = \\frac{R - 273}{373 - 273}\\)\n\\(\\frac{R \\theta - 5.67}{1,88} = \\frac{R - 273}{100}\\)\n\\(\\beta\\) Gas thermometer\n\\(\\frac{a}{b} = \\frac{c}{d}; \\frac{G\\theta - (7.13) x 10^4}{(9.74 - 7.13) \\times 10^4} = \\frac{G - 273}{373 - 273}\\)\n\\(\\frac{G\\theta - 7.13}{2.16} = \\frac{G - 273}{100}\\)\n\n\n', '(a) A body is said to be in stable equillibrium if it tends to return to its original position after being slightly displaced.~(b)\n\ni.e. V.R = Number of pulleys in the system\nV.R = 3\n\n(c) \\(P_1 = 4.5 \\times 10^5\\)Pa; \\(_2 = 4.6 \\times 10^5\\)Pa\nT\\(_1\\) = 20\\(^{o}\\) + 273 = 293k\nT\\(_2\\) = ?\n\\(\\frac{P_1}{T_1} = \\frac{P_2}{T_2}; \\frac{4.5 \\times 10^5}{293} = \\frac{4.6 \\times 10^5}{T_2}\\)\n\\(T_2 = \\frac{4.6 \\times 10^5 \\times 293}{4.5 \\times 10^5}\\)\n\\(T_2\\) = 299.5k or 26.5\\(^o\\)C\n\n(d) \\(\\alpha\\) Resistance thermometer\n\\(\\frac{a}{b} = \\frac{c}{d}; \\frac{R \\theta - 5.67}{7.75 - 5.67} = \\frac{R - 273}{373 - 273}\\)\n\\(\\frac{R \\theta - 5.67}{1,88} = \\frac{R - 273}{100}\\)\n\\(\\beta\\) Gas thermometer\n\\(\\frac{a}{b} = \\frac{c}{d}; \\frac{G\\theta - (7.13) x 10^4}{(9.74 - 7.13) \\times 10^4} = \\frac{G - 273}{373 - 273}\\)\n\\(\\frac{G\\theta - 7.13}{2.16} = \\frac{G - 273}{100}\\)\n\n~(c) \\(P_1 = 4.5 \\times 10^5\\)Pa; \\(_2 = 4.6 \\times 10^5\\)Pa\nT\\(_1\\) = 20\\(^{o}\\) + 273 = 293k\nT\\(_2\\) = ?\n\\(\\frac{P_1}{T_1} = \\frac{P_2}{T_2}; \\frac{4.5 \\times 10^5}{293} = \\frac{4.6 \\times 10^5}{T_2}\\)\n\\(T_2 = \\frac{4.6 \\times 10^5 \\times 293}{4.5 \\times 10^5}\\)\n\\(T_2\\) = 299.5k or 26.5\\(^o\\)C\n\n(d) \\(\\alpha\\) Resistance thermometer\n\\(\\frac{a}{b} = \\frac{c}{d}; \\frac{R \\theta - 5.67}{7.75 - 5.67} = \\frac{R - 273}{373 - 273}\\)\n\\(\\frac{R \\theta - 5.67}{1,88} = \\frac{R - 273}{100}\\)\n\\(\\beta\\) Gas thermometer\n\\(\\frac{a}{b} = \\frac{c}{d}; \\frac{G\\theta - (7.13) x 10^4}{(9.74 - 7.13) \\times 10^4} = \\frac{G - 273}{373 - 273}\\)\n\\(\\frac{G\\theta - 7.13}{2.16} = \\frac{G - 273}{100}\\)\n\n~(d) \\(\\alpha\\) Resistance thermometer\n\\(\\frac{a}{b} = \\frac{c}{d}; \\frac{R \\theta - 5.67}{7.75 - 5.67} = \\frac{R - 273}{373 - 273}\\)\n\\(\\frac{R \\theta - 5.67}{1,88} = \\frac{R - 273}{100}\\)\n\\(\\beta\\) Gas thermometer\n\\(\\frac{a}{b} = \\frac{c}{d}; \\frac{G\\theta - (7.13) x 10^4}{(9.74 - 7.13) \\times 10^4} = \\frac{G - 273}{373 - 273}\\)\n\\(\\frac{G\\theta - 7.13}{2.16} = \\frac{G - 273}{100}\\)\n\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(22, '2015', '\n\nQuestion\n\n(a) What is a wavefront?\n(b) (i) State two practical uses of glass prisms.\n(ii) List two factors that determine the deviation of a ray of light travelling from air into a triangular glass prism.\n(iii) Sketch a graph to illustrate the variation of the angle of deviation d, with that of incidence, i, for a ray of light travelling from air into a triangular glass prism. Indicate on the graph the point at which the angle of incidence equal to the angle of emergence e.\n\n(c) (i) Draw and label a diagram of an astronomical telescope in normal adjustment.\n(ii) The angular magnification of an astronomical telescope in normal adjustment is 5. If the focal length of the objective is 100cm, calculate the:\n(\\(\\alpha\\)) focal length of the eyepiece;\n(\\(\\beta\\)) length of the telescope.\n\n', '', '(a) What is a wavefront?\n(b) (i) State two practical uses of glass prisms.\n(ii) List two factors that determine the deviation of a ray of light travelling from air into a triangular glass prism.\n(iii) Sketch a graph to illustrate the variation of the angle of deviation d, with that of incidence, i, for a ray of light travelling from air into a triangular glass prism. Indicate on the graph the point at which the angle of incidence equal to the angle of emergence e.\n\n(c) (i) Draw and label a diagram of an astronomical telescope in normal adjustment.\n(ii) The angular magnification of an astronomical telescope in normal adjustment is 5. If the focal length of the objective is 100cm, calculate the:\n(\\(\\alpha\\)) focal length of the eyepiece;\n(\\(\\beta\\)) length of the telescope.', '(a) What is a wavefront?~(b) (i) State two practical uses of glass prisms.\n(ii) List two factors that determine the deviation of a ray of light travelling from air into a triangular glass prism.\n(iii) Sketch a graph to illustrate the variation of the angle of deviation d, with that of incidence, i, for a ray of light travelling from air into a triangular glass prism. Indicate on the graph the point at which the angle of incidence equal to the angle of emergence e.\n\n(c) (i) Draw and label a diagram of an astronomical telescope in normal adjustment.\n(ii) The angular magnification of an astronomical telescope in normal adjustment is 5. If the focal length of the objective is 100cm, calculate the:\n(\\(\\alpha\\)) focal length of the eyepiece;\n(\\(\\beta\\)) length of the telescope.~(c) (i) Draw and label a diagram of an astronomical telescope in normal adjustment.\n(ii) The angular magnification of an astronomical telescope in normal adjustment is 5. If the focal length of the objective is 100cm, calculate the:\n(\\(\\alpha\\)) focal length of the eyepiece;\n(\\(\\beta\\)) length of the telescope.', NULL, '\n\nExplanation\n(a) A Wavefront is a surface over which particles vibrate in the same phase.\n\n(b)(i) Glass prisms are used practically for\n(I) changing direction of light I dispersion\n(II) viewing distant objects (Binoculars)\n(III) determing refractive index of glass\n(IV) viewing objects over obstacles (periscope)\n\n(ii) The factors include:\n(I) Angle of incidence\n(II) refractive index of the prism\n(III) refracting angle of the prism\n\n(iii)\n\n\n\n(c)\n\n\n\n\n\n(ii) (\\(alpha\\)) Magnifiation = \\(\\frac{F_o}{F_e}\\); 5 = \\(\\frac{100}{Fe}\\)\nfe = 20cm\n(\\(\\beta\\)) length of telescope = \\(F_o + F_e = 100 + 20\\)\n= 120cm\n\n', '', 'Explanation\n(a) A Wavefront is a surface over which particles vibrate in the same phase.\n\n(b)(i) Glass prisms are used practically for\n(I) changing direction of light I dispersion\n(II) viewing distant objects (Binoculars)\n(III) determing refractive index of glass\n(IV) viewing objects over obstacles (periscope)\n\n(ii) The factors include:\n(I) Angle of incidence\n(II) refractive index of the prism\n(III) refracting angle of the prism\n\n(iii)\n\n\n\n(c)\n\n\n\n\n\n(ii) (\\(alpha\\)) Magnifiation = \\(\\frac{F_o}{F_e}\\); 5 = \\(\\frac{100}{Fe}\\)\nfe = 20cm\n(\\(\\beta\\)) length of telescope = \\(F_o + F_e = 100 + 20\\)\n= 120cm\n\n', '(a) A Wavefront is a surface over which particles vibrate in the same phase.\n~(b)(i) Glass prisms are used practically for\n(I) changing direction of light I dispersion\n(II) viewing distant objects (Binoculars)\n(III) determing refractive index of glass\n(IV) viewing objects over obstacles (periscope)\n\n(ii) The factors include:\n(I) Angle of incidence\n(II) refractive index of the prism\n(III) refracting angle of the prism\n\n(iii)\n\n\n\n(c)\n\n\n\n\n\n(ii) (\\(alpha\\)) Magnifiation = \\(\\frac{F_o}{F_e}\\); 5 = \\(\\frac{100}{Fe}\\)\nfe = 20cm\n(\\(\\beta\\)) length of telescope = \\(F_o + F_e = 100 + 20\\)\n= 120cm\n~(c)\n\n\n\n\n\n(ii) (\\(alpha\\)) Magnifiation = \\(\\frac{F_o}{F_e}\\); 5 = \\(\\frac{100}{Fe}\\)\nfe = 20cm\n(\\(\\beta\\)) length of telescope = \\(F_o + F_e = 100 + 20\\)\n= 120cm\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(23, '2015', '\n\nQuestion\n\n(i) What is dielectric?\n(ii) A parallel plate capacitor consists of two plates each of area 9.6 x 10\\(^{-2}\\)m\\(^2\\), separated by a dielectric of thickness 2.25 x 10\\(^{-3}\\) and dielectric constant 900. Calculate the capacitance of the capacitor. [ \\(\\varepsilon_o\\) = permittivity of free space = 8.85 x 10\\(^{-12}\\)Fm\\(^{-1}\\)]\n(b) (i) Which of the following devices has a higher resistance; an ammeter or a voltmeter? Give a reason for your answer.\n(ii)\n\n\nThe resistance of the voltmeter in the circuit diagram illustrated above is 800\\(\\Omega\\), Calculate the voltmeter reading.\n(c)\n\nA battery of negligible internal resistance is connected to a set of resistors as illustrated in the circuit diagram above. Determine the equivalent resistance of the circuit.\n\n\n\n\n(i) What is dielectric?\n(ii) A parallel plate capacitor consists of two plates each of area 9.6 x 10\\(^{-2}\\)m\\(^2\\), separated by a dielectric of thickness 2.25 x 10\\(^{-3}\\) and dielectric constant 900. Calculate the capacitance of the capacitor. [ \\(\\varepsilon_o\\) = permittivity of free space = 8.85 x 10\\(^{-12}\\)Fm\\(^{-1}\\)]\n(b) (i) Which of the following devices has a higher resistance; an ammeter or a voltmeter? Give a reason for your answer.\n(ii)\n\n\nThe resistance of the voltmeter in the circuit diagram illustrated above is 800\\(\\Omega\\), Calculate the voltmeter reading.\n(c)\n\nA battery of negligible internal resistance is connected to a set of resistors as illustrated in the circuit diagram above. Determine the equivalent resistance of the circuit.\n\n', '', '(i) What is dielectric?\n(ii) A parallel plate capacitor consists of two plates each of area 9.6 x 10\\(^{-2}\\)m\\(^2\\), separated by a dielectric of thickness 2.25 x 10\\(^{-3}\\) and dielectric constant 900. Calculate the capacitance of the capacitor. [ \\(\\varepsilon_o\\) = permittivity of free space = 8.85 x 10\\(^{-12}\\)Fm\\(^{-1}\\)]\n(b) (i) Which of the following devices has a higher resistance; an ammeter or a voltmeter? Give a reason for your answer.\n(ii)\n\n\nThe resistance of the voltmeter in the circuit diagram illustrated above is 800\\(\\Omega\\), Calculate the voltmeter reading.\n(c)\n\nA battery of negligible internal resistance is connected to a set of resistors as illustrated in the circuit diagram above. Determine the equivalent resistance of the circuit.\n\n\n\n\n(i) What is dielectric?\n(ii) A parallel plate capacitor consists of two plates each of area 9.6 x 10\\(^{-2}\\)m\\(^2\\), separated by a dielectric of thickness 2.25 x 10\\(^{-3}\\) and dielectric constant 900. Calculate the capacitance of the capacitor. [ \\(\\varepsilon_o\\) = permittivity of free space = 8.85 x 10\\(^{-12}\\)Fm\\(^{-1}\\)]\n(b) (i) Which of the following devices has a higher resistance; an ammeter or a voltmeter? Give a reason for your answer.\n(ii)\n\n\nThe resistance of the voltmeter in the circuit diagram illustrated above is 800\\(\\Omega\\), Calculate the voltmeter reading.\n(c)\n\nA battery of negligible internal resistance is connected to a set of resistors as illustrated in the circuit diagram above. Determine the equivalent resistance of the circuit.', '(i) What is dielectric?\n(ii) A parallel plate capacitor consists of two plates each of area 9.6 x 10\\(^{-2}\\)m\\(^2\\), separated by a dielectric of thickness 2.25 x 10\\(^{-3}\\) and dielectric constant 900. Calculate the capacitance of the capacitor. [ \\(\\varepsilon_o\\) = permittivity of free space = 8.85 x 10\\(^{-12}\\)Fm\\(^{-1}\\)]\n(b) (i) Which of the following devices has a higher resistance; an ammeter or a voltmeter? Give a reason for your answer.\n(ii)\n\n\nThe resistance of the voltmeter in the circuit diagram illustrated above is 800\\(\\Omega\\), Calculate the voltmeter reading.\n(c)\n\nA battery of negligible internal resistance is connected to a set of resistors as illustrated in the circuit diagram above. Determine the equivalent resistance of the circuit.\n\n\n\n\n(i) What is dielectric?\n(ii) A parallel plate capacitor consists of two plates each of area 9.6 x 10\\(^{-2}\\)m\\(^2\\), separated by a dielectric of thickness 2.25 x 10\\(^{-3}\\) and dielectric constant 900. Calculate the capacitance of the capacitor. [ \\(\\varepsilon_o\\) = permittivity of free space = 8.85 x 10\\(^{-12}\\)Fm\\(^{-1}\\)]~(b) (i) Which of the following devices has a higher resistance; an ammeter or a voltmeter? Give a reason for your answer.\n(ii)\n\n\nThe resistance of the voltmeter in the circuit diagram illustrated above is 800\\(\\Omega\\), Calculate the voltmeter reading.\n(c)\n\nA battery of negligible internal resistance is connected to a set of resistors as illustrated in the circuit diagram above. Determine the equivalent resistance of the circuit.\n\n\n\n\n(i) What is dielectric?\n(ii) A parallel plate capacitor consists of two plates each of area 9.6 x 10\\(^{-2}\\)m\\(^2\\), separated by a dielectric of thickness 2.25 x 10\\(^{-3}\\) and dielectric constant 900. Calculate the capacitance of the capacitor. [ \\(\\varepsilon_o\\) = permittivity of free space = 8.85 x 10\\(^{-12}\\)Fm\\(^{-1}\\)]\n(b) (i) Which of the following devices has a higher resistance; an ammeter or a voltmeter? Give a reason for your answer.\n(ii)\n\n\nThe resistance of the voltmeter in the circuit diagram illustrated above is 800\\(\\Omega\\), Calculate the voltmeter reading.\n(c)\n\nA battery of negligible internal resistance is connected to a set of resistors as illustrated in the circuit diagram above. Determine the equivalent resistance of the circuit.~(c)\n\nA battery of negligible internal resistance is connected to a set of resistors as illustrated in the circuit diagram above. Determine the equivalent resistance of the circuit.\n\n\n\n\n(i) What is dielectric?\n(ii) A parallel plate capacitor consists of two plates each of area 9.6 x 10\\(^{-2}\\)m\\(^2\\), separated by a dielectric of thickness 2.25 x 10\\(^{-3}\\) and dielectric constant 900. Calculate the capacitance of the capacitor. [ \\(\\varepsilon_o\\) = permittivity of free space = 8.85 x 10\\(^{-12}\\)Fm\\(^{-1}\\)]\n(b) (i) Which of the following devices has a higher resistance; an ammeter or a voltmeter? Give a reason for your answer.\n(ii)\n\n\nThe resistance of the voltmeter in the circuit diagram illustrated above is 800\\(\\Omega\\), Calculate the voltmeter reading.\n(c)\n\nA battery of negligible internal resistance is connected to a set of resistors as illustrated in the circuit diagram above. Determine the equivalent resistance of the circuit.', NULL, '\n\nExplanation\n(a)(i) A dielectric is an insulating material (solid, liquid or gas) separating the two plates of a capacitor.\n(ii) Using C = \\(\\frac{\\xi_1.\\xi_oA}{d} = \\frac{900 \\times 8.85 \\times 10^{-12} \\times 9.6 \\times 10^{-3}}{2.25 \\times 10^{-4}}\\)\n= 3.4 x 10\\(^{-7}\\)\n\n(b)(i) A voltmeter has a higher resistance. Reason: A voltmeter is connected inparallel in a circuit to allow passage of minimal current through it; hence the circuit current is not significantly affected.\n(ii)\n\n\\(\\frac{1}{R_1} = \\frac{1}{400} + \\frac{1}{400}\\) \\(R_1 = 200\\Omega\\)\n\\(\\frac{1}{R_2} = \\frac{1}{800} + \\frac{1}{800}\\) \\(R_1 = 400\\Omega\\)\nRT= 200 + 400 = 600\\(\\Omega\\)\nI = \\(\\frac{V_T}{R_T} = \\frac{6}{600}\\) = 0.01A\nVoltmeter reading\n= Pd across R\\(_2\\)\n= 0.01A\nVoltmeter reading = Pd across R\\(_2\\)\n= 0.01 x 400 = 4v\n\n(c)\n\n\\(\\frac{1}{R_p} = \\frac{1}{2} + \\frac{1}{2} = \\frac{1 + 1}{2} = \\frac{2}{2}\\)\nNote: This catadation is for the two resistors in parallel.\nR\\(_T = 2 + 1 + 2 = 5\\Omega\\)\n\n', '', 'Explanation\n(a)(i) A dielectric is an insulating material (solid, liquid or gas) separating the two plates of a capacitor.\n(ii) Using C = \\(\\frac{\\xi_1.\\xi_oA}{d} = \\frac{900 \\times 8.85 \\times 10^{-12} \\times 9.6 \\times 10^{-3}}{2.25 \\times 10^{-4}}\\)\n= 3.4 x 10\\(^{-7}\\)\n\n(b)(i) A voltmeter has a higher resistance. Reason: A voltmeter is connected inparallel in a circuit to allow passage of minimal current through it; hence the circuit current is not significantly affected.\n(ii)\n\n\\(\\frac{1}{R_1} = \\frac{1}{400} + \\frac{1}{400}\\) \\(R_1 = 200\\Omega\\)\n\\(\\frac{1}{R_2} = \\frac{1}{800} + \\frac{1}{800}\\) \\(R_1 = 400\\Omega\\)\nRT= 200 + 400 = 600\\(\\Omega\\)\nI = \\(\\frac{V_T}{R_T} = \\frac{6}{600}\\) = 0.01A\nVoltmeter reading\n= Pd across R\\(_2\\)\n= 0.01A\nVoltmeter reading = Pd across R\\(_2\\)\n= 0.01 x 400 = 4v\n\n(c)\n\n\\(\\frac{1}{R_p} = \\frac{1}{2} + \\frac{1}{2} = \\frac{1 + 1}{2} = \\frac{2}{2}\\)\nNote: This catadation is for the two resistors in parallel.\nR\\(_T = 2 + 1 + 2 = 5\\Omega\\)\n\n', '(a)(i) A dielectric is an insulating material (solid, liquid or gas) separating the two plates of a capacitor.\n(ii) Using C = \\(\\frac{\\xi_1.\\xi_oA}{d} = \\frac{900 \\times 8.85 \\times 10^{-12} \\times 9.6 \\times 10^{-3}}{2.25 \\times 10^{-4}}\\)\n= 3.4 x 10\\(^{-7}\\)\n~(b)(i) A voltmeter has a higher resistance. Reason: A voltmeter is connected inparallel in a circuit to allow passage of minimal current through it; hence the circuit current is not significantly affected.\n(ii)\n\n\\(\\frac{1}{R_1} = \\frac{1}{400} + \\frac{1}{400}\\) \\(R_1 = 200\\Omega\\)\n\\(\\frac{1}{R_2} = \\frac{1}{800} + \\frac{1}{800}\\) \\(R_1 = 400\\Omega\\)\nRT= 200 + 400 = 600\\(\\Omega\\)\nI = \\(\\frac{V_T}{R_T} = \\frac{6}{600}\\) = 0.01A\nVoltmeter reading\n= Pd across R\\(_2\\)\n= 0.01A\nVoltmeter reading = Pd across R\\(_2\\)\n= 0.01 x 400 = 4v\n\n(c)\n\n\\(\\frac{1}{R_p} = \\frac{1}{2} + \\frac{1}{2} = \\frac{1 + 1}{2} = \\frac{2}{2}\\)\nNote: This catadation is for the two resistors in parallel.\nR\\(_T = 2 + 1 + 2 = 5\\Omega\\)\n~(c)\n\n\\(\\frac{1}{R_p} = \\frac{1}{2} + \\frac{1}{2} = \\frac{1 + 1}{2} = \\frac{2}{2}\\)\nNote: This catadation is for the two resistors in parallel.\nR\\(_T = 2 + 1 + 2 = 5\\Omega\\)\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(24, '2015', '\n\nQuestion\n\n(a) (i) What is nuclear fission?\n(ii) State the function of each of the following ma-terials in a nuclear fision reactor:\n(\\(\\alpha\\)) graphite, (\\(\\beta\\)) boron rods; (\\(\\gamma\\)) liquid sodium\n(b) The table below gives some of the energy levels of a hydrogen atom.\n\n\n\nn\n1\n2\n3\n4\n5\n\\(\\infty\\)\n\n\nE\\(_n\\)/\\_eV\\)\n-13.60\n-3.39\n-1.51\n-0.85\n-0.54\n0.00\n\n\n\n\n(i) Draw the energy level diagram for the atom.\n(ii) Determine the wavelength of the photon emitted when the atom goes from the energy state n = 3 to the ground state. [h = 6.6 x 10\\(^{-34}\\) Js, c = 3 x 10\\(^{8}\\)ms\\(^{-1}\\), e = 1.6 x 10\\(^{-19}\\)]\n(c) A piece of ancient bone from an excavation site showed \\(^{14}_{6}\\)C activity of 9.5 disintegrations per minute per 1.0x10\\(^{-3}\\)kg. If a bone specimen from a living creature shows \\(^{14}_{6}\\)C activity of 12.0 disintegrations per minute per 1.0 x 1 0\\(^{-3}\\), determine the age of the ancient bone. [Half - life of \\(^{14}_{6}\\)C = 5572 years].\n\n\n(a) (i) What is nuclear fission?\n(ii) State the function of each of the following ma-terials in a nuclear fision reactor:\n(\\(\\alpha\\)) graphite, (\\(\\beta\\)) boron rods; (\\(\\gamma\\)) liquid sodium\n(b) The table below gives some of the energy levels of a hydrogen atom.\n\n\n\nn\n1\n2\n3\n4\n5\n\\(\\infty\\)\n\n\nE\\(_n\\)/\\_eV\\)\n-13.60\n-3.39\n-1.51\n-0.85\n-0.54\n0.00\n\n\n\n', '', '(a) (i) What is nuclear fission?\n(ii) State the function of each of the following ma-terials in a nuclear fision reactor:\n(\\(\\alpha\\)) graphite, (\\(\\beta\\)) boron rods; (\\(\\gamma\\)) liquid sodium\n(b) The table below gives some of the energy levels of a hydrogen atom.\n\n\n\nn\n1\n2\n3\n4\n5\n\\(\\infty\\)\n\n\nE\\(_n\\)/\\_eV\\)\n-13.60\n-3.39\n-1.51\n-0.85\n-0.54\n0.00\n\n\n\n\n(i) Draw the energy level diagram for the atom.\n(ii) Determine the wavelength of the photon emitted when the atom goes from the energy state n = 3 to the ground state. [h = 6.6 x 10\\(^{-34}\\) Js, c = 3 x 10\\(^{8}\\)ms\\(^{-1}\\), e = 1.6 x 10\\(^{-19}\\)]\n(c) A piece of ancient bone from an excavation site showed \\(^{14}_{6}\\)C activity of 9.5 disintegrations per minute per 1.0x10\\(^{-3}\\)kg. If a bone specimen from a living creature shows \\(^{14}_{6}\\)C activity of 12.0 disintegrations per minute per 1.0 x 1 0\\(^{-3}\\), determine the age of the ancient bone. [Half - life of \\(^{14}_{6}\\)C = 5572 years].\n\n\n(a) (i) What is nuclear fission?\n(ii) State the function of each of the following ma-terials in a nuclear fision reactor:\n(\\(\\alpha\\)) graphite, (\\(\\beta\\)) boron rods; (\\(\\gamma\\)) liquid sodium\n(b) The table below gives some of the energy levels of a hydrogen atom.\n\n\n\nn\n1\n2\n3\n4\n5\n\\(\\infty\\)\n\n\nE\\(_n\\)/\\_eV\\)\n-13.60\n-3.39\n-1.51\n-0.85\n-0.54\n0.00\n\n', '(a) (i) What is nuclear fission?\n(ii) State the function of each of the following ma-terials in a nuclear fision reactor:\n(\\(\\alpha\\)) graphite, (\\(\\beta\\)) boron rods; (\\(\\gamma\\)) liquid sodium\n(b) The table below gives some of the energy levels of a hydrogen atom.\n\n\n\nn\n1\n2\n3\n4\n5\n\\(\\infty\\)\n\n\nE\\(_n\\)/\\_eV\\)\n-13.60\n-3.39\n-1.51\n-0.85\n-0.54\n0.00\n\n\n\n\n(i) Draw the energy level diagram for the atom.\n(ii) Determine the wavelength of the photon emitted when the atom goes from the energy state n = 3 to the ground state. [h = 6.6 x 10\\(^{-34}\\) Js, c = 3 x 10\\(^{8}\\)ms\\(^{-1}\\), e = 1.6 x 10\\(^{-19}\\)]\n(c) A piece of ancient bone from an excavation site showed \\(^{14}_{6}\\)C activity of 9.5 disintegrations per minute per 1.0x10\\(^{-3}\\)kg. If a bone specimen from a living creature shows \\(^{14}_{6}\\)C activity of 12.0 disintegrations per minute per 1.0 x 1 0\\(^{-3}\\), determine the age of the ancient bone. [Half - life of \\(^{14}_{6}\\)C = 5572 years].\n\n\n(a) (i) What is nuclear fission?\n(ii) State the function of each of the following ma-terials in a nuclear fision reactor:\n(\\(\\alpha\\)) graphite, (\\(\\beta\\)) boron rods; (\\(\\gamma\\)) liquid sodium~(b) The table below gives some of the energy levels of a hydrogen atom.\n\n\n\nn\n1\n2\n3\n4\n5\n\\(\\infty\\)\n\n\nE\\(_n\\)/\\_eV\\)\n-13.60\n-3.39\n-1.51\n-0.85\n-0.54\n0.00\n\n\n\n\n(i) Draw the energy level diagram for the atom.\n(ii) Determine the wavelength of the photon emitted when the atom goes from the energy state n = 3 to the ground state. [h = 6.6 x 10\\(^{-34}\\) Js, c = 3 x 10\\(^{8}\\)ms\\(^{-1}\\), e = 1.6 x 10\\(^{-19}\\)]\n(c) A piece of ancient bone from an excavation site showed \\(^{14}_{6}\\)C activity of 9.5 disintegrations per minute per 1.0x10\\(^{-3}\\)kg. If a bone specimen from a living creature shows \\(^{14}_{6}\\)C activity of 12.0 disintegrations per minute per 1.0 x 1 0\\(^{-3}\\), determine the age of the ancient bone. [Half - life of \\(^{14}_{6}\\)C = 5572 years].\n\n\n(a) (i) What is nuclear fission?\n(ii) State the function of each of the following ma-terials in a nuclear fision reactor:\n(\\(\\alpha\\)) graphite, (\\(\\beta\\)) boron rods; (\\(\\gamma\\)) liquid sodium\n(b) The table below gives some of the energy levels of a hydrogen atom.\n\n\n\nn\n1\n2\n3\n4\n5\n\\(\\infty\\)\n\n\nE\\(_n\\)/\\_eV\\)\n-13.60\n-3.39\n-1.51\n-0.85\n-0.54\n0.00\n\n~(c) A piece of ancient bone from an excavation site showed \\(^{14}_{6}\\)C activity of 9.5 disintegrations per minute per 1.0x10\\(^{-3}\\)kg. If a bone specimen from a living creature shows \\(^{14}_{6}\\)C activity of 12.0 disintegrations per minute per 1.0 x 1 0\\(^{-3}\\), determine the age of the ancient bone. [Half - life of \\(^{14}_{6}\\)C = 5572 years].\n\n\n(a) (i) What is nuclear fission?\n(ii) State the function of each of the following ma-terials in a nuclear fision reactor:\n(\\(\\alpha\\)) graphite, (\\(\\beta\\)) boron rods; (\\(\\gamma\\)) liquid sodium\n(b) The table below gives some of the energy levels of a hydrogen atom.\n\n\n\nn\n1\n2\n3\n4\n5\n\\(\\infty\\)\n\n\nE\\(_n\\)/\\_eV\\)\n-13.60\n-3.39\n-1.51\n-0.85\n-0.54\n0.00\n\n', NULL, '\n\nExplanation\n(a)(i) Nuclear Fission is the splitting up of a heavy nucleus into lighter nuclei with release of a huge amount of energy and neutrons.\n(ii) (\\(\\alpha\\)) graphite; (\\(\\beta\\)) Boron rods; (\\(\\gamma\\) )Liquid sodium\n\n\n\n\nMaterial\n\n\nFunction\n\n\n\n\nGraphite\n\n\nSlows down high energetic neutrons in the reaction chamber.\n\n\n\n\nBoron rods\n\n\nTo absorb excess neutrons so as to control fission rate\n\n\n\n\nLiquid sodium\n\n\nExtraction / conduction of excess heat from the reaction chamber.\n\n\n\n\n\n(b)(i) E\\(_n\\), n = \\(\\alpha\\) ---------------------------------- 0.eV\nE\\(_4\\), n = 5 ---------------------------------------------------- - 0.54eV\nE\\(_3\\), n = 4 ---------------------------------------------------- - 0.85eV\nE\\(_2\\), n = 3 ---------------------------------------------------- 1.51eV\nE\\(_1\\), n = 2 ------------------------------------------------------ 3.39eV\n\n(ii) Using \\(\\lambda\\) = \\(\\frac{hc}{\\Delta E}\\)\n= \\(\\frac{6.6 \\times 10^{-34} \\times 3.0 \\times 10^8}{(-1.51 - ( - 13.60)) \\times 1.6 \\times 10^{-19}}\\)\n= 1.02 x 10\\(^{-7}\\)m\n\n(c) \\(\\lambda\\) = \\(\\frac{0.693}{ T^{\\frac{1}{2}}}\\) = \\(\\frac{0.693}{5572}\\)\n= 0.0001244 years\\(^{-1}\\)\nN = Noe\\(^{- \\lambda t}\\)\ne\\(^{0.0001244t}\\) = 1.2632\n0.0001244t = 0.234\nt = 1878 years\n\n\n\n\n', '', 'Explanation\n(a)(i) Nuclear Fission is the splitting up of a heavy nucleus into lighter nuclei with release of a huge amount of energy and neutrons.\n(ii) (\\(\\alpha\\)) graphite; (\\(\\beta\\)) Boron rods; (\\(\\gamma\\) )Liquid sodium\n\n\n\n\nMaterial\n\n\nFunction\n\n\n\n\nGraphite\n\n\nSlows down high energetic neutrons in the reaction chamber.\n\n\n\n\nBoron rods\n\n\nTo absorb excess neutrons so as to control fission rate\n\n\n\n\nLiquid sodium\n\n\nExtraction / conduction of excess heat from the reaction chamber.\n\n\n\n\n\n(b)(i) E\\(_n\\), n = \\(\\alpha\\) ---------------------------------- 0.eV\nE\\(_4\\), n = 5 ---------------------------------------------------- - 0.54eV\nE\\(_3\\), n = 4 ---------------------------------------------------- - 0.85eV\nE\\(_2\\), n = 3 ---------------------------------------------------- 1.51eV\nE\\(_1\\), n = 2 ------------------------------------------------------ 3.39eV\n\n(ii) Using \\(\\lambda\\) = \\(\\frac{hc}{\\Delta E}\\)\n= \\(\\frac{6.6 \\times 10^{-34} \\times 3.0 \\times 10^8}{(-1.51 - ( - 13.60)) \\times 1.6 \\times 10^{-19}}\\)\n= 1.02 x 10\\(^{-7}\\)m\n\n(c) \\(\\lambda\\) = \\(\\frac{0.693}{ T^{\\frac{1}{2}}}\\) = \\(\\frac{0.693}{5572}\\)\n= 0.0001244 years\\(^{-1}\\)\nN = Noe\\(^{- \\lambda t}\\)\ne\\(^{0.0001244t}\\) = 1.2632\n0.0001244t = 0.234\nt = 1878 years\n\n\n\n\n', '(a)(i) Nuclear Fission is the splitting up of a heavy nucleus into lighter nuclei with release of a huge amount of energy and neutrons.\n(ii) (\\(\\alpha\\)) graphite; (\\(\\beta\\)) Boron rods; (\\(\\gamma\\) )Liquid sodium\n\n\n\n\nMaterial\n\n\nFunction\n\n\n\n\nGraphite\n\n\nSlows down high energetic neutrons in the reaction chamber.\n\n\n\n\nBoron rods\n\n\nTo absorb excess neutrons so as to control fission rate\n\n\n\n\nLiquid sodium\n\n\nExtraction / conduction of excess heat from the reaction chamber.\n\n\n\n\n~(b)(i) E\\(_n\\), n = \\(\\alpha\\) ---------------------------------- 0.eV\nE\\(_4\\), n = 5 ---------------------------------------------------- - 0.54eV\nE\\(_3\\), n = 4 ---------------------------------------------------- - 0.85eV\nE\\(_2\\), n = 3 ---------------------------------------------------- 1.51eV\nE\\(_1\\), n = 2 ------------------------------------------------------ 3.39eV\n\n(ii) Using \\(\\lambda\\) = \\(\\frac{hc}{\\Delta E}\\)\n= \\(\\frac{6.6 \\times 10^{-34} \\times 3.0 \\times 10^8}{(-1.51 - ( - 13.60)) \\times 1.6 \\times 10^{-19}}\\)\n= 1.02 x 10\\(^{-7}\\)m\n\n(c) \\(\\lambda\\) = \\(\\frac{0.693}{ T^{\\frac{1}{2}}}\\) = \\(\\frac{0.693}{5572}\\)\n= 0.0001244 years\\(^{-1}\\)\nN = Noe\\(^{- \\lambda t}\\)\ne\\(^{0.0001244t}\\) = 1.2632\n0.0001244t = 0.234\nt = 1878 years\n\n\n\n~(c) \\(\\lambda\\) = \\(\\frac{0.693}{ T^{\\frac{1}{2}}}\\) = \\(\\frac{0.693}{5572}\\)\n= 0.0001244 years\\(^{-1}\\)\nN = Noe\\(^{- \\lambda t}\\)\ne\\(^{0.0001244t}\\) = 1.2632\n0.0001244t = 0.234\nt = 1878 years\n\n\n\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(25, '2015', '\n\nQuestion\n\nState the dimension of; (a) impulse; (ii) acceleration; (iii) work\n\n\n', 'State the dimension of; (a) impulse; (ii) acceleration; (iii) work\n', '', '', NULL, '\n\nExplanation\nQuality\nDimensions\n\n\n\na) Impulse\nb) Acceleration\nc) Work\n\n\nMLT\\(^{-1}\\)\nLT\\(^{-2}\\)\nML\\(^2\\)T\\(^{-2}\\)\n\n\n\n\n\n\n', 'Quality\nDimensions\n\n\n\na) Impulse\nb) Acceleration\nc) Work\n\n\nMLT\\(^{-1}\\)\nLT\\(^{-2}\\)\nML\\(^2\\)T\\(^{-2}\\)\n\n\n\n\n\n\n', '', '', 'WAEC-PK-1', 'WAEC-PHY', 1),
(26, '2015', '\n\nQuestion\n\nA projectile is fired with a velocity of 20 ms\\(^{-1}\\) at an angle of 40° to the horizontal. Determine the components of the velocity of the projectile at its maximum height.\n\n\n', 'A projectile is fired with a velocity of 20 ms\\(^{-1}\\) at an angle of 40° to the horizontal. Determine the components of the velocity of the projectile at its maximum height.\n', '', '', NULL, '\n\nExplanation\nThe components of velocity at maximum height are:\nVy = 0\nVx = UCos \\(\\theta\\) = 20 Cos 40°\n=15.32ms\\(^{-1}\\)\n\n\n', 'The components of velocity at maximum height are:\nVy = 0\nVx = UCos \\(\\theta\\) = 20 Cos 40°\n=15.32ms\\(^{-1}\\)\n\n\n', '', '', 'WAEC-PK-1', 'WAEC-PHY', 1),
(27, '2015', '\n\nQuestion\n\nState three different materials that can be used to demonstrate Brownian motion.\n\n', 'State three different materials that can be used to demonstrate Brownian motion.', '', '', NULL, '\n\nExplanation\nMaterials used to demonstrate Brownian motion are: Pollen grains, dust particles, smoke particles, potassium permanganate cystals, carbon particles, etc.\n\n', 'Materials used to demonstrate Brownian motion are: Pollen grains, dust particles, smoke particles, potassium permanganate cystals, carbon particles, etc.\n\n', '', '', 'WAEC-PK-1', 'WAEC-PHY', 1),
(28, '2015', '\n\nQuestion\n\nAn electron enters perpendicularly into a uniform magnetic field which has a flux density of 0.12 T This results in a magnetic force of 9.6 x 10\\(^{-2}\\) N on the electron. Calculate the speed of the electron as it enters the magnetic field. (e =1.6 x \\(10^{19}\\) C)\n\n', 'An electron enters perpendicularly into a uniform magnetic field which has a flux density of 0.12 T This results in a magnetic force of 9.6 x 10\\(^{-2}\\) N on the electron. Calculate the speed of the electron as it enters the magnetic field. (e =1.6 x \\(10^{19}\\) C)', '', '', NULL, '\n\nExplanation\nF = evB sin \\(\\theta\\)\nV = \\(\\frac{F}{evB sin \\theta} = \\frac{9.6 \\times 10^{12}}{1.6 \\times 10^{-19} \\times 0.12 sin 90^o}\\)\n= \\(\\frac{9.6 \\times 10^{-12}}{1.6 x 10^{-19} \\times 0.12 \\times 1}\\)\n= 5.0 x 10\\(^8 ms{-1}\\)\n\n\n\n', 'F = evB sin \\(\\theta\\)\nV = \\(\\frac{F}{evB sin \\theta} = \\frac{9.6 \\times 10^{12}}{1.6 \\times 10^{-19} \\times 0.12 sin 90^o}\\)\n= \\(\\frac{9.6 \\times 10^{-12}}{1.6 x 10^{-19} \\times 0.12 \\times 1}\\)\n= 5.0 x 10\\(^8 ms{-1}\\)\n\n\n\n', '', '', 'WAEC-PK-1', 'WAEC-PHY', 1),
(29, '2015', '\n\nQuestion\n\nList three uses of rockets.\n\n', 'List three uses of rockets.', '', '', NULL, '\n\nExplanation\nUses rockets are:\n— for space travel\n— for space exploration\n— for warfare\n— for launching of artificial satellites into orbits\n— for entertainment, that is fireworks\n\n\n', 'Uses rockets are:\n— for space travel\n— for space exploration\n— for warfare\n— for launching of artificial satellites into orbits\n— for entertainment, that is fireworks\n\n\n', '', '', 'WAEC-PK-1', 'WAEC-PHY', 1),
(30, '2015', '\n\nQuestion\n\n(a) What is doping?\n(b) Explain how doping improves the conductivity of a semiconductor\n\n', '', '(a) What is doping?\n(b) Explain how doping improves the conductivity of a semiconductor', '(a) What is doping?~(b) Explain how doping improves the conductivity of a semiconductor', NULL, '\n\nExplanation\n(a) Doping is the introduction of impurity atoms into a semicondoctor\n(b) Doping increases the conductivity of a semicondoctor by increasing the number of chargers or electron or holes.\n\n', '', 'Explanation\n(a) Doping is the introduction of impurity atoms into a semicondoctor\n(b) Doping increases the conductivity of a semicondoctor by increasing the number of chargers or electron or holes.\n\n', '(a) Doping is the introduction of impurity atoms into a semicondoctor~(b) Doping increases the conductivity of a semicondoctor by increasing the number of chargers or electron or holes.\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(31, '2015', '\n\nQuestion\n\nThe diagram above illustrates a cathode ray tube. Identify the components X, Y, and Z.\n\n\n', 'The diagram above illustrates a cathode ray tube. Identify the components X, Y, and Z.\n', '', '', NULL, '\n\nExplanation\nThe cathode ray tube components are:\nX = Filament or heater or cathode\nY = Anode\nZ = Plates or deflectors.\n\n', 'The cathode ray tube components are:\nX = Filament or heater or cathode\nY = Anode\nZ = Plates or deflectors.\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
(32, '2015', '\n\nQuestion\n\n(a) Explain the term net force.\n(b) Define the principle of conservation of linear momentum and state one example of it.\n(c) A ball of mass 200 g released from a height of 2.0 m hits a horizontal floor and rebounds to a height of 1.8 in. Calculate the impulse received by the floor. (g = 10 ms\\(^{-2}\\)).\n(d) A body of mass 20 g performs a simple harmonic motion at a frequency of 5 Hz. At a distance of 10 cm from the mean position, its velocity is 200 cms\\(^{-1}\\). Calculate its:\n(i) maximum displacement from the mean position;\n(ii) maximum velocity;\n(iii) maximum potential energy. (g = 10 ms\\(^{-2}\\) \\(\\pi\\) = 3.14)\n\n', '', '(a) Explain the term net force.\n(b) Define the principle of conservation of linear momentum and state one example of it.\n(c) A ball of mass 200 g released from a height of 2.0 m hits a horizontal floor and rebounds to a height of 1.8 in. Calculate the impulse received by the floor. (g = 10 ms\\(^{-2}\\)).\n(d) A body of mass 20 g performs a simple harmonic motion at a frequency of 5 Hz. At a distance of 10 cm from the mean position, its velocity is 200 cms\\(^{-1}\\). Calculate its:\n(i) maximum displacement from the mean position;\n(ii) maximum velocity;\n(iii) maximum potential energy. (g = 10 ms\\(^{-2}\\) \\(\\pi\\) = 3.14)', '(a) Explain the term net force.~(b) Define the principle of conservation of linear momentum and state one example of it.\n(c) A ball of mass 200 g released from a height of 2.0 m hits a horizontal floor and rebounds to a height of 1.8 in. Calculate the impulse received by the floor. (g = 10 ms\\(^{-2}\\)).\n(d) A body of mass 20 g performs a simple harmonic motion at a frequency of 5 Hz. At a distance of 10 cm from the mean position, its velocity is 200 cms\\(^{-1}\\). Calculate its:\n(i) maximum displacement from the mean position;\n(ii) maximum velocity;\n(iii) maximum potential energy. (g = 10 ms\\(^{-2}\\) \\(\\pi\\) = 3.14)~(c) A ball of mass 200 g released from a height of 2.0 m hits a horizontal floor and rebounds to a height of 1.8 in. Calculate the impulse received by the floor. (g = 10 ms\\(^{-2}\\)).\n(d) A body of mass 20 g performs a simple harmonic motion at a frequency of 5 Hz. At a distance of 10 cm from the mean position, its velocity is 200 cms\\(^{-1}\\). Calculate its:\n(i) maximum displacement from the mean position;\n(ii) maximum velocity;\n(iii) maximum potential energy. (g = 10 ms\\(^{-2}\\) \\(\\pi\\) = 3.14)~(d) A body of mass 20 g performs a simple harmonic motion at a frequency of 5 Hz. At a distance of 10 cm from the mean position, its velocity is 200 cms\\(^{-1}\\). Calculate its:\n(i) maximum displacement from the mean position;\n(ii) maximum velocity;\n(iii) maximum potential energy. (g = 10 ms\\(^{-2}\\) \\(\\pi\\) = 3.14)', NULL, '\n\nExplanation\n(a) Net force is the effective or resultant force resulting from the actions of a system of forces on a body.\n\n(b) The principle ofconservation of linear momentum explains that in an isolated or closed system of colliding bodies the total linear momentum in a fixed direction remains constant, e.g. the recoil of gun, colliding trolleys, rocket propulsion, etc.\n\n(c) V\\(_1^2\\) = \\(U_1^2 + 2gh_1\\) = 0 + 2 x 10 x 2\n= 40\n\\(V_1 = \\sqrt{40}\\) = 6.325m/s\n\\(v_2^2 = u_2^2 - 2gh_2\\)\nO = \\(U^2_2\\) - 2 x 10 x 1.8\nu = \\(\\sqrt{36}\\)\n= 6ms\\(^{-1}\\)\nImpulse Change in momentun\nmv\\(_1\\) - (-mu\\(_2\\)) = m (v\\(_1\\) + u\\(_2\\))\n= 0.2(6.325 + 6.000)\n= 2.46 Ns\n\n(d) v = \\(\\omega \\sqrt{r^2 - y^2}\\)\n2\\(\\pi\\)f\\(\\sqrt{r^2 - y^2}\\)\n2 = 2 x 3.14 x 5 x \\(\\sqrt{r^2 - 0.1^2}\\)\nr = 0.12m\n(ii) Vmax = \\(\\omega\\)r = 2\\(\\pi\\)fr\n= 2 x 3.142 x 5 x 0.12\n= 3,77 m/s\n(iii) P.E = \\(\\frac{1}{2}m \\omega^2e^2 = \\times \\frac{1}{2}m(2 \\pi f)^2r^2\\)\n= \\(\\frac{1}{2} \\times 0.02(2 \\times 3.142 \\times 5)^2 \\times (0.12)^2\\)\n= 1.42 x 10\\(^{-1}J\\)\n\n', '', 'Explanation\n(a) Net force is the effective or resultant force resulting from the actions of a system of forces on a body.\n\n(b) The principle ofconservation of linear momentum explains that in an isolated or closed system of colliding bodies the total linear momentum in a fixed direction remains constant, e.g. the recoil of gun, colliding trolleys, rocket propulsion, etc.\n\n(c) V\\(_1^2\\) = \\(U_1^2 + 2gh_1\\) = 0 + 2 x 10 x 2\n= 40\n\\(V_1 = \\sqrt{40}\\) = 6.325m/s\n\\(v_2^2 = u_2^2 - 2gh_2\\)\nO = \\(U^2_2\\) - 2 x 10 x 1.8\nu = \\(\\sqrt{36}\\)\n= 6ms\\(^{-1}\\)\nImpulse Change in momentun\nmv\\(_1\\) - (-mu\\(_2\\)) = m (v\\(_1\\) + u\\(_2\\))\n= 0.2(6.325 + 6.000)\n= 2.46 Ns\n\n(d) v = \\(\\omega \\sqrt{r^2 - y^2}\\)\n2\\(\\pi\\)f\\(\\sqrt{r^2 - y^2}\\)\n2 = 2 x 3.14 x 5 x \\(\\sqrt{r^2 - 0.1^2}\\)\nr = 0.12m\n(ii) Vmax = \\(\\omega\\)r = 2\\(\\pi\\)fr\n= 2 x 3.142 x 5 x 0.12\n= 3,77 m/s\n(iii) P.E = \\(\\frac{1}{2}m \\omega^2e^2 = \\times \\frac{1}{2}m(2 \\pi f)^2r^2\\)\n= \\(\\frac{1}{2} \\times 0.02(2 \\times 3.142 \\times 5)^2 \\times (0.12)^2\\)\n= 1.42 x 10\\(^{-1}J\\)\n\n', '(a) Net force is the effective or resultant force resulting from the actions of a system of forces on a body.\n~(b) The principle ofconservation of linear momentum explains that in an isolated or closed system of colliding bodies the total linear momentum in a fixed direction remains constant, e.g. the recoil of gun, colliding trolleys, rocket propulsion, etc.\n\n(c) V\\(_1^2\\) = \\(U_1^2 + 2gh_1\\) = 0 + 2 x 10 x 2\n= 40\n\\(V_1 = \\sqrt{40}\\) = 6.325m/s\n\\(v_2^2 = u_2^2 - 2gh_2\\)\nO = \\(U^2_2\\) - 2 x 10 x 1.8\nu = \\(\\sqrt{36}\\)\n= 6ms\\(^{-1}\\)\nImpulse Change in momentun\nmv\\(_1\\) - (-mu\\(_2\\)) = m (v\\(_1\\) + u\\(_2\\))\n= 0.2(6.325 + 6.000)\n= 2.46 Ns\n\n(d) v = \\(\\omega \\sqrt{r^2 - y^2}\\)\n2\\(\\pi\\)f\\(\\sqrt{r^2 - y^2}\\)\n2 = 2 x 3.14 x 5 x \\(\\sqrt{r^2 - 0.1^2}\\)\nr = 0.12m\n(ii) Vmax = \\(\\omega\\)r = 2\\(\\pi\\)fr\n= 2 x 3.142 x 5 x 0.12\n= 3,77 m/s\n(iii) P.E = \\(\\frac{1}{2}m \\omega^2e^2 = \\times \\frac{1}{2}m(2 \\pi f)^2r^2\\)\n= \\(\\frac{1}{2} \\times 0.02(2 \\times 3.142 \\times 5)^2 \\times (0.12)^2\\)\n= 1.42 x 10\\(^{-1}J\\)\n~(c) V\\(_1^2\\) = \\(U_1^2 + 2gh_1\\) = 0 + 2 x 10 x 2\n= 40\n\\(V_1 = \\sqrt{40}\\) = 6.325m/s\n\\(v_2^2 = u_2^2 - 2gh_2\\)\nO = \\(U^2_2\\) - 2 x 10 x 1.8\nu = \\(\\sqrt{36}\\)\n= 6ms\\(^{-1}\\)\nImpulse Change in momentun\nmv\\(_1\\) - (-mu\\(_2\\)) = m (v\\(_1\\) + u\\(_2\\))\n= 0.2(6.325 + 6.000)\n= 2.46 Ns\n\n(d) v = \\(\\omega \\sqrt{r^2 - y^2}\\)\n2\\(\\pi\\)f\\(\\sqrt{r^2 - y^2}\\)\n2 = 2 x 3.14 x 5 x \\(\\sqrt{r^2 - 0.1^2}\\)\nr = 0.12m\n(ii) Vmax = \\(\\omega\\)r = 2\\(\\pi\\)fr\n= 2 x 3.142 x 5 x 0.12\n= 3,77 m/s\n(iii) P.E = \\(\\frac{1}{2}m \\omega^2e^2 = \\times \\frac{1}{2}m(2 \\pi f)^2r^2\\)\n= \\(\\frac{1}{2} \\times 0.02(2 \\times 3.142 \\times 5)^2 \\times (0.12)^2\\)\n= 1.42 x 10\\(^{-1}J\\)\n~(d) v = \\(\\omega \\sqrt{r^2 - y^2}\\)\n2\\(\\pi\\)f\\(\\sqrt{r^2 - y^2}\\)\n2 = 2 x 3.14 x 5 x \\(\\sqrt{r^2 - 0.1^2}\\)\nr = 0.12m\n(ii) Vmax = \\(\\omega\\)r = 2\\(\\pi\\)fr\n= 2 x 3.142 x 5 x 0.12\n= 3,77 m/s\n(iii) P.E = \\(\\frac{1}{2}m \\omega^2e^2 = \\times \\frac{1}{2}m(2 \\pi f)^2r^2\\)\n= \\(\\frac{1}{2} \\times 0.02(2 \\times 3.142 \\times 5)^2 \\times (0.12)^2\\)\n= 1.42 x 10\\(^{-1}J\\)\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(33, '2015', '\n\nQuestion\n\n(a) Explain the terms:\n(i) thermal equilibrium;\n(ii) fundamental interval.\n\n(b) List two uses of the hydraulic press.\n\n(c) Name the material used to reset the steel index in the Six\'s maximum and minimum thermometer.\n\n(d)(i) A nursing mother prepared her baby\'s milk mixture at 85°C, in a feeding bottle. In order to cool it to 40°C, she immersed the bottle in an aluminium bowl of heat capacity 90 JK\\(^{-1}\\) containing 500 g of water at 26°C. If the mass of the mixture is 300g, calculate the specific heat capacity of the mixture. [Neglect heat losses and heat capacity of the bottle; specific heat capacity of water = 4200 J kg\\(^{-1}\\) K\\(^{-1}\\)]\n(ii) (\\(\\alpha\\)) Name two ways through which the bottle losses heat.\n(\\(\\beta\\)) Name two industrial processes in which heat exchanger is used.\n\n', '', '(a) Explain the terms:\n(i) thermal equilibrium;\n(ii) fundamental interval.\n\n(b) List two uses of the hydraulic press.\n\n(c) Name the material used to reset the steel index in the Six\'s maximum and minimum thermometer.\n\n(d)(i) A nursing mother prepared her baby\'s milk mixture at 85°C, in a feeding bottle. In order to cool it to 40°C, she immersed the bottle in an aluminium bowl of heat capacity 90 JK\\(^{-1}\\) containing 500 g of water at 26°C. If the mass of the mixture is 300g, calculate the specific heat capacity of the mixture. [Neglect heat losses and heat capacity of the bottle; specific heat capacity of water = 4200 J kg\\(^{-1}\\) K\\(^{-1}\\)]\n(ii) (\\(\\alpha\\)) Name two ways through which the bottle losses heat.\n(\\(\\beta\\)) Name two industrial processes in which heat exchanger is used.', '(a) Explain the terms:\n(i) thermal equilibrium;\n(ii) fundamental interval.\n~(b) List two uses of the hydraulic press.\n\n(c) Name the material used to reset the steel index in the Six\'s maximum and minimum thermometer.\n\n(d)(i) A nursing mother prepared her baby\'s milk mixture at 85°C, in a feeding bottle. In order to cool it to 40°C, she immersed the bottle in an aluminium bowl of heat capacity 90 JK\\(^{-1}\\) containing 500 g of water at 26°C. If the mass of the mixture is 300g, calculate the specific heat capacity of the mixture. [Neglect heat losses and heat capacity of the bottle; specific heat capacity of water = 4200 J kg\\(^{-1}\\) K\\(^{-1}\\)]\n(ii) (\\(\\alpha\\)) Name two ways through which the bottle losses heat.\n(\\(\\beta\\)) Name two industrial processes in which heat exchanger is used.~(c) Name the material used to reset the steel index in the Six\'s maximum and minimum thermometer.\n\n(d)(i) A nursing mother prepared her baby\'s milk mixture at 85°C, in a feeding bottle. In order to cool it to 40°C, she immersed the bottle in an aluminium bowl of heat capacity 90 JK\\(^{-1}\\) containing 500 g of water at 26°C. If the mass of the mixture is 300g, calculate the specific heat capacity of the mixture. [Neglect heat losses and heat capacity of the bottle; specific heat capacity of water = 4200 J kg\\(^{-1}\\) K\\(^{-1}\\)]\n(ii) (\\(\\alpha\\)) Name two ways through which the bottle losses heat.\n(\\(\\beta\\)) Name two industrial processes in which heat exchanger is used.~(d)(i) A nursing mother prepared her baby\'s milk mixture at 85°C, in a feeding bottle. In order to cool it to 40°C, she immersed the bottle in an aluminium bowl of heat capacity 90 JK\\(^{-1}\\) containing 500 g of water at 26°C. If the mass of the mixture is 300g, calculate the specific heat capacity of the mixture. [Neglect heat losses and heat capacity of the bottle; specific heat capacity of water = 4200 J kg\\(^{-1}\\) K\\(^{-1}\\)]\n(ii) (\\(\\alpha\\)) Name two ways through which the bottle losses heat.\n(\\(\\beta\\)) Name two industrial processes in which heat exchanger is used.', NULL, '\n\nExplanation\n(a)(i) Thermal equilibrium is a condition in which two bodies in contact have no net flow of heat between them.\n(ii) Fundamental interval is the difference in temperature between the lower fixed point, i.e. temperature of melting ice at standard atmospheric pressure and that upper fixed point, i.e. temperature of steam at standard atmospheric pressure.\n\n(b) Uses of hydraulic press are:-\n-Compressing metal sheets;\n-Compressing soft materials into bales;\n-Compressing paper in printing industries;\n-Lifting of objects.\n\n(c) The material used to reset the steel index in the six\'s maximum and minimum thermometer is magnet.\n\n(d)(i) Heat lost by mixture = Heat gained by water + Heat gained by aluminium.\nM\\(_mC_w\\Delta \\theta_m\\) + C\\(_a\\Delta\\theta_a\\)\n0.3 x C\\(_m\\) x (85 - 40) = 0.5 x 4200 (40 - 26) + 90 (40 - 26)\n13.5 C\\(_m\\) = 29400 + 1260\nC\\(_m\\) = \\(\\frac{30660}{13.5}\\)\n= 2271.1 Jkg\\(^{-1}\\)K\\(^{-1}\\)\n\n(ii) (\\(\\alpha\\)) The bottle losses heat through conduction and convection. (\\(\\beta\\)) Heat exchanger in: Incubation, refriaeration, air-conditioning, sewage treatment, cooling byradiators, generation of electricity, etc.\n\n', '', 'Explanation\n(a)(i) Thermal equilibrium is a condition in which two bodies in contact have no net flow of heat between them.\n(ii) Fundamental interval is the difference in temperature between the lower fixed point, i.e. temperature of melting ice at standard atmospheric pressure and that upper fixed point, i.e. temperature of steam at standard atmospheric pressure.\n\n(b) Uses of hydraulic press are:-\n-Compressing metal sheets;\n-Compressing soft materials into bales;\n-Compressing paper in printing industries;\n-Lifting of objects.\n\n(c) The material used to reset the steel index in the six\'s maximum and minimum thermometer is magnet.\n\n(d)(i) Heat lost by mixture = Heat gained by water + Heat gained by aluminium.\nM\\(_mC_w\\Delta \\theta_m\\) + C\\(_a\\Delta\\theta_a\\)\n0.3 x C\\(_m\\) x (85 - 40) = 0.5 x 4200 (40 - 26) + 90 (40 - 26)\n13.5 C\\(_m\\) = 29400 + 1260\nC\\(_m\\) = \\(\\frac{30660}{13.5}\\)\n= 2271.1 Jkg\\(^{-1}\\)K\\(^{-1}\\)\n\n(ii) (\\(\\alpha\\)) The bottle losses heat through conduction and convection. (\\(\\beta\\)) Heat exchanger in: Incubation, refriaeration, air-conditioning, sewage treatment, cooling byradiators, generation of electricity, etc.\n\n', '(a)(i) Thermal equilibrium is a condition in which two bodies in contact have no net flow of heat between them.\n(ii) Fundamental interval is the difference in temperature between the lower fixed point, i.e. temperature of melting ice at standard atmospheric pressure and that upper fixed point, i.e. temperature of steam at standard atmospheric pressure.\n~(b) Uses of hydraulic press are:-\n-Compressing metal sheets;\n-Compressing soft materials into bales;\n-Compressing paper in printing industries;\n-Lifting of objects.\n\n(c) The material used to reset the steel index in the six\'s maximum and minimum thermometer is magnet.\n\n(d)(i) Heat lost by mixture = Heat gained by water + Heat gained by aluminium.\nM\\(_mC_w\\Delta \\theta_m\\) + C\\(_a\\Delta\\theta_a\\)\n0.3 x C\\(_m\\) x (85 - 40) = 0.5 x 4200 (40 - 26) + 90 (40 - 26)\n13.5 C\\(_m\\) = 29400 + 1260\nC\\(_m\\) = \\(\\frac{30660}{13.5}\\)\n= 2271.1 Jkg\\(^{-1}\\)K\\(^{-1}\\)\n\n(ii) (\\(\\alpha\\)) The bottle losses heat through conduction and convection. (\\(\\beta\\)) Heat exchanger in: Incubation, refriaeration, air-conditioning, sewage treatment, cooling byradiators, generation of electricity, etc.\n~(c) The material used to reset the steel index in the six\'s maximum and minimum thermometer is magnet.\n\n(d)(i) Heat lost by mixture = Heat gained by water + Heat gained by aluminium.\nM\\(_mC_w\\Delta \\theta_m\\) + C\\(_a\\Delta\\theta_a\\)\n0.3 x C\\(_m\\) x (85 - 40) = 0.5 x 4200 (40 - 26) + 90 (40 - 26)\n13.5 C\\(_m\\) = 29400 + 1260\nC\\(_m\\) = \\(\\frac{30660}{13.5}\\)\n= 2271.1 Jkg\\(^{-1}\\)K\\(^{-1}\\)\n\n(ii) (\\(\\alpha\\)) The bottle losses heat through conduction and convection. (\\(\\beta\\)) Heat exchanger in: Incubation, refriaeration, air-conditioning, sewage treatment, cooling byradiators, generation of electricity, etc.\n~(d)(i) Heat lost by mixture = Heat gained by water + Heat gained by aluminium.\nM\\(_mC_w\\Delta \\theta_m\\) + C\\(_a\\Delta\\theta_a\\)\n0.3 x C\\(_m\\) x (85 - 40) = 0.5 x 4200 (40 - 26) + 90 (40 - 26)\n13.5 C\\(_m\\) = 29400 + 1260\nC\\(_m\\) = \\(\\frac{30660}{13.5}\\)\n= 2271.1 Jkg\\(^{-1}\\)K\\(^{-1}\\)\n\n(ii) (\\(\\alpha\\)) The bottle losses heat through conduction and convection. (\\(\\beta\\)) Heat exchanger in: Incubation, refriaeration, air-conditioning, sewage treatment, cooling byradiators, generation of electricity, etc.\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(34, '2015', '\n\nQuestion\n\n(a) Define critical angle.\n(b) How are anti-nodes created in a stationary wave?\n(c) The angle of minimum deviation of an equilateral triangular glass prism is 46.2°. Calculate the refractive index of the glass.\n(d) An illuminated object is placed in front of a concave mirror and the position of a screen is adjusted in front of the mirror but no image is obtained on the screen. Give two possible reasons for this observation.\n(e) An illuminated object is placed at a distance of 75 cm from a converging lens of focal length 30 cm.\n(i) Determine the image distance.\n(ii) If the lens is replaced by another converging lens, the object has to be moved 25 cm further away to have its sharp image on the screen. Determine the focal length of the second lens.\n\n', '', '(a) Define critical angle.\n(b) How are anti-nodes created in a stationary wave?\n(c) The angle of minimum deviation of an equilateral triangular glass prism is 46.2°. Calculate the refractive index of the glass.\n(d) An illuminated object is placed in front of a concave mirror and the position of a screen is adjusted in front of the mirror but no image is obtained on the screen. Give two possible reasons for this observation.\n(e) An illuminated object is placed at a distance of 75 cm from a converging lens of focal length 30 cm.\n(i) Determine the image distance.\n(ii) If the lens is replaced by another converging lens, the object has to be moved 25 cm further away to have its sharp image on the screen. Determine the focal length of the second lens.', '(a) Define critical angle.~(b) How are anti-nodes created in a stationary wave?\n(c) The angle of minimum deviation of an equilateral triangular glass prism is 46.2°. Calculate the refractive index of the glass.\n(d) An illuminated object is placed in front of a concave mirror and the position of a screen is adjusted in front of the mirror but no image is obtained on the screen. Give two possible reasons for this observation.\n(e) An illuminated object is placed at a distance of 75 cm from a converging lens of focal length 30 cm.\n(i) Determine the image distance.\n(ii) If the lens is replaced by another converging lens, the object has to be moved 25 cm further away to have its sharp image on the screen. Determine the focal length of the second lens.~(c) The angle of minimum deviation of an equilateral triangular glass prism is 46.2°. Calculate the refractive index of the glass.\n(d) An illuminated object is placed in front of a concave mirror and the position of a screen is adjusted in front of the mirror but no image is obtained on the screen. Give two possible reasons for this observation.\n(e) An illuminated object is placed at a distance of 75 cm from a converging lens of focal length 30 cm.\n(i) Determine the image distance.\n(ii) If the lens is replaced by another converging lens, the object has to be moved 25 cm further away to have its sharp image on the screen. Determine the focal length of the second lens.~(d) An illuminated object is placed in front of a concave mirror and the position of a screen is adjusted in front of the mirror but no image is obtained on the screen. Give two possible reasons for this observation.\n(e) An illuminated object is placed at a distance of 75 cm from a converging lens of focal length 30 cm.\n(i) Determine the image distance.\n(ii) If the lens is replaced by another converging lens, the object has to be moved 25 cm further away to have its sharp image on the screen. Determine the focal length of the second lens.~(e) An illuminated object is placed at a distance of 75 cm from a converging lens of focal length 30 cm.\n(i) Determine the image distance.\n(ii) If the lens is replaced by another converging lens, the object has to be moved 25 cm further away to have its sharp image on the screen. Determine the focal length of the second lens.', NULL, '\n\nExplanation\n(a) Critical angle is defined as the angle of incidence in the denser medium for which the angle of refraction in the less dense medium is 90°.\n\n(b) Antinodes are created in a stationary wave when an incident and its reflected waves are superposed and 180° out of phase.\n\n(c) n = \\(\\frac{Sin\\frac{(A + Dm)}{2}}{\\sin (\\frac{A}{2})}\\)\n= \\(\\frac{Sin\\frac{(60 + 46.2)}{2}}{\\sin (\\frac{60}{2})}\\)\n= \\(\\frac{sin 53.1}{sin 30}\\)\n= 1.6\n(d) No image is obtained on the screen because\n- Object, screen and mirror are not coaxial;\n- Object is placed at focal point of mirror;\n- Object is placed between the focus and the pole;\n- Screen is located at a distance less than or greater than actual image distance.\n\n(e) (i) \\(\\frac{1}{u} + \\frac{1}{v} = \\frac{1}{f}\\)\n\\(\\frac{1}{v} = \\frac{1}{f} - \\frac{1}{u} = \\frac{1}{30} - \\frac{1}{7}\\)\n= \\(\\frac{5 - 2}{150} = \\frac{3}{150}\\)\nU = 75 + 25 = 100cm\n\\(\\frac{1}{u} + \\frac{1}{v} = \\frac{1}{f}\\)\n\\(\\frac{1}{100} + \\frac{1}{50} = \\frac{1}{50} = \\frac{1}{50}\\)\nf = 50cm\n\n\n\n', '', 'Explanation\n(a) Critical angle is defined as the angle of incidence in the denser medium for which the angle of refraction in the less dense medium is 90°.\n\n(b) Antinodes are created in a stationary wave when an incident and its reflected waves are superposed and 180° out of phase.\n\n(c) n = \\(\\frac{Sin\\frac{(A + Dm)}{2}}{\\sin (\\frac{A}{2})}\\)\n= \\(\\frac{Sin\\frac{(60 + 46.2)}{2}}{\\sin (\\frac{60}{2})}\\)\n= \\(\\frac{sin 53.1}{sin 30}\\)\n= 1.6\n(d) No image is obtained on the screen because\n- Object, screen and mirror are not coaxial;\n- Object is placed at focal point of mirror;\n- Object is placed between the focus and the pole;\n- Screen is located at a distance less than or greater than actual image distance.\n\n(e) (i) \\(\\frac{1}{u} + \\frac{1}{v} = \\frac{1}{f}\\)\n\\(\\frac{1}{v} = \\frac{1}{f} - \\frac{1}{u} = \\frac{1}{30} - \\frac{1}{7}\\)\n= \\(\\frac{5 - 2}{150} = \\frac{3}{150}\\)\nU = 75 + 25 = 100cm\n\\(\\frac{1}{u} + \\frac{1}{v} = \\frac{1}{f}\\)\n\\(\\frac{1}{100} + \\frac{1}{50} = \\frac{1}{50} = \\frac{1}{50}\\)\nf = 50cm\n\n\n\n', '(a) Critical angle is defined as the angle of incidence in the denser medium for which the angle of refraction in the less dense medium is 90°.\n~(b) Antinodes are created in a stationary wave when an incident and its reflected waves are superposed and 180° out of phase.\n\n(c) n = \\(\\frac{Sin\\frac{(A + Dm)}{2}}{\\sin (\\frac{A}{2})}\\)\n= \\(\\frac{Sin\\frac{(60 + 46.2)}{2}}{\\sin (\\frac{60}{2})}\\)\n= \\(\\frac{sin 53.1}{sin 30}\\)\n= 1.6\n(d) No image is obtained on the screen because\n- Object, screen and mirror are not coaxial;\n- Object is placed at focal point of mirror;\n- Object is placed between the focus and the pole;\n- Screen is located at a distance less than or greater than actual image distance.\n\n(e) (i) \\(\\frac{1}{u} + \\frac{1}{v} = \\frac{1}{f}\\)\n\\(\\frac{1}{v} = \\frac{1}{f} - \\frac{1}{u} = \\frac{1}{30} - \\frac{1}{7}\\)\n= \\(\\frac{5 - 2}{150} = \\frac{3}{150}\\)\nU = 75 + 25 = 100cm\n\\(\\frac{1}{u} + \\frac{1}{v} = \\frac{1}{f}\\)\n\\(\\frac{1}{100} + \\frac{1}{50} = \\frac{1}{50} = \\frac{1}{50}\\)\nf = 50cm\n\n\n~(c) n = \\(\\frac{Sin\\frac{(A + Dm)}{2}}{\\sin (\\frac{A}{2})}\\)\n= \\(\\frac{Sin\\frac{(60 + 46.2)}{2}}{\\sin (\\frac{60}{2})}\\)\n= \\(\\frac{sin 53.1}{sin 30}\\)\n= 1.6\n(d) No image is obtained on the screen because\n- Object, screen and mirror are not coaxial;\n- Object is placed at focal point of mirror;\n- Object is placed between the focus and the pole;\n- Screen is located at a distance less than or greater than actual image distance.\n\n(e) (i) \\(\\frac{1}{u} + \\frac{1}{v} = \\frac{1}{f}\\)\n\\(\\frac{1}{v} = \\frac{1}{f} - \\frac{1}{u} = \\frac{1}{30} - \\frac{1}{7}\\)\n= \\(\\frac{5 - 2}{150} = \\frac{3}{150}\\)\nU = 75 + 25 = 100cm\n\\(\\frac{1}{u} + \\frac{1}{v} = \\frac{1}{f}\\)\n\\(\\frac{1}{100} + \\frac{1}{50} = \\frac{1}{50} = \\frac{1}{50}\\)\nf = 50cm\n\n\n~(d) No image is obtained on the screen because\n- Object, screen and mirror are not coaxial;\n- Object is placed at focal point of mirror;\n- Object is placed between the focus and the pole;\n- Screen is located at a distance less than or greater than actual image distance.\n\n(e) (i) \\(\\frac{1}{u} + \\frac{1}{v} = \\frac{1}{f}\\)\n\\(\\frac{1}{v} = \\frac{1}{f} - \\frac{1}{u} = \\frac{1}{30} - \\frac{1}{7}\\)\n= \\(\\frac{5 - 2}{150} = \\frac{3}{150}\\)\nU = 75 + 25 = 100cm\n\\(\\frac{1}{u} + \\frac{1}{v} = \\frac{1}{f}\\)\n\\(\\frac{1}{100} + \\frac{1}{50} = \\frac{1}{50} = \\frac{1}{50}\\)\nf = 50cm\n\n\n~(e) (i) \\(\\frac{1}{u} + \\frac{1}{v} = \\frac{1}{f}\\)\n\\(\\frac{1}{v} = \\frac{1}{f} - \\frac{1}{u} = \\frac{1}{30} - \\frac{1}{7}\\)\n= \\(\\frac{5 - 2}{150} = \\frac{3}{150}\\)\nU = 75 + 25 = 100cm\n\\(\\frac{1}{u} + \\frac{1}{v} = \\frac{1}{f}\\)\n\\(\\frac{1}{100} + \\frac{1}{50} = \\frac{1}{50} = \\frac{1}{50}\\)\nf = 50cm\n\n\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(35, '2015', '\n\nQuestion\n\n(a) Explain briefly dielectric strength.\n(b) An electromagnetic wave has its wavelength shorter than those of radiowave and microwave but longer than that of visible light.\n(i) Identify the wave.\n(ii) Name one suitable detector for the wave.\n(iii) Name one source of the wave.\n(c) An oil drop carrying a charge of 1.0 x 10\\(^{-19 }\\)C is found to remain at rest in a uniform electric field of intensity 1200 NC\\(^{-1}\\). Calculate the weight of the oil drop.\n(d) An RLC series circuit consists of a 100\\(\\Omega\\) resistor, 0.05 H inductor and a 25 \\(\\mu\\) capacitor. A 220 V, 50 Hz mains voltage is applied across the circuit. Calculate the:\n(i) impedance;\n(ii) current. (\\(\\pi\\) = 3.14)\n\n', '', '(a) Explain briefly dielectric strength.\n(b) An electromagnetic wave has its wavelength shorter than those of radiowave and microwave but longer than that of visible light.\n(i) Identify the wave.\n(ii) Name one suitable detector for the wave.\n(iii) Name one source of the wave.\n(c) An oil drop carrying a charge of 1.0 x 10\\(^{-19 }\\)C is found to remain at rest in a uniform electric field of intensity 1200 NC\\(^{-1}\\). Calculate the weight of the oil drop.\n(d) An RLC series circuit consists of a 100\\(\\Omega\\) resistor, 0.05 H inductor and a 25 \\(\\mu\\) capacitor. A 220 V, 50 Hz mains voltage is applied across the circuit. Calculate the:\n(i) impedance;\n(ii) current. (\\(\\pi\\) = 3.14)', '(a) Explain briefly dielectric strength.~(b) An electromagnetic wave has its wavelength shorter than those of radiowave and microwave but longer than that of visible light.\n(i) Identify the wave.\n(ii) Name one suitable detector for the wave.\n(iii) Name one source of the wave.\n(c) An oil drop carrying a charge of 1.0 x 10\\(^{-19 }\\)C is found to remain at rest in a uniform electric field of intensity 1200 NC\\(^{-1}\\). Calculate the weight of the oil drop.\n(d) An RLC series circuit consists of a 100\\(\\Omega\\) resistor, 0.05 H inductor and a 25 \\(\\mu\\) capacitor. A 220 V, 50 Hz mains voltage is applied across the circuit. Calculate the:\n(i) impedance;\n(ii) current. (\\(\\pi\\) = 3.14)~(c) An oil drop carrying a charge of 1.0 x 10\\(^{-19 }\\)C is found to remain at rest in a uniform electric field of intensity 1200 NC\\(^{-1}\\). Calculate the weight of the oil drop.\n(d) An RLC series circuit consists of a 100\\(\\Omega\\) resistor, 0.05 H inductor and a 25 \\(\\mu\\) capacitor. A 220 V, 50 Hz mains voltage is applied across the circuit. Calculate the:\n(i) impedance;\n(ii) current. (\\(\\pi\\) = 3.14)~(d) An RLC series circuit consists of a 100\\(\\Omega\\) resistor, 0.05 H inductor and a 25 \\(\\mu\\) capacitor. A 220 V, 50 Hz mains voltage is applied across the circuit. Calculate the:\n(i) impedance;\n(ii) current. (\\(\\pi\\) = 3.14)', NULL, '\n\nExplanation\na) Dielectric strength is the maximum potential gradient or electric field strength an insulator can withstand without breaking down. It is measured in Vmm\\(^{-1}\\) or Vm\\(^{-1}\\).\n(b)(i) The wave is infra-red wave.\n(ii) Suitable detector can be -Infra-red camera;\n-Thermopile;\n-Bolometer or\n-Photographic film.\n(iii) Source of the wave can be - Hot bodies or sun\n\n(c) E = \\(\\frac{F}{q} = \\frac{W}{q}\\)\nW =mF = Eq\n= 1200 x 1.0 x 10\\(^{-19}\\)\n= 1200 x 1.0 x 10\\(^{-19}\\)N\n\n\n(d) X\\(_L\\) = 1.2\\(\\pi fL\\)\n= 2 x 3.142 x 50 x 0.50 = 15.7\\(\\Omega\\)\nX\\(_c\\) \\(\\frac{1}{2 \\pi fL}\\)\n= \\(\\frac{1}{2 \\times 3.14 \\times 50 \\times 25 \\times 10^{-16}}\\)\n= 127.4 \\(\\Omega\\)\n(i) Z = \\(\\sqrt{R^2 + (X_1 - X_c)^2}\\)\n= \\(\\sqrt{100^2 + (127.4 - 15.7)^2}\\)\n= 149.9\\(\\Omega\\)\n(ii) I \\(\\frac{V}{Z} = \\frac{220}{149.9}\\)\n= 1.47A\n\n\n', '', 'Explanation\na) Dielectric strength is the maximum potential gradient or electric field strength an insulator can withstand without breaking down. It is measured in Vmm\\(^{-1}\\) or Vm\\(^{-1}\\).\n(b)(i) The wave is infra-red wave.\n(ii) Suitable detector can be -Infra-red camera;\n-Thermopile;\n-Bolometer or\n-Photographic film.\n(iii) Source of the wave can be - Hot bodies or sun\n\n(c) E = \\(\\frac{F}{q} = \\frac{W}{q}\\)\nW =mF = Eq\n= 1200 x 1.0 x 10\\(^{-19}\\)\n= 1200 x 1.0 x 10\\(^{-19}\\)N\n\n\n(d) X\\(_L\\) = 1.2\\(\\pi fL\\)\n= 2 x 3.142 x 50 x 0.50 = 15.7\\(\\Omega\\)\nX\\(_c\\) \\(\\frac{1}{2 \\pi fL}\\)\n= \\(\\frac{1}{2 \\times 3.14 \\times 50 \\times 25 \\times 10^{-16}}\\)\n= 127.4 \\(\\Omega\\)\n(i) Z = \\(\\sqrt{R^2 + (X_1 - X_c)^2}\\)\n= \\(\\sqrt{100^2 + (127.4 - 15.7)^2}\\)\n= 149.9\\(\\Omega\\)\n(ii) I \\(\\frac{V}{Z} = \\frac{220}{149.9}\\)\n= 1.47A\n\n\n', 'a) Dielectric strength is the maximum potential gradient or electric field strength an insulator can withstand without breaking down. It is measured in Vmm\\(^{-1}\\) or Vm\\(^{-1}\\).~(b)(i) The wave is infra-red wave.\n(ii) Suitable detector can be -Infra-red camera;\n-Thermopile;\n-Bolometer or\n-Photographic film.\n(iii) Source of the wave can be - Hot bodies or sun\n\n(c) E = \\(\\frac{F}{q} = \\frac{W}{q}\\)\nW =mF = Eq\n= 1200 x 1.0 x 10\\(^{-19}\\)\n= 1200 x 1.0 x 10\\(^{-19}\\)N\n\n\n(d) X\\(_L\\) = 1.2\\(\\pi fL\\)\n= 2 x 3.142 x 50 x 0.50 = 15.7\\(\\Omega\\)\nX\\(_c\\) \\(\\frac{1}{2 \\pi fL}\\)\n= \\(\\frac{1}{2 \\times 3.14 \\times 50 \\times 25 \\times 10^{-16}}\\)\n= 127.4 \\(\\Omega\\)\n(i) Z = \\(\\sqrt{R^2 + (X_1 - X_c)^2}\\)\n= \\(\\sqrt{100^2 + (127.4 - 15.7)^2}\\)\n= 149.9\\(\\Omega\\)\n(ii) I \\(\\frac{V}{Z} = \\frac{220}{149.9}\\)\n= 1.47A\n\n~(c) E = \\(\\frac{F}{q} = \\frac{W}{q}\\)\nW =mF = Eq\n= 1200 x 1.0 x 10\\(^{-19}\\)\n= 1200 x 1.0 x 10\\(^{-19}\\)N\n\n\n(d) X\\(_L\\) = 1.2\\(\\pi fL\\)\n= 2 x 3.142 x 50 x 0.50 = 15.7\\(\\Omega\\)\nX\\(_c\\) \\(\\frac{1}{2 \\pi fL}\\)\n= \\(\\frac{1}{2 \\times 3.14 \\times 50 \\times 25 \\times 10^{-16}}\\)\n= 127.4 \\(\\Omega\\)\n(i) Z = \\(\\sqrt{R^2 + (X_1 - X_c)^2}\\)\n= \\(\\sqrt{100^2 + (127.4 - 15.7)^2}\\)\n= 149.9\\(\\Omega\\)\n(ii) I \\(\\frac{V}{Z} = \\frac{220}{149.9}\\)\n= 1.47A\n\n~(d) X\\(_L\\) = 1.2\\(\\pi fL\\)\n= 2 x 3.142 x 50 x 0.50 = 15.7\\(\\Omega\\)\nX\\(_c\\) \\(\\frac{1}{2 \\pi fL}\\)\n= \\(\\frac{1}{2 \\times 3.14 \\times 50 \\times 25 \\times 10^{-16}}\\)\n= 127.4 \\(\\Omega\\)\n(i) Z = \\(\\sqrt{R^2 + (X_1 - X_c)^2}\\)\n= \\(\\sqrt{100^2 + (127.4 - 15.7)^2}\\)\n= 149.9\\(\\Omega\\)\n(ii) I \\(\\frac{V}{Z} = \\frac{220}{149.9}\\)\n= 1.47A\n\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(36, '2015', '\n\nQuestion\n\n(a) Explain the following terms:\n(i) mass defect;\n(ii) binding energy of a nucleus.\n\n(b)(i) Assuming the wave nature of an electron, what is the effect of decreasing the speed of a photoelectron on its; (\\(\\alpha\\)) wavelength? (\\(\\beta\\)) energy?\n(ii) A particle of friasS 4.4 x 10\\(^{-23}\\) kg moves with a velocity of 10\\(^5\\)ms\\(^{-1}\\). Calculate its wavelength. (h = 6.6 x 10\\(^{-34}\\) Js)\n\nThe diagram above shows part of a radioactive decay series. Use it to answer the following questions.\n(i) Name a pair of isotopes.\n(ii) Name the isotopes with which the series starts.\n(iii) Write down a nuclear equation for two ekgmples of each of: (\\(\\alpha\\)) alpha decay; (\\(\\beta\\)) beta decay.\n\n', '', '(a) Explain the following terms:\n(i) mass defect;\n(ii) binding energy of a nucleus.\n\n(b)(i) Assuming the wave nature of an electron, what is the effect of decreasing the speed of a photoelectron on its; (\\(\\alpha\\)) wavelength? (\\(\\beta\\)) energy?\n(ii) A particle of friasS 4.4 x 10\\(^{-23}\\) kg moves with a velocity of 10\\(^5\\)ms\\(^{-1}\\). Calculate its wavelength. (h = 6.6 x 10\\(^{-34}\\) Js)\n\nThe diagram above shows part of a radioactive decay series. Use it to answer the following questions.\n(i) Name a pair of isotopes.\n(ii) Name the isotopes with which the series starts.\n(iii) Write down a nuclear equation for two ekgmples of each of: (\\(\\alpha\\)) alpha decay; (\\(\\beta\\)) beta decay.', '(a) Explain the following terms:\n(i) mass defect;\n(ii) binding energy of a nucleus.\n~(b)(i) Assuming the wave nature of an electron, what is the effect of decreasing the speed of a photoelectron on its; (\\(\\alpha\\)) wavelength? (\\(\\beta\\)) energy?\n(ii) A particle of friasS 4.4 x 10\\(^{-23}\\) kg moves with a velocity of 10\\(^5\\)ms\\(^{-1}\\). Calculate its wavelength. (h = 6.6 x 10\\(^{-34}\\) Js)\n\nThe diagram above shows part of a radioactive decay series. Use it to answer the following questions.\n(i) Name a pair of isotopes.\n(ii) Name the isotopes with which the series starts.\n(iii) Write down a nuclear equation for two ekgmples of each of: (\\(\\alpha\\)) alpha decay; (\\(\\beta\\)) beta decay.', NULL, '\n\nExplanation\n(a)(i) Mass defect is the difference between the mass of a nucleus and the total mass of its nucleons. It expressed in atomic mass unit (a.m.u.). It is also a measure of binding energy.\nMass defect = total mass of nucleus\n– mass of nucleons.\n(ii) Binding energy is the work done or energy needed to separate the nucleons in a nucleus. It is measured in eV or Joules (J).\n\n(b)(i) The effect of decreasing the speed of a photoelectron on: (\\(\\alpha\\)) wavelength --> wavelength increases. (\\(\\beta\\)) energy ---> energy decreases.\n(ii) \\(\\lambda = \\frac{h}{mv}\\)\n= \\(\\frac{6.6 \\times 10^{-34}}{4.4 \\times 10^{-2} \\times 10^{-16}}\\)m\n(c) Pairs of isotopes in the figure are;\n\\(^{208}_{82}Pb\\) and \\(^{212}_{82}Pb\\) or \\(^{212}_{84}Po\\) and \\(^{216}_{84}Po\\) or\n\\(^{224}_{88}Ra\\) and \\(^{288}_{88}Ra\\) or \\(^{288}_{90}Th\\) and \\(^{232}_{90}Th\\)\n(ii) The series started with isotopes \\(^{232}_{90}Th\\)\n(iii) Nuclear equation for;\n\\(\\alpha\\)Alpha decay\n\\(^{232}_{90}Th\\) \\(\\to\\) \\(^{228}_{88}Ra\\) + \\(^{4}_{2}He\\) + Energy;\n\\(^{228}_{90}Th\\) \\(\\to\\) \\(^{22}_{88}Ra\\) + \\(^{4}_{2}He\\) + Energy;\n\\(^{224}_{88}Ra\\) \\(\\to\\) \\(^{220}_{86}Rn\\) + \\(^{4}_{2}He\\) + Energy;\n\\(^{212}_{84}Po\\) \\(\\to\\) \\(^{208}_{82}Pb\\) + \\(^{4}_{2}He\\) + Energy;\n\\(^{216}_{84}Po\\) \\(\\to\\) \\(^{212}_{82}Pb\\) + \\(^{4}_{2}He\\) + Energy;\n\\(^{212}_{84}Rn\\) \\(\\to\\) \\(^{208}_{82}Po\\) + \\(^{4}_{2}He\\) + Energy;\n\\(^{220}_{84}Rn\\) \\(\\to\\) \\(^{216}_{84}Po\\) + \\(^{4}_{2}He\\) + Energy;\n\\(^{216}_{84}Rn\\) \\(\\to\\) \\(^{212}_{82}Pb\\) + \\(^{4}_{2}He\\) + Energy;\n\n(\\(\\beta\\)) Beta decay\n\\(^{224}_{88}Ra\\) \\(\\to\\) \\(^{224}_{89}Ac\\) + \\(^{o}_{-1}e\\) + Energy;\n\\(^{228}_{88}Ac\\) \\(\\to\\) \\(^{228}_{90}Ac\\) + \\(^{o}_{-1}e\\) + Energy;\n\\(^{212}_{82}Pb\\) \\(\\to\\) \\(^{212}_{83}Bi\\) + \\(^{o}_{-1}e\\) + Energy;\n\\(^{212}_{83}Ra\\) \\(\\to\\) \\(^{212}_{84}Ac\\) + \\(^{o}_{-1}e\\) + Energy;\n\n', '', 'Explanation\n(a)(i) Mass defect is the difference between the mass of a nucleus and the total mass of its nucleons. It expressed in atomic mass unit (a.m.u.). It is also a measure of binding energy.\nMass defect = total mass of nucleus\n– mass of nucleons.\n(ii) Binding energy is the work done or energy needed to separate the nucleons in a nucleus. It is measured in eV or Joules (J).\n\n(b)(i) The effect of decreasing the speed of a photoelectron on: (\\(\\alpha\\)) wavelength --> wavelength increases. (\\(\\beta\\)) energy ---> energy decreases.\n(ii) \\(\\lambda = \\frac{h}{mv}\\)\n= \\(\\frac{6.6 \\times 10^{-34}}{4.4 \\times 10^{-2} \\times 10^{-16}}\\)m\n(c) Pairs of isotopes in the figure are;\n\\(^{208}_{82}Pb\\) and \\(^{212}_{82}Pb\\) or \\(^{212}_{84}Po\\) and \\(^{216}_{84}Po\\) or\n\\(^{224}_{88}Ra\\) and \\(^{288}_{88}Ra\\) or \\(^{288}_{90}Th\\) and \\(^{232}_{90}Th\\)\n(ii) The series started with isotopes \\(^{232}_{90}Th\\)\n(iii) Nuclear equation for;\n\\(\\alpha\\)Alpha decay\n\\(^{232}_{90}Th\\) \\(\\to\\) \\(^{228}_{88}Ra\\) + \\(^{4}_{2}He\\) + Energy;\n\\(^{228}_{90}Th\\) \\(\\to\\) \\(^{22}_{88}Ra\\) + \\(^{4}_{2}He\\) + Energy;\n\\(^{224}_{88}Ra\\) \\(\\to\\) \\(^{220}_{86}Rn\\) + \\(^{4}_{2}He\\) + Energy;\n\\(^{212}_{84}Po\\) \\(\\to\\) \\(^{208}_{82}Pb\\) + \\(^{4}_{2}He\\) + Energy;\n\\(^{216}_{84}Po\\) \\(\\to\\) \\(^{212}_{82}Pb\\) + \\(^{4}_{2}He\\) + Energy;\n\\(^{212}_{84}Rn\\) \\(\\to\\) \\(^{208}_{82}Po\\) + \\(^{4}_{2}He\\) + Energy;\n\\(^{220}_{84}Rn\\) \\(\\to\\) \\(^{216}_{84}Po\\) + \\(^{4}_{2}He\\) + Energy;\n\\(^{216}_{84}Rn\\) \\(\\to\\) \\(^{212}_{82}Pb\\) + \\(^{4}_{2}He\\) + Energy;\n\n(\\(\\beta\\)) Beta decay\n\\(^{224}_{88}Ra\\) \\(\\to\\) \\(^{224}_{89}Ac\\) + \\(^{o}_{-1}e\\) + Energy;\n\\(^{228}_{88}Ac\\) \\(\\to\\) \\(^{228}_{90}Ac\\) + \\(^{o}_{-1}e\\) + Energy;\n\\(^{212}_{82}Pb\\) \\(\\to\\) \\(^{212}_{83}Bi\\) + \\(^{o}_{-1}e\\) + Energy;\n\\(^{212}_{83}Ra\\) \\(\\to\\) \\(^{212}_{84}Ac\\) + \\(^{o}_{-1}e\\) + Energy;\n\n', '(a)(i) Mass defect is the difference between the mass of a nucleus and the total mass of its nucleons. It expressed in atomic mass unit (a.m.u.). It is also a measure of binding energy.\nMass defect = total mass of nucleus\n– mass of nucleons.\n(ii) Binding energy is the work done or energy needed to separate the nucleons in a nucleus. It is measured in eV or Joules (J).\n~(b)(i) The effect of decreasing the speed of a photoelectron on: (\\(\\alpha\\)) wavelength --> wavelength increases. (\\(\\beta\\)) energy ---> energy decreases.\n(ii) \\(\\lambda = \\frac{h}{mv}\\)\n= \\(\\frac{6.6 \\times 10^{-34}}{4.4 \\times 10^{-2} \\times 10^{-16}}\\)m\n(c) Pairs of isotopes in the figure are;\n\\(^{208}_{82}Pb\\) and \\(^{212}_{82}Pb\\) or \\(^{212}_{84}Po\\) and \\(^{216}_{84}Po\\) or\n\\(^{224}_{88}Ra\\) and \\(^{288}_{88}Ra\\) or \\(^{288}_{90}Th\\) and \\(^{232}_{90}Th\\)\n(ii) The series started with isotopes \\(^{232}_{90}Th\\)\n(iii) Nuclear equation for;\n\\(\\alpha\\)Alpha decay\n\\(^{232}_{90}Th\\) \\(\\to\\) \\(^{228}_{88}Ra\\) + \\(^{4}_{2}He\\) + Energy;\n\\(^{228}_{90}Th\\) \\(\\to\\) \\(^{22}_{88}Ra\\) + \\(^{4}_{2}He\\) + Energy;\n\\(^{224}_{88}Ra\\) \\(\\to\\) \\(^{220}_{86}Rn\\) + \\(^{4}_{2}He\\) + Energy;\n\\(^{212}_{84}Po\\) \\(\\to\\) \\(^{208}_{82}Pb\\) + \\(^{4}_{2}He\\) + Energy;\n\\(^{216}_{84}Po\\) \\(\\to\\) \\(^{212}_{82}Pb\\) + \\(^{4}_{2}He\\) + Energy;\n\\(^{212}_{84}Rn\\) \\(\\to\\) \\(^{208}_{82}Po\\) + \\(^{4}_{2}He\\) + Energy;\n\\(^{220}_{84}Rn\\) \\(\\to\\) \\(^{216}_{84}Po\\) + \\(^{4}_{2}He\\) + Energy;\n\\(^{216}_{84}Rn\\) \\(\\to\\) \\(^{212}_{82}Pb\\) + \\(^{4}_{2}He\\) + Energy;\n\n(\\(\\beta\\)) Beta decay\n\\(^{224}_{88}Ra\\) \\(\\to\\) \\(^{224}_{89}Ac\\) + \\(^{o}_{-1}e\\) + Energy;\n\\(^{228}_{88}Ac\\) \\(\\to\\) \\(^{228}_{90}Ac\\) + \\(^{o}_{-1}e\\) + Energy;\n\\(^{212}_{82}Pb\\) \\(\\to\\) \\(^{212}_{83}Bi\\) + \\(^{o}_{-1}e\\) + Energy;\n\\(^{212}_{83}Ra\\) \\(\\to\\) \\(^{212}_{84}Ac\\) + \\(^{o}_{-1}e\\) + Energy;\n~(c) Pairs of isotopes in the figure are;\n\\(^{208}_{82}Pb\\) and \\(^{212}_{82}Pb\\) or \\(^{212}_{84}Po\\) and \\(^{216}_{84}Po\\) or\n\\(^{224}_{88}Ra\\) and \\(^{288}_{88}Ra\\) or \\(^{288}_{90}Th\\) and \\(^{232}_{90}Th\\)\n(ii) The series started with isotopes \\(^{232}_{90}Th\\)\n(iii) Nuclear equation for;\n\\(\\alpha\\)Alpha decay\n\\(^{232}_{90}Th\\) \\(\\to\\) \\(^{228}_{88}Ra\\) + \\(^{4}_{2}He\\) + Energy;\n\\(^{228}_{90}Th\\) \\(\\to\\) \\(^{22}_{88}Ra\\) + \\(^{4}_{2}He\\) + Energy;\n\\(^{224}_{88}Ra\\) \\(\\to\\) \\(^{220}_{86}Rn\\) + \\(^{4}_{2}He\\) + Energy;\n\\(^{212}_{84}Po\\) \\(\\to\\) \\(^{208}_{82}Pb\\) + \\(^{4}_{2}He\\) + Energy;\n\\(^{216}_{84}Po\\) \\(\\to\\) \\(^{212}_{82}Pb\\) + \\(^{4}_{2}He\\) + Energy;\n\\(^{212}_{84}Rn\\) \\(\\to\\) \\(^{208}_{82}Po\\) + \\(^{4}_{2}He\\) + Energy;\n\\(^{220}_{84}Rn\\) \\(\\to\\) \\(^{216}_{84}Po\\) + \\(^{4}_{2}He\\) + Energy;\n\\(^{216}_{84}Rn\\) \\(\\to\\) \\(^{212}_{82}Pb\\) + \\(^{4}_{2}He\\) + Energy;\n\n(\\(\\beta\\)) Beta decay\n\\(^{224}_{88}Ra\\) \\(\\to\\) \\(^{224}_{89}Ac\\) + \\(^{o}_{-1}e\\) + Energy;\n\\(^{228}_{88}Ac\\) \\(\\to\\) \\(^{228}_{90}Ac\\) + \\(^{o}_{-1}e\\) + Energy;\n\\(^{212}_{82}Pb\\) \\(\\to\\) \\(^{212}_{83}Bi\\) + \\(^{o}_{-1}e\\) + Energy;\n\\(^{212}_{83}Ra\\) \\(\\to\\) \\(^{212}_{84}Ac\\) + \\(^{o}_{-1}e\\) + Energy;\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(37, '2015', '\n\nQuestion\n\n(a) What is a projectile?\n(b) Give the reason why the horizontal component of the velocity of a projectile remains the same at every point of its flight.\n\n\n\n\n\n\n\n\n(a) What is a projectile?\n(b) Give the reason why the horizontal component of the velocity of a projectile remains the same at every point of its flight.\n\n\n\n\n\n', '', '(a) What is a projectile?\n(b) Give the reason why the horizontal component of the velocity of a projectile remains the same at every point of its flight.\n\n\n\n\n\n\n\n\n(a) What is a projectile?\n(b) Give the reason why the horizontal component of the velocity of a projectile remains the same at every point of its flight.\n\n\n\n', '(a) What is a projectile?\n(b) Give the reason why the horizontal component of the velocity of a projectile remains the same at every point of its flight.\n\n\n\n\n\n\n\n\n(a) What is a projectile?~(b) Give the reason why the horizontal component of the velocity of a projectile remains the same at every point of its flight.\n\n\n\n\n\n\n\n\n(a) What is a projectile?\n(b) Give the reason why the horizontal component of the velocity of a projectile remains the same at every point of its flight.\n\n\n\n', NULL, '\n\nExplanation\n(a) Projectile is an object thrown into the air, and left to move under the influence of gravity.\n(b) the acceleration of free fall due to gravity does not act in the horizontal direction.\n\n', '', 'Explanation\n(a) Projectile is an object thrown into the air, and left to move under the influence of gravity.\n(b) the acceleration of free fall due to gravity does not act in the horizontal direction.\n\n', '(a) Projectile is an object thrown into the air, and left to move under the influence of gravity.~(b) the acceleration of free fall due to gravity does not act in the horizontal direction.\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(38, '2015', '\n\nQuestion\n\nA body is projected at an angle of 30° to the horizontal with a velocity of 150 ms\\(^{-1.}\\). Calculate the time it takes to reach the greatest height [Take g = 10 ms\\(^2\\) and neglect air resistance]\n\n', 'A body is projected at an angle of 30° to the horizontal with a velocity of 150 ms\\(^{-1.}\\). Calculate the time it takes to reach the greatest height [Take g = 10 ms\\(^2\\) and neglect air resistance]', '', '', NULL, '\n\nExplanation\nYoung\'s Modulus = \\(\\frac{Fl}{Ae}\\)\n= \\(\\frac{1000 \\times 2.00}{\\frac{1}{3.14 \\times 1000} \\times \\frac{1}{1000} \\times \\frac{25}{1000}}\\)\n= 2.55 x 10\\(^{10}\\)NM\\(^{-2}\\)\n\n', 'Young\'s Modulus = \\(\\frac{Fl}{Ae}\\)\n= \\(\\frac{1000 \\times 2.00}{\\frac{1}{3.14 \\times 1000} \\times \\frac{1}{1000} \\times \\frac{25}{1000}}\\)\n= 2.55 x 10\\(^{10}\\)NM\\(^{-2}\\)\n\n', '', '', 'WAEC-PK-1', 'WAEC-PHY', 1),
(39, '2015', '\n\nQuestion\n\nA wire of length 2.00 m and radius 1.0 mm is stretched by 25.0 mm on application of a force of 103 N. Calculate the Young\'s modulus for the wire.\n\n', 'A wire of length 2.00 m and radius 1.0 mm is stretched by 25.0 mm on application of a force of 103 N. Calculate the Young\'s modulus for the wire.', '', '', NULL, '\n\nExplanation\nt = \\(\\frac{Usin\\theta}{g} = \\frac{150 sin 30^o}{10} = 7.5s\\)\n\n\n', 't = \\(\\frac{Usin\\theta}{g} = \\frac{150 sin 30^o}{10} = 7.5s\\)\n\n\n', '', '', 'WAEC-PK-1', 'WAEC-PHY', 1),
(40, '2015', '\n\nQuestion\n\nState three ways of increasing the rate of cooling of a cup of hot tea.\n\n', 'State three ways of increasing the rate of cooling of a cup of hot tea.', '', '', NULL, '\n\nExplanation\nWays of increasing the cooling rate of a cup of tea are:\n(i) Stirring the tea continuously.\n(ii) Increasing the rate of evaporation.\n(iii) Reducing the ambient temperature.\n\n', '', 'Explanation\nWays of increasing the cooling rate of a cup of tea are:\n(i) Stirring the tea continuously.\n(ii) Increasing the rate of evaporation.\n(iii) Reducing the ambient temperature.\n\n', '', 'WAEC-PK-1', 'WAEC-PHY', 1),
(41, '2015', '\n\nQuestion\n\n(a) What is doping in a semi conductor?\n(b) Draw the symbol for OR gate. \n\n\n\n\n(a) What is doping in a semi conductor?\n(b) Draw the symbol for OR gate. \n\n', '', '(a) What is doping in a semi conductor?\n(b) Draw the symbol for OR gate. \n\n\n\n\n(a) What is doping in a semi conductor?\n(b) Draw the symbol for OR gate. ', '(a) What is doping in a semi conductor?\n(b) Draw the symbol for OR gate. \n\n\n\n\n(a) What is doping in a semi conductor?~(b) Draw the symbol for OR gate. \n\n\n\n\n(a) What is doping in a semi conductor?\n(b) Draw the symbol for OR gate. ', NULL, '\n\nExplanation\n(a) Doping in a semi conductor is the addition of impurity to a pure semi condutor to increase its electrical conductivity through increase in the number of holes and electrons or charge carriers.\n\n\n\n', '(a) Doping in a semi conductor is the addition of impurity to a pure semi condutor to increase its electrical conductivity through increase in the number of holes and electrons or charge carriers.\n\n\n\n', '', '', 'WAEC-PK-1', 'WAEC-PHY', 1),
(42, '2015', '\n\nQuestion\n\nThe circuit above consists of an a. c. voltage input, a diode, a resistor and a voltmeter.\n(a) Identify the circuit.\n(b) Draw the waveform for the output voltage.\n\n', '', 'The circuit above consists of an a. c. voltage input, a diode, a resistor and a voltmeter.\n(a) Identify the circuit.\n(b) Draw the waveform for the output voltage.', '(a) Identify the circuit.~(b) Draw the waveform for the output voltage.', NULL, '\n\nExplanation\n(a) The circuit is half-wave rectifier.\n(b)\n\n\n', '', 'Explanation\n(a) The circuit is half-wave rectifier.\n(b)\n\n\n', '(a) The circuit is half-wave rectifier.~(b)\n\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(43, '2015', '\n\nQuestion\n\n(a) Write the de Broglie equation.\n(b) Explain the significance of the equation.\n\n', '', '(a) Write the de Broglie equation.\n(b) Explain the significance of the equation.', '(a) Write the de Broglie equation.~(b) Explain the significance of the equation.', NULL, '\n\nExplanation\na) \\(\\lambda = \\frac{h}{p} or \\alpha = \\frac{h}{mV}\\)\nb) The significance of the equation is tha, a particle of mass (m) moving with a velocity (v) has an associated wavelength (\\(\\lambda\\)). The wavelength decreases with increase in momentum or velocity.\n\n', 'a) \\(\\lambda = \\frac{h}{p} or \\alpha = \\frac{h}{mV}\\)\nb) The significance of the equation is tha, a particle of mass (m) moving with a velocity (v) has an associated wavelength (\\(\\lambda\\)). The wavelength decreases with increase in momentum or velocity.\n\n', '', '', 'WAEC-PK-1', 'WAEC-PHY', 1);
INSERT INTO `exam_simulation2` (`id`, `examYear`, `questions`, `parent_que`, `child_que`, `gen_child_que_desc`, `imageDir`, `explanation`, `parent_que_expl`, `child_que_exp`, `gen_child_que_expl`, `package_id`, `subject_id`, `status`) VALUES
(44, '2015', '\n\nQuestion\n\n(a) (i) Define uniform acceleration.\n(ii) Write an equation that relates linear velocity, angular velocity and radius\nof path in circular motion.\n(b) Two forces 30 N and 40 N act at right angles to each other. Determine by scale drawing, the magnitude and direction of the resultant force,.using a scale of 1 cm to 5 N.\n(c) Explain why ships are usually refilled with sand and water after they have been emptied of their cargo.\n(d) A crate of drinks of mass 20 kg is placed on a plane inclined at 30° to the horizontal. If the crate slides down with a constant speed, calculate the:\n(i) co-efficient of kinetic friction;\n(ii) magnitude of the frictional force acting on the crate. [ g = 10 ms\\(^{-2}\\)]\n\n', '', '(a) (i) Define uniform acceleration.\n(ii) Write an equation that relates linear velocity, angular velocity and radius\nof path in circular motion.\n(b) Two forces 30 N and 40 N act at right angles to each other. Determine by scale drawing, the magnitude and direction of the resultant force,.using a scale of 1 cm to 5 N.\n(c) Explain why ships are usually refilled with sand and water after they have been emptied of their cargo.\n(d) A crate of drinks of mass 20 kg is placed on a plane inclined at 30° to the horizontal. If the crate slides down with a constant speed, calculate the:\n(i) co-efficient of kinetic friction;\n(ii) magnitude of the frictional force acting on the crate. [ g = 10 ms\\(^{-2}\\)]', '(a) (i) Define uniform acceleration.\n(ii) Write an equation that relates linear velocity, angular velocity and radius\nof path in circular motion.~(b) Two forces 30 N and 40 N act at right angles to each other. Determine by scale drawing, the magnitude and direction of the resultant force,.using a scale of 1 cm to 5 N.\n(c) Explain why ships are usually refilled with sand and water after they have been emptied of their cargo.\n(d) A crate of drinks of mass 20 kg is placed on a plane inclined at 30° to the horizontal. If the crate slides down with a constant speed, calculate the:\n(i) co-efficient of kinetic friction;\n(ii) magnitude of the frictional force acting on the crate. [ g = 10 ms\\(^{-2}\\)]~(c) Explain why ships are usually refilled with sand and water after they have been emptied of their cargo.\n(d) A crate of drinks of mass 20 kg is placed on a plane inclined at 30° to the horizontal. If the crate slides down with a constant speed, calculate the:\n(i) co-efficient of kinetic friction;\n(ii) magnitude of the frictional force acting on the crate. [ g = 10 ms\\(^{-2}\\)]~(d) A crate of drinks of mass 20 kg is placed on a plane inclined at 30° to the horizontal. If the crate slides down with a constant speed, calculate the:\n(i) co-efficient of kinetic friction;\n(ii) magnitude of the frictional force acting on the crate. [ g = 10 ms\\(^{-2}\\)]', NULL, '\n\nExplanation\n(i) Uniform acceleration is a constant time rate of increase in velocity.\n(ii) Linear velocity = angular velocity x radius of path V = \\(\\omega\\)r\n(b) Scale: 1 cm to 5 N\n30N = 6cm and 40N = 8cm\n\n(c) A floating ship is balanced by its weight and the upthrust on it. Hence when off-loaded the upthrust becomes greater than the weight of the ship, therefore when refilled with sand and water, equilibrium is restored for stability.\n\nd)(i) \\(\\mu = tan \\theta = tan 30^o = 0.5774\\)\n(ii) F = \\(\\mu g cos \\theta\\) = 0.5774 x 20 x 10 x Cos 30\\(^o\\) = 100N\n\n\n\n', '', 'Explanation\n(i) Uniform acceleration is a constant time rate of increase in velocity.\n(ii) Linear velocity = angular velocity x radius of path V = \\(\\omega\\)r\n(b) Scale: 1 cm to 5 N\n30N = 6cm and 40N = 8cm\n\n(c) A floating ship is balanced by its weight and the upthrust on it. Hence when off-loaded the upthrust becomes greater than the weight of the ship, therefore when refilled with sand and water, equilibrium is restored for stability.\n\nd)(i) \\(\\mu = tan \\theta = tan 30^o = 0.5774\\)\n(ii) F = \\(\\mu g cos \\theta\\) = 0.5774 x 20 x 10 x Cos 30\\(^o\\) = 100N\n\n\n\n', '(i) Uniform acceleration is a constant time rate of increase in velocity.\n(ii) Linear velocity = angular velocity x radius of path V = \\(\\omega\\)r~(b) Scale: 1 cm to 5 N\n30N = 6cm and 40N = 8cm\n\n(c) A floating ship is balanced by its weight and the upthrust on it. Hence when off-loaded the upthrust becomes greater than the weight of the ship, therefore when refilled with sand and water, equilibrium is restored for stability.\n\nd)(i) \\(\\mu = tan \\theta = tan 30^o = 0.5774\\)\n(ii) F = \\(\\mu g cos \\theta\\) = 0.5774 x 20 x 10 x Cos 30\\(^o\\) = 100N\n\n\n~(c) A floating ship is balanced by its weight and the upthrust on it. Hence when off-loaded the upthrust becomes greater than the weight of the ship, therefore when refilled with sand and water, equilibrium is restored for stability.\n\nd)(i) \\(\\mu = tan \\theta = tan 30^o = 0.5774\\)\n(ii) F = \\(\\mu g cos \\theta\\) = 0.5774 x 20 x 10 x Cos 30\\(^o\\) = 100N\n\n\n~d)(i) \\(\\mu = tan \\theta = tan 30^o = 0.5774\\)\n(ii) F = \\(\\mu g cos \\theta\\) = 0.5774 x 20 x 10 x Cos 30\\(^o\\) = 100N\n\n\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(45, '2015', '\n\nQuestion\n\n(a) Define specific latent heat of vaporization.\n(b) (i) What are renewable energy sources?\n(ii) List four renewable energy sources.\n(c) Explain why tomatoes keep longer when kept in a moist jute bag in a clay pot.\n(d) A box has a volume of 0.28 m\\(^3\\) and is 70% filled with iron fillings at 25°C. Calculate the:\n(i) total mass of the iron fillings;\n(ii) energy required to melt 10% of the iron fillings. [Density of iron = 8.00 x 10\\(^3\\) kgm\\(^{-3}\\); specific latent heat of fusion of iron = 1.38 x 10\\(^5\\) Jkg\\(^{-1}\\), specific heat capacity of iron = 460 Jkg\\(^{-1}\\) k\\(^{-1}\\), melting point of iron = 1500°C] 10.\n\n\n', '', '(a) Define specific latent heat of vaporization.\n(b) (i) What are renewable energy sources?\n(ii) List four renewable energy sources.\n(c) Explain why tomatoes keep longer when kept in a moist jute bag in a clay pot.\n(d) A box has a volume of 0.28 m\\(^3\\) and is 70% filled with iron fillings at 25°C. Calculate the:\n(i) total mass of the iron fillings;\n(ii) energy required to melt 10% of the iron fillings. [Density of iron = 8.00 x 10\\(^3\\) kgm\\(^{-3}\\); specific latent heat of fusion of iron = 1.38 x 10\\(^5\\) Jkg\\(^{-1}\\), specific heat capacity of iron = 460 Jkg\\(^{-1}\\) k\\(^{-1}\\), melting point of iron = 1500°C] 10.\n', '(a) Define specific latent heat of vaporization.~(b) (i) What are renewable energy sources?\n(ii) List four renewable energy sources.\n(c) Explain why tomatoes keep longer when kept in a moist jute bag in a clay pot.\n(d) A box has a volume of 0.28 m\\(^3\\) and is 70% filled with iron fillings at 25°C. Calculate the:\n(i) total mass of the iron fillings;\n(ii) energy required to melt 10% of the iron fillings. [Density of iron = 8.00 x 10\\(^3\\) kgm\\(^{-3}\\); specific latent heat of fusion of iron = 1.38 x 10\\(^5\\) Jkg\\(^{-1}\\), specific heat capacity of iron = 460 Jkg\\(^{-1}\\) k\\(^{-1}\\), melting point of iron = 1500°C] 10.\n~(c) Explain why tomatoes keep longer when kept in a moist jute bag in a clay pot.\n(d) A box has a volume of 0.28 m\\(^3\\) and is 70% filled with iron fillings at 25°C. Calculate the:\n(i) total mass of the iron fillings;\n(ii) energy required to melt 10% of the iron fillings. [Density of iron = 8.00 x 10\\(^3\\) kgm\\(^{-3}\\); specific latent heat of fusion of iron = 1.38 x 10\\(^5\\) Jkg\\(^{-1}\\), specific heat capacity of iron = 460 Jkg\\(^{-1}\\) k\\(^{-1}\\), melting point of iron = 1500°C] 10.\n~(d) A box has a volume of 0.28 m\\(^3\\) and is 70% filled with iron fillings at 25°C. Calculate the:\n(i) total mass of the iron fillings;\n(ii) energy required to melt 10% of the iron fillings. [Density of iron = 8.00 x 10\\(^3\\) kgm\\(^{-3}\\); specific latent heat of fusion of iron = 1.38 x 10\\(^5\\) Jkg\\(^{-1}\\), specific heat capacity of iron = 460 Jkg\\(^{-1}\\) k\\(^{-1}\\), melting point of iron = 1500°C] 10.\n', NULL, '\n\nExplanation\na) Specific latent heat of vaporization in the quantity of heat required to change a unit mass of a liquid at its boiling point to gas.\n\n(b)(i) Renewable energy sources are sources of energy that cannot be exhausted.\n(ii) Renewable energy sources are sun or solar, wave, wind, hydroelectric, tidal, biomass or vegetable oils, geothermal, etc.\n\n(c) Clay pot is porous, therefore molecules of water in the moist jute bag will extract latent heat of vaporization from the tomatoes which will escape through the pores of the pot, thereby causing cooling of the tomatoes.\n(i) Mass = density x volume = 8.00 x 10\\(^3\\) x \\(\\frac{70}{80}\\) x 0.28 = 1568 kg.\n(ii) Energy required Q = MC (\\(\\theta_2 - \\theta_1\\)) + mL\n= \\(\\frac{10}{100}\\) x 1568 x 460(1500 - 25) + \\(\\frac{10}{100}\\) x 1568 x 1.38 x 10\\(^5\\)\n= 106, 388, 800 + 21,638, 400\n= 128,027,200J or 1.28 x 10\\(^8\\)J\n\n\n', '', 'Explanation\na) Specific latent heat of vaporization in the quantity of heat required to change a unit mass of a liquid at its boiling point to gas.\n\n(b)(i) Renewable energy sources are sources of energy that cannot be exhausted.\n(ii) Renewable energy sources are sun or solar, wave, wind, hydroelectric, tidal, biomass or vegetable oils, geothermal, etc.\n\n(c) Clay pot is porous, therefore molecules of water in the moist jute bag will extract latent heat of vaporization from the tomatoes which will escape through the pores of the pot, thereby causing cooling of the tomatoes.\n(i) Mass = density x volume = 8.00 x 10\\(^3\\) x \\(\\frac{70}{80}\\) x 0.28 = 1568 kg.\n(ii) Energy required Q = MC (\\(\\theta_2 - \\theta_1\\)) + mL\n= \\(\\frac{10}{100}\\) x 1568 x 460(1500 - 25) + \\(\\frac{10}{100}\\) x 1568 x 1.38 x 10\\(^5\\)\n= 106, 388, 800 + 21,638, 400\n= 128,027,200J or 1.28 x 10\\(^8\\)J\n\n\n', 'a) Specific latent heat of vaporization in the quantity of heat required to change a unit mass of a liquid at its boiling point to gas.\n~(b)(i) Renewable energy sources are sources of energy that cannot be exhausted.\n(ii) Renewable energy sources are sun or solar, wave, wind, hydroelectric, tidal, biomass or vegetable oils, geothermal, etc.\n\n(c) Clay pot is porous, therefore molecules of water in the moist jute bag will extract latent heat of vaporization from the tomatoes which will escape through the pores of the pot, thereby causing cooling of the tomatoes.\n(i) Mass = density x volume = 8.00 x 10\\(^3\\) x \\(\\frac{70}{80}\\) x 0.28 = 1568 kg.\n(ii) Energy required Q = MC (\\(\\theta_2 - \\theta_1\\)) + mL\n= \\(\\frac{10}{100}\\) x 1568 x 460(1500 - 25) + \\(\\frac{10}{100}\\) x 1568 x 1.38 x 10\\(^5\\)\n= 106, 388, 800 + 21,638, 400\n= 128,027,200J or 1.28 x 10\\(^8\\)J\n\n~(c) Clay pot is porous, therefore molecules of water in the moist jute bag will extract latent heat of vaporization from the tomatoes which will escape through the pores of the pot, thereby causing cooling of the tomatoes.\n(i) Mass = density x volume = 8.00 x 10\\(^3\\) x \\(\\frac{70}{80}\\) x 0.28 = 1568 kg.\n(ii) Energy required Q = MC (\\(\\theta_2 - \\theta_1\\)) + mL\n= \\(\\frac{10}{100}\\) x 1568 x 460(1500 - 25) + \\(\\frac{10}{100}\\) x 1568 x 1.38 x 10\\(^5\\)\n= 106, 388, 800 + 21,638, 400\n= 128,027,200J or 1.28 x 10\\(^8\\)J\n\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(46, '2015', '\n\nQuestion\n\n(a)(i) what is resonance?\n(ii) State two eXwilples of resonance.\n(iii) Differentiate between loudness and intensity of sound.\n\n(b) When a ray is refracted through a rectangular glass prism, which of the following properties of the ray will change? Wavelength, frequency and speed.\n(c)\n\n(i) Copy the diagram above in your answer booklet.\n(iii) On the copied diagram, sketch the pattern of the waves immediately after passing through the opening.\n(d) A diverging lens of focal length 18.0 in is used to view a shark that is 90.0 in away from the lens. If the image formed is 1.0 m long, calculate the:\n(i) image distance;\n(ii) length of the shark. \n\n\n', '', '(a)(i) what is resonance?\n(ii) State two eXwilples of resonance.\n(iii) Differentiate between loudness and intensity of sound.\n\n(b) When a ray is refracted through a rectangular glass prism, which of the following properties of the ray will change? Wavelength, frequency and speed.\n(c)\n\n(i) Copy the diagram above in your answer booklet.\n(iii) On the copied diagram, sketch the pattern of the waves immediately after passing through the opening.\n(d) A diverging lens of focal length 18.0 in is used to view a shark that is 90.0 in away from the lens. If the image formed is 1.0 m long, calculate the:\n(i) image distance;\n(ii) length of the shark. \n', '(a)(i) what is resonance?\n(ii) State two eXwilples of resonance.\n(iii) Differentiate between loudness and intensity of sound.\n~(b) When a ray is refracted through a rectangular glass prism, which of the following properties of the ray will change? Wavelength, frequency and speed.\n(c)\n\n(i) Copy the diagram above in your answer booklet.\n(iii) On the copied diagram, sketch the pattern of the waves immediately after passing through the opening.\n(d) A diverging lens of focal length 18.0 in is used to view a shark that is 90.0 in away from the lens. If the image formed is 1.0 m long, calculate the:\n(i) image distance;\n(ii) length of the shark. \n~(c)\n\n(i) Copy the diagram above in your answer booklet.\n(iii) On the copied diagram, sketch the pattern of the waves immediately after passing through the opening.\n(d) A diverging lens of focal length 18.0 in is used to view a shark that is 90.0 in away from the lens. If the image formed is 1.0 m long, calculate the:\n(i) image distance;\n(ii) length of the shark. \n~(d) A diverging lens of focal length 18.0 in is used to view a shark that is 90.0 in away from the lens. If the image formed is 1.0 m long, calculate the:\n(i) image distance;\n(ii) length of the shark. \n', NULL, '\n\nExplanation\n(a)(i) Resonance is a phenomenon which occurs whenever a body is set in oscillation at its own natural frequency as a result of impulses received from another vibrating body, vibrating with the same frequency.\n(ii) Examples of resonance are:\n(1) Loud sound from vibrating prongs of a tuning fork.\n(2) Bridge vibrating under uniform matching of soldiers.\n(3) Repeated jumping of a diver at the free end of a diving board causes the board to resonate.\n(4) Rattling of car engine at high speed causing vibrating of the body of the car. (5) X\\(_{c}\\) equals X\\(_L\\) in an L-C-R series circuit for tuning a radio set.\n(iii)\n\n\n\n\nLoudness\n\n\nIntensity\n\n\n\n\n- The magnitude of the sensation resulting from sound reaching the ear.\n-Sensation that depends on the individual hearing ability.\n\n\n- The time rate of flow of energy per unit area perpendicular to the direction of propagation of the sound.\n- Depends on physical quantities of energy, time and area\n\n\n\n\n\n(b) The following properties will change in speed and wavelength.\n\n(c)\n\n\n(d) \\(\\frac{1}{u} + \\frac{1}{v} = \\frac{1}{f}\\)\n\\(\\frac{1}{v} = \\frac{1}{f} - \\frac{1}{u} = \\frac{1}{-18} - \\frac{1}{90}\\)\n= \\(\\frac{-5 -1}{90}\\)\nv = -15m\nM = \\(\\frac{u}{v} = \\frac{h_i}{h_o}\\)\n\\(h_o = \\frac{hi \\times u}{v} = \\frac{1 \\times 90}{1.5}\\)\n= 6.0 m\n\n', '', 'Explanation\n(a)(i) Resonance is a phenomenon which occurs whenever a body is set in oscillation at its own natural frequency as a result of impulses received from another vibrating body, vibrating with the same frequency.\n(ii) Examples of resonance are:\n(1) Loud sound from vibrating prongs of a tuning fork.\n(2) Bridge vibrating under uniform matching of soldiers.\n(3) Repeated jumping of a diver at the free end of a diving board causes the board to resonate.\n(4) Rattling of car engine at high speed causing vibrating of the body of the car. (5) X\\(_{c}\\) equals X\\(_L\\) in an L-C-R series circuit for tuning a radio set.\n(iii)\n\n\n\n\nLoudness\n\n\nIntensity\n\n\n\n\n- The magnitude of the sensation resulting from sound reaching the ear.\n-Sensation that depends on the individual hearing ability.\n\n\n- The time rate of flow of energy per unit area perpendicular to the direction of propagation of the sound.\n- Depends on physical quantities of energy, time and area\n\n\n\n\n\n(b) The following properties will change in speed and wavelength.\n\n(c)\n\n\n(d) \\(\\frac{1}{u} + \\frac{1}{v} = \\frac{1}{f}\\)\n\\(\\frac{1}{v} = \\frac{1}{f} - \\frac{1}{u} = \\frac{1}{-18} - \\frac{1}{90}\\)\n= \\(\\frac{-5 -1}{90}\\)\nv = -15m\nM = \\(\\frac{u}{v} = \\frac{h_i}{h_o}\\)\n\\(h_o = \\frac{hi \\times u}{v} = \\frac{1 \\times 90}{1.5}\\)\n= 6.0 m\n\n', '(a)(i) Resonance is a phenomenon which occurs whenever a body is set in oscillation at its own natural frequency as a result of impulses received from another vibrating body, vibrating with the same frequency.\n(ii) Examples of resonance are:\n(1) Loud sound from vibrating prongs of a tuning fork.\n(2) Bridge vibrating under uniform matching of soldiers.\n(3) Repeated jumping of a diver at the free end of a diving board causes the board to resonate.\n(4) Rattling of car engine at high speed causing vibrating of the body of the car. (5) X\\(_{c}\\) equals X\\(_L\\) in an L-C-R series circuit for tuning a radio set.\n(iii)\n\n\n\n\nLoudness\n\n\nIntensity\n\n\n\n\n- The magnitude of the sensation resulting from sound reaching the ear.\n-Sensation that depends on the individual hearing ability.\n\n\n- The time rate of flow of energy per unit area perpendicular to the direction of propagation of the sound.\n- Depends on physical quantities of energy, time and area\n\n\n\n\n~(b) The following properties will change in speed and wavelength.\n\n(c)\n\n\n(d) \\(\\frac{1}{u} + \\frac{1}{v} = \\frac{1}{f}\\)\n\\(\\frac{1}{v} = \\frac{1}{f} - \\frac{1}{u} = \\frac{1}{-18} - \\frac{1}{90}\\)\n= \\(\\frac{-5 -1}{90}\\)\nv = -15m\nM = \\(\\frac{u}{v} = \\frac{h_i}{h_o}\\)\n\\(h_o = \\frac{hi \\times u}{v} = \\frac{1 \\times 90}{1.5}\\)\n= 6.0 m\n~(c)\n\n\n(d) \\(\\frac{1}{u} + \\frac{1}{v} = \\frac{1}{f}\\)\n\\(\\frac{1}{v} = \\frac{1}{f} - \\frac{1}{u} = \\frac{1}{-18} - \\frac{1}{90}\\)\n= \\(\\frac{-5 -1}{90}\\)\nv = -15m\nM = \\(\\frac{u}{v} = \\frac{h_i}{h_o}\\)\n\\(h_o = \\frac{hi \\times u}{v} = \\frac{1 \\times 90}{1.5}\\)\n= 6.0 m\n~(d) \\(\\frac{1}{u} + \\frac{1}{v} = \\frac{1}{f}\\)\n\\(\\frac{1}{v} = \\frac{1}{f} - \\frac{1}{u} = \\frac{1}{-18} - \\frac{1}{90}\\)\n= \\(\\frac{-5 -1}{90}\\)\nv = -15m\nM = \\(\\frac{u}{v} = \\frac{h_i}{h_o}\\)\n\\(h_o = \\frac{hi \\times u}{v} = \\frac{1 \\times 90}{1.5}\\)\n= 6.0 m\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(47, '2015', '\n\nQuestion\n\n(a) Define:(i) the coulomb; (ii) resistance.\n(b) State the three effects of an electric current.\n(c) State the standard international colour convention for the insulating material covering the following electrical wires in a three-pin plug: (i) live; (ii) neutral; (iii) earth.\n(d)\n\nA water melon of mass 5.0 kg is suspended on a uniform rod of mass 4.0 kg and 4.0 in long as illustrated in the diagram above. If the rod is in equilibrium by the action of the force between the charges +q and -q, calculate the:\n(i) anti-clockwise moment;\n(ii) value of q. [ g = 10 ms\\(^{-2}\\)]\n\n', '', '(a) Define:(i) the coulomb; (ii) resistance.\n(b) State the three effects of an electric current.\n(c) State the standard international colour convention for the insulating material covering the following electrical wires in a three-pin plug: (i) live; (ii) neutral; (iii) earth.\n(d)\n\nA water melon of mass 5.0 kg is suspended on a uniform rod of mass 4.0 kg and 4.0 in long as illustrated in the diagram above. If the rod is in equilibrium by the action of the force between the charges +q and -q, calculate the:\n(i) anti-clockwise moment;\n(ii) value of q. [ g = 10 ms\\(^{-2}\\)]', '(a) Define:(i) the coulomb; (ii) resistance.~(b) State the three effects of an electric current.\n(c) State the standard international colour convention for the insulating material covering the following electrical wires in a three-pin plug: (i) live; (ii) neutral; (iii) earth.\n(d)\n\nA water melon of mass 5.0 kg is suspended on a uniform rod of mass 4.0 kg and 4.0 in long as illustrated in the diagram above. If the rod is in equilibrium by the action of the force between the charges +q and -q, calculate the:\n(i) anti-clockwise moment;\n(ii) value of q. [ g = 10 ms\\(^{-2}\\)]~(c) State the standard international colour convention for the insulating material covering the following electrical wires in a three-pin plug: (i) live; (ii) neutral; (iii) earth.\n(d)\n\nA water melon of mass 5.0 kg is suspended on a uniform rod of mass 4.0 kg and 4.0 in long as illustrated in the diagram above. If the rod is in equilibrium by the action of the force between the charges +q and -q, calculate the:\n(i) anti-clockwise moment;\n(ii) value of q. [ g = 10 ms\\(^{-2}\\)]~(d)\n\nA water melon of mass 5.0 kg is suspended on a uniform rod of mass 4.0 kg and 4.0 in long as illustrated in the diagram above. If the rod is in equilibrium by the action of the force between the charges +q and -q, calculate the:\n(i) anti-clockwise moment;\n(ii) value of q. [ g = 10 ms\\(^{-2}\\)]', NULL, '\n\nExplanation\n(a)(i) The coulomb is defined as the quantity of electric charge passing through a point in a conductor when a steady current of one ampere passes for one second.\n(ii) Resistance is the opposition offered to the flow or passage of an electric current in a conductor.\n\n(b) The effects of electric current are:\n(i) Magnetic (ii) Chemical\n(iii) Heating or thermal.\n\n(c) The standard international colour for insulating cover for a three in plug are:\n(i) Live is brown or red colour.\n(ii) Neutral is blue or black colour.\n(iii) Earth is green/yellow/green and yellow colour.\n\n(d) Clock wise moment = 4gl\\(_2\\) + \\(\\frac{kq^2l_3}{12}\\)\n\\(\\frac{4 \\times 10 \\times 1 + 8.99 \\times 10^9 \\times q^2 \\times 3}{(2.5)^2}\\) = 40 + 4.315 x 10\\(^{9}\\) x \\(q^{2}\\)\nAnti-clockwise moment = 5gl\\(_1\\) = 5 x 10 x 1 = 50 Nm\nBut sum of clockwise moment = Anti-clockwise moment 40 + 4.315 x 10\\(^9\\) q\\(^2\\) = 50\nq = \\(\\sqrt{\\frac{10}{4.315 \\times 10^9}}\\)\n= 4.814 x 10\\(^{-5}\\)C\n\n', '', 'Explanation\n(a)(i) The coulomb is defined as the quantity of electric charge passing through a point in a conductor when a steady current of one ampere passes for one second.\n(ii) Resistance is the opposition offered to the flow or passage of an electric current in a conductor.\n\n(b) The effects of electric current are:\n(i) Magnetic (ii) Chemical\n(iii) Heating or thermal.\n\n(c) The standard international colour for insulating cover for a three in plug are:\n(i) Live is brown or red colour.\n(ii) Neutral is blue or black colour.\n(iii) Earth is green/yellow/green and yellow colour.\n\n(d) Clock wise moment = 4gl\\(_2\\) + \\(\\frac{kq^2l_3}{12}\\)\n\\(\\frac{4 \\times 10 \\times 1 + 8.99 \\times 10^9 \\times q^2 \\times 3}{(2.5)^2}\\) = 40 + 4.315 x 10\\(^{9}\\) x \\(q^{2}\\)\nAnti-clockwise moment = 5gl\\(_1\\) = 5 x 10 x 1 = 50 Nm\nBut sum of clockwise moment = Anti-clockwise moment 40 + 4.315 x 10\\(^9\\) q\\(^2\\) = 50\nq = \\(\\sqrt{\\frac{10}{4.315 \\times 10^9}}\\)\n= 4.814 x 10\\(^{-5}\\)C\n\n', '(a)(i) The coulomb is defined as the quantity of electric charge passing through a point in a conductor when a steady current of one ampere passes for one second.\n(ii) Resistance is the opposition offered to the flow or passage of an electric current in a conductor.\n~(b) The effects of electric current are:\n(i) Magnetic (ii) Chemical\n(iii) Heating or thermal.\n\n(c) The standard international colour for insulating cover for a three in plug are:\n(i) Live is brown or red colour.\n(ii) Neutral is blue or black colour.\n(iii) Earth is green/yellow/green and yellow colour.\n\n(d) Clock wise moment = 4gl\\(_2\\) + \\(\\frac{kq^2l_3}{12}\\)\n\\(\\frac{4 \\times 10 \\times 1 + 8.99 \\times 10^9 \\times q^2 \\times 3}{(2.5)^2}\\) = 40 + 4.315 x 10\\(^{9}\\) x \\(q^{2}\\)\nAnti-clockwise moment = 5gl\\(_1\\) = 5 x 10 x 1 = 50 Nm\nBut sum of clockwise moment = Anti-clockwise moment 40 + 4.315 x 10\\(^9\\) q\\(^2\\) = 50\nq = \\(\\sqrt{\\frac{10}{4.315 \\times 10^9}}\\)\n= 4.814 x 10\\(^{-5}\\)C\n~(c) The standard international colour for insulating cover for a three in plug are:\n(i) Live is brown or red colour.\n(ii) Neutral is blue or black colour.\n(iii) Earth is green/yellow/green and yellow colour.\n\n(d) Clock wise moment = 4gl\\(_2\\) + \\(\\frac{kq^2l_3}{12}\\)\n\\(\\frac{4 \\times 10 \\times 1 + 8.99 \\times 10^9 \\times q^2 \\times 3}{(2.5)^2}\\) = 40 + 4.315 x 10\\(^{9}\\) x \\(q^{2}\\)\nAnti-clockwise moment = 5gl\\(_1\\) = 5 x 10 x 1 = 50 Nm\nBut sum of clockwise moment = Anti-clockwise moment 40 + 4.315 x 10\\(^9\\) q\\(^2\\) = 50\nq = \\(\\sqrt{\\frac{10}{4.315 \\times 10^9}}\\)\n= 4.814 x 10\\(^{-5}\\)C\n~(d) Clock wise moment = 4gl\\(_2\\) + \\(\\frac{kq^2l_3}{12}\\)\n\\(\\frac{4 \\times 10 \\times 1 + 8.99 \\times 10^9 \\times q^2 \\times 3}{(2.5)^2}\\) = 40 + 4.315 x 10\\(^{9}\\) x \\(q^{2}\\)\nAnti-clockwise moment = 5gl\\(_1\\) = 5 x 10 x 1 = 50 Nm\nBut sum of clockwise moment = Anti-clockwise moment 40 + 4.315 x 10\\(^9\\) q\\(^2\\) = 50\nq = \\(\\sqrt{\\frac{10}{4.315 \\times 10^9}}\\)\n= 4.814 x 10\\(^{-5}\\)C\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(48, '2015', '\n\nQuestion\n\n(a) Define nucleon number\n(b) A radioactive isotope of Americium (Am —241) decays into a nucleus of Neptunium (Np — 237) and an alpha (\\(\\alpha\\)) particle as shown in the nuclear equation below. \\(^{241}_{95}Am\\) \\(\\to\\) \\(^{237}_{c} + ^b_a \\alpha\\)\n(i) State the number of neutrons in the nucleus of Americium — 241.\n(ii) Determine the values of a,b and c.\n(c)(i) Why are y-rays not deflected by electromagnetic field?\n(ii) State two properties of gamma rays that make them suitable for sterilizing medical equipment.\n(d) A sample of radioactive substance was found to be left with h of its initial count rate after 110 years. Calculate its decay constant.\n\n', '', '(a) Define nucleon number\n(b) A radioactive isotope of Americium (Am —241) decays into a nucleus of Neptunium (Np — 237) and an alpha (\\(\\alpha\\)) particle as shown in the nuclear equation below. \\(^{241}_{95}Am\\) \\(\\to\\) \\(^{237}_{c} + ^b_a \\alpha\\)\n(i) State the number of neutrons in the nucleus of Americium — 241.\n(ii) Determine the values of a,b and c.\n(c)(i) Why are y-rays not deflected by electromagnetic field?\n(ii) State two properties of gamma rays that make them suitable for sterilizing medical equipment.\n(d) A sample of radioactive substance was found to be left with h of its initial count rate after 110 years. Calculate its decay constant.', '(a) Define nucleon number~(b) A radioactive isotope of Americium (Am —241) decays into a nucleus of Neptunium (Np — 237) and an alpha (\\(\\alpha\\)) particle as shown in the nuclear equation below. \\(^{241}_{95}Am\\) \\(\\to\\) \\(^{237}_{c} + ^b_a \\alpha\\)\n(i) State the number of neutrons in the nucleus of Americium — 241.\n(ii) Determine the values of a,b and c.\n(c)(i) Why are y-rays not deflected by electromagnetic field?\n(ii) State two properties of gamma rays that make them suitable for sterilizing medical equipment.\n(d) A sample of radioactive substance was found to be left with h of its initial count rate after 110 years. Calculate its decay constant.~(c)(i) Why are y-rays not deflected by electromagnetic field?\n(ii) State two properties of gamma rays that make them suitable for sterilizing medical equipment.\n(d) A sample of radioactive substance was found to be left with h of its initial count rate after 110 years. Calculate its decay constant.~(d) A sample of radioactive substance was found to be left with h of its initial count rate after 110 years. Calculate its decay constant.', NULL, '\n\nExplanation\n(a) Nucleon number is the total number of protons and neutrons in the nucleus of an atom.\n\n(b)(i) Number of neutrons in the nucleus of Americium = 241 - 90 =146.\n(ii) a = 2\nb = 4\nc = 95 - 2\n= 93.\n\n(c)(i) Y-ray are electrically neutral or carry no charge.\n(ii) properties of gamma rays that makes them suitable for medical sterilization of equipment are:\n(1) Possession of strong penetrating power.\n(2) Posession of low ionizing power.\n\n(d) \\(\\frac{N}{N_o} = (\\frac{1}{2})^a = \\frac{1}{32} = (\\frac{1}{2})^a\\) = (\\(\\frac{1}{2})^5\\)\nBut T\\(_\\frac{1}{2}\\) = \\(\\frac{t}{n} = \\frac{110}{5}\\) = 22 years\nHence \\(\\lambda\\) = \\(\\frac{0.693}{T_{\\frac{1}{2}}} = \\frac{0,693}{22}\\)\nDecay constant \\(\\lambda\\) 3.2 x 10\\(^{-2}\\) per year\n\n\n', '', 'Explanation\n(a) Nucleon number is the total number of protons and neutrons in the nucleus of an atom.\n\n(b)(i) Number of neutrons in the nucleus of Americium = 241 - 90 =146.\n(ii) a = 2\nb = 4\nc = 95 - 2\n= 93.\n\n(c)(i) Y-ray are electrically neutral or carry no charge.\n(ii) properties of gamma rays that makes them suitable for medical sterilization of equipment are:\n(1) Possession of strong penetrating power.\n(2) Posession of low ionizing power.\n\n(d) \\(\\frac{N}{N_o} = (\\frac{1}{2})^a = \\frac{1}{32} = (\\frac{1}{2})^a\\) = (\\(\\frac{1}{2})^5\\)\nBut T\\(_\\frac{1}{2}\\) = \\(\\frac{t}{n} = \\frac{110}{5}\\) = 22 years\nHence \\(\\lambda\\) = \\(\\frac{0.693}{T_{\\frac{1}{2}}} = \\frac{0,693}{22}\\)\nDecay constant \\(\\lambda\\) 3.2 x 10\\(^{-2}\\) per year\n\n\n', '(a) Nucleon number is the total number of protons and neutrons in the nucleus of an atom.\n~(b)(i) Number of neutrons in the nucleus of Americium = 241 - 90 =146.\n(ii) a = 2\nb = 4\nc = 95 - 2\n= 93.\n\n(c)(i) Y-ray are electrically neutral or carry no charge.\n(ii) properties of gamma rays that makes them suitable for medical sterilization of equipment are:\n(1) Possession of strong penetrating power.\n(2) Posession of low ionizing power.\n\n(d) \\(\\frac{N}{N_o} = (\\frac{1}{2})^a = \\frac{1}{32} = (\\frac{1}{2})^a\\) = (\\(\\frac{1}{2})^5\\)\nBut T\\(_\\frac{1}{2}\\) = \\(\\frac{t}{n} = \\frac{110}{5}\\) = 22 years\nHence \\(\\lambda\\) = \\(\\frac{0.693}{T_{\\frac{1}{2}}} = \\frac{0,693}{22}\\)\nDecay constant \\(\\lambda\\) 3.2 x 10\\(^{-2}\\) per year\n\n~(c)(i) Y-ray are electrically neutral or carry no charge.\n(ii) properties of gamma rays that makes them suitable for medical sterilization of equipment are:\n(1) Possession of strong penetrating power.\n(2) Posession of low ionizing power.\n\n(d) \\(\\frac{N}{N_o} = (\\frac{1}{2})^a = \\frac{1}{32} = (\\frac{1}{2})^a\\) = (\\(\\frac{1}{2})^5\\)\nBut T\\(_\\frac{1}{2}\\) = \\(\\frac{t}{n} = \\frac{110}{5}\\) = 22 years\nHence \\(\\lambda\\) = \\(\\frac{0.693}{T_{\\frac{1}{2}}} = \\frac{0,693}{22}\\)\nDecay constant \\(\\lambda\\) 3.2 x 10\\(^{-2}\\) per year\n\n~(d) \\(\\frac{N}{N_o} = (\\frac{1}{2})^a = \\frac{1}{32} = (\\frac{1}{2})^a\\) = (\\(\\frac{1}{2})^5\\)\nBut T\\(_\\frac{1}{2}\\) = \\(\\frac{t}{n} = \\frac{110}{5}\\) = 22 years\nHence \\(\\lambda\\) = \\(\\frac{0.693}{T_{\\frac{1}{2}}} = \\frac{0,693}{22}\\)\nDecay constant \\(\\lambda\\) 3.2 x 10\\(^{-2}\\) per year\n\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(49, '2015', '\n\nQuestion\n\nA particle is projected horizontally at 10 ms\\(^{-2}\\) from the top of a tower 20 M high. Calculate the horizontal distance travelled by the particle when it hits the level ground. [g= 10 ms\\(^{-2}\\)\n\n\n\n\nA particle is projected horizontally at 10 ms\\(^{-2}\\) from the top of a tower 20 M high. Calculate the horizontal distance travelled by the particle when it hits the level ground. [g= 10 ms\\(^{-2}\\)\n\n', 'A particle is projected horizontally at 10 ms\\(^{-2}\\) from the top of a tower 20 M high. Calculate the horizontal distance travelled by the particle when it hits the level ground. [g= 10 ms\\(^{-2}\\)\n\n\n\n\nA particle is projected horizontally at 10 ms\\(^{-2}\\) from the top of a tower 20 M high. Calculate the horizontal distance travelled by the particle when it hits the level ground. [g= 10 ms\\(^{-2}\\)', '', '', NULL, '\n\nExplanation\nR = U\\(\\sqrt{\\frac{2H}{g}} = 10\\sqrt{\\frac{2 \\times 20}{10}}\\) = 20\n\n', 'R = U\\(\\sqrt{\\frac{2H}{g}} = 10\\sqrt{\\frac{2 \\times 20}{10}}\\) = 20\n\n', '', '', 'WAEC-PK-1', 'WAEC-PHY', 1),
(50, '2015', '\n\nQuestion\n\nA tennis ball projected at an angle 0 attains a range R = 78. If the velocity imparted to the ball by the racket is 30 ms\\(^{-1}\\), calculate O. [ g = 10 ms\\(^{-2}\\)]\n\n\n', 'A tennis ball projected at an angle 0 attains a range R = 78. If the velocity imparted to the ball by the racket is 30 ms\\(^{-1}\\), calculate O. [ g = 10 ms\\(^{-2}\\)]\n', '', '', NULL, '\n\nExplanation\nR = \\(\\frac{U^2 sin 2\\theta}{g}\\), 78 = \\(\\frac{30^2 sin 2\\theta}{10}\\)\n\\(sin2\\theta = \\frac{780}{900}\\) = 0.8667, \\(2\\theta = sin^{-1}0.8667 = 60^o\\)\n\\(\\theta \\frac{60^o}{2} = 30^o\\)\n\n', 'R = \\(\\frac{U^2 sin 2\\theta}{g}\\), 78 = \\(\\frac{30^2 sin 2\\theta}{10}\\)\n\\(sin2\\theta = \\frac{780}{900}\\) = 0.8667, \\(2\\theta = sin^{-1}0.8667 = 60^o\\)\n\\(\\theta \\frac{60^o}{2} = 30^o\\)\n\n', '', '', 'WAEC-PK-1', 'WAEC-PHY', 1),
(51, '2015', '\n\nQuestion\n\nAn electron moves with a speed of 2.00 x 10\\(^7\\) ms\\(^{-1}\\) in an orbit in a uniform magnetic field of 1.20 x 10\\(^{-3}\\) T. Calculate the radius of the orbit. [Mass of an electron = 9.11 x 10\\(^{-3}\\) kg; charge on an electron = 1.61 x 10\\(^{-19}\\)C]\n\n\n', 'An electron moves with a speed of 2.00 x 10\\(^7\\) ms\\(^{-1}\\) in an orbit in a uniform magnetic field of 1.20 x 10\\(^{-3}\\) T. Calculate the radius of the orbit. [Mass of an electron = 9.11 x 10\\(^{-3}\\) kg; charge on an electron = 1.61 x 10\\(^{-19}\\)C]\n', '', '', NULL, '\n\nExplanation\n\\(\\frac{Mv^2}{r}\\) = qvB\nr = \\(\\frac{MV^2}{qVB} = \\frac{9.11 \\times 10^{31} \\times (2.00 \\times 10^7)^2}{1.61 \\times 10^{-19} \\times 2.00 \\times 10^7 \\times 1.2 \\times 10^{-3}}\\)\n= 9.43 x 10\\(^{-2}\\)\n\n\n', '\\(\\frac{Mv^2}{r}\\) = qvB\nr = \\(\\frac{MV^2}{qVB} = \\frac{9.11 \\times 10^{31} \\times (2.00 \\times 10^7)^2}{1.61 \\times 10^{-19} \\times 2.00 \\times 10^7 \\times 1.2 \\times 10^{-3}}\\)\n= 9.43 x 10\\(^{-2}\\)\n\n\n', '', '', 'WAEC-PK-1', 'WAEC-PHY', 1),
(52, '2015', '\n\nQuestion\n\nA metallic bar 50 cm long has a uniform cross-sectional area of 4.0 cm\\(^2\\). If a tensile force of 35 kN produces an extension of 0.25 mm, calculate the value of Young\'s modulus\n\n', 'A metallic bar 50 cm long has a uniform cross-sectional area of 4.0 cm\\(^2\\). If a tensile force of 35 kN produces an extension of 0.25 mm, calculate the value of Young\'s modulus', '', '', NULL, '\n\nExplanation\nE \\(\\frac{Fl}{Ae}\\)\n= \\(\\frac{35 \\times 10^3 \\times 50 \\times 10^{-2}}{4 \\times 10^{-4} \\times 0.25 \\times 10 ^{-3}}\\)\n= 1.75 x 10\\(^{11}\\)Nm\\(^{-2}\\)\n\n\n', 'E \\(\\frac{Fl}{Ae}\\)\n= \\(\\frac{35 \\times 10^3 \\times 50 \\times 10^{-2}}{4 \\times 10^{-4} \\times 0.25 \\times 10 ^{-3}}\\)\n= 1.75 x 10\\(^{11}\\)Nm\\(^{-2}\\)\n\n\n', '', '', 'WAEC-PK-1', 'WAEC-PHY', 1),
(53, '2015', '\n\nQuestion\n\n(a) Explain how a gas can be made to conduct electricity.\n(b) Name the electric charge carriers in gases.\n\n', '', '(a) Explain how a gas can be made to conduct electricity.\n(b) Name the electric charge carriers in gases.', '(a) Explain how a gas can be made to conduct electricity.~(b) Name the electric charge carriers in gases.', NULL, '\n\nExplanation\n(a) Under very low pressure and very high voltage in at enclosed discharge tube ionized gas molecules or atom can conduct electricity.\n(b) Electric charge carriers in gases are ions and electrons.\n\n', '', 'Explanation\n(a) Under very low pressure and very high voltage in at enclosed discharge tube ionized gas molecules or atom can conduct electricity.\n(b) Electric charge carriers in gases are ions and electrons.\n\n', '(a) Under very low pressure and very high voltage in at enclosed discharge tube ionized gas molecules or atom can conduct electricity.~(b) Electric charge carriers in gases are ions and electrons.\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(54, '2015', '\n\nQuestion\n\nThe diagram above represents the graph of electron energy against the frequency of the radiation incident on a metal surface. Interpret the: (a) slope of the graph; (b) intercept, OC; (c) intercept, OK.\n\n\n', 'The diagram above represents the graph of electron energy against the frequency of the radiation incident on a metal surface. Interpret the: (a) slope of the graph; (b) intercept, OC; (c) intercept, OK.\n', '', '', NULL, '\n\nExplanation\n(a) Graph slope represents Planck\'s constant.\n(b) Intercept OC represents work function of the metal.\n(c) Intercept OK represents the threshold frequency F\\(_o\\) of the incident radiation\n\n', '', 'Explanation\n(a) Graph slope represents Planck\'s constant.\n(b) Intercept OC represents work function of the metal.\n(c) Intercept OK represents the threshold frequency F\\(_o\\) of the incident radiation\n\n', '(a) Graph slope represents Planck\'s constant.~(b) Intercept OC represents work function of the metal.\n(c) Intercept OK represents the threshold frequency F\\(_o\\) of the incident radiation\n~(c) Intercept OK represents the threshold frequency F\\(_o\\) of the incident radiation\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(55, '2015', '\n\nQuestion\n\n(a) State two conditions under which photo-electrons can be emitted from the surface of a metal.\n(b) List two particle characteristics of electromagnetic waves.\n\n', '', '(a) State two conditions under which photo-electrons can be emitted from the surface of a metal.\n(b) List two particle characteristics of electromagnetic waves.', '(a) State two conditions under which photo-electrons can be emitted from the surface of a metal.~(b) List two particle characteristics of electromagnetic waves.', NULL, '\n\nExplanation\nConditions for photo-electrons emission are:\n(i) the energy of the incident photons radiation mus\', be greater than the work function of the metal.\n(ii) the wavelength of incident radiation must be less than the threshold wavelength for metal surface.\n(iii) the frequency of incident radiation must be greater than the threshold, frequency.\n\n(b) Particle characteristics of electroniagnetic waves are:\n(i) Thermionic emission\n(ii) Photoelectric effect\n(iii) Compton effect\n(iv) Emission and absorption of incandescence light.\n(v) Radiations from heated black bodies.\n\n', '', 'Explanation\nConditions for photo-electrons emission are:\n(i) the energy of the incident photons radiation mus\', be greater than the work function of the metal.\n(ii) the wavelength of incident radiation must be less than the threshold wavelength for metal surface.\n(iii) the frequency of incident radiation must be greater than the threshold, frequency.\n\n(b) Particle characteristics of electroniagnetic waves are:\n(i) Thermionic emission\n(ii) Photoelectric effect\n(iii) Compton effect\n(iv) Emission and absorption of incandescence light.\n(v) Radiations from heated black bodies.\n\n', '(i) the energy of the incident photons radiation mus\', be greater than the work function of the metal.\n(ii) the wavelength of incident radiation must be less than the threshold wavelength for metal surface.\n(iii) the frequency of incident radiation must be greater than the threshold, frequency.\n~(b) Particle characteristics of electroniagnetic waves are:\n(i) Thermionic emission\n(ii) Photoelectric effect\n(iii) Compton effect\n(iv) Emission and absorption of incandescence light.\n(v) Radiations from heated black bodies.\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
(56, '2015', '\n\nQuestion\n\n(a) Give two examples each of:\n(i) rotational motion;\n(ii) linear motion.\n(b) Describe a laboratory experiment to determine the density of an irregularly shaped solid.\n(c) State Newton\'s second law of motion\n(d) Explain the term inertia.\n(e)\n\nThe diagram above illustrates a body of mass 5.0 kg being pulled by a horizontal force F. If the body accelerates at 2.0 ms\\(^{-2}\\) and experiences a frictional force of 5 N, calculate the:\n(i) net force on it;\n(ii) magnitude of F;\n(iii) coefficient of kinetic friction. [ g = 10 ms\\(^{-2}\\)]\n\n\n\n\n(a) Give two examples each of:\n(i) rotational motion;\n(ii) linear motion.\n(b) Describe a laboratory experiment to determine the density of an irregularly shaped solid.\n(c) State Newton\'s second law of motion\n(d) Explain the term inertia.\n(e)\n\nThe diagram above illustrates a body of mass 5.0 kg being pulled by a horizontal force F. If the body accelerates at 2.0 ms\\(^{-2}\\) and experiences a frictional force of 5 N, calculate the:\n(i) net force on it;\n(ii) magnitude of F;\n(iii) coefficient of kinetic friction. [ g = 10 ms\\(^{-2}\\)]\n\n', '', '(a) Give two examples each of:\n(i) rotational motion;\n(ii) linear motion.\n(b) Describe a laboratory experiment to determine the density of an irregularly shaped solid.\n(c) State Newton\'s second law of motion\n(d) Explain the term inertia.\n(e)\n\nThe diagram above illustrates a body of mass 5.0 kg being pulled by a horizontal force F. If the body accelerates at 2.0 ms\\(^{-2}\\) and experiences a frictional force of 5 N, calculate the:\n(i) net force on it;\n(ii) magnitude of F;\n(iii) coefficient of kinetic friction. [ g = 10 ms\\(^{-2}\\)]\n\n\n\n\n(a) Give two examples each of:\n(i) rotational motion;\n(ii) linear motion.\n(b) Describe a laboratory experiment to determine the density of an irregularly shaped solid.\n(c) State Newton\'s second law of motion\n(d) Explain the term inertia.\n(e)\n\nThe diagram above illustrates a body of mass 5.0 kg being pulled by a horizontal force F. If the body accelerates at 2.0 ms\\(^{-2}\\) and experiences a frictional force of 5 N, calculate the:\n(i) net force on it;\n(ii) magnitude of F;\n(iii) coefficient of kinetic friction. [ g = 10 ms\\(^{-2}\\)]', '(a) Give two examples each of:\n(i) rotational motion;\n(ii) linear motion.\n(b) Describe a laboratory experiment to determine the density of an irregularly shaped solid.\n(c) State Newton\'s second law of motion\n(d) Explain the term inertia.\n(e)\n\nThe diagram above illustrates a body of mass 5.0 kg being pulled by a horizontal force F. If the body accelerates at 2.0 ms\\(^{-2}\\) and experiences a frictional force of 5 N, calculate the:\n(i) net force on it;\n(ii) magnitude of F;\n(iii) coefficient of kinetic friction. [ g = 10 ms\\(^{-2}\\)]\n\n\n\n\n(a) Give two examples each of:\n(i) rotational motion;\n(ii) linear motion.~(b) Describe a laboratory experiment to determine the density of an irregularly shaped solid.\n(c) State Newton\'s second law of motion\n(d) Explain the term inertia.\n(e)\n\nThe diagram above illustrates a body of mass 5.0 kg being pulled by a horizontal force F. If the body accelerates at 2.0 ms\\(^{-2}\\) and experiences a frictional force of 5 N, calculate the:\n(i) net force on it;\n(ii) magnitude of F;\n(iii) coefficient of kinetic friction. [ g = 10 ms\\(^{-2}\\)]\n\n\n\n\n(a) Give two examples each of:\n(i) rotational motion;\n(ii) linear motion.\n(b) Describe a laboratory experiment to determine the density of an irregularly shaped solid.\n(c) State Newton\'s second law of motion\n(d) Explain the term inertia.\n(e)\n\nThe diagram above illustrates a body of mass 5.0 kg being pulled by a horizontal force F. If the body accelerates at 2.0 ms\\(^{-2}\\) and experiences a frictional force of 5 N, calculate the:\n(i) net force on it;\n(ii) magnitude of F;\n(iii) coefficient of kinetic friction. [ g = 10 ms\\(^{-2}\\)]~(c) State Newton\'s second law of motion\n(d) Explain the term inertia.\n(e)\n\nThe diagram above illustrates a body of mass 5.0 kg being pulled by a horizontal force F. If the body accelerates at 2.0 ms\\(^{-2}\\) and experiences a frictional force of 5 N, calculate the:\n(i) net force on it;\n(ii) magnitude of F;\n(iii) coefficient of kinetic friction. [ g = 10 ms\\(^{-2}\\)]\n\n\n\n\n(a) Give two examples each of:\n(i) rotational motion;\n(ii) linear motion.\n(b) Describe a laboratory experiment to determine the density of an irregularly shaped solid.\n(c) State Newton\'s second law of motion\n(d) Explain the term inertia.\n(e)\n\nThe diagram above illustrates a body of mass 5.0 kg being pulled by a horizontal force F. If the body accelerates at 2.0 ms\\(^{-2}\\) and experiences a frictional force of 5 N, calculate the:\n(i) net force on it;\n(ii) magnitude of F;\n(iii) coefficient of kinetic friction. [ g = 10 ms\\(^{-2}\\)]~(d) Explain the term inertia.\n(e)\n\nThe diagram above illustrates a body of mass 5.0 kg being pulled by a horizontal force F. If the body accelerates at 2.0 ms\\(^{-2}\\) and experiences a frictional force of 5 N, calculate the:\n(i) net force on it;\n(ii) magnitude of F;\n(iii) coefficient of kinetic friction. [ g = 10 ms\\(^{-2}\\)]\n\n\n\n\n(a) Give two examples each of:\n(i) rotational motion;\n(ii) linear motion.\n(b) Describe a laboratory experiment to determine the density of an irregularly shaped solid.\n(c) State Newton\'s second law of motion\n(d) Explain the term inertia.\n(e)\n\nThe diagram above illustrates a body of mass 5.0 kg being pulled by a horizontal force F. If the body accelerates at 2.0 ms\\(^{-2}\\) and experiences a frictional force of 5 N, calculate the:\n(i) net force on it;\n(ii) magnitude of F;\n(iii) coefficient of kinetic friction. [ g = 10 ms\\(^{-2}\\)]~(e)\n\nThe diagram above illustrates a body of mass 5.0 kg being pulled by a horizontal force F. If the body accelerates at 2.0 ms\\(^{-2}\\) and experiences a frictional force of 5 N, calculate the:\n(i) net force on it;\n(ii) magnitude of F;\n(iii) coefficient of kinetic friction. [ g = 10 ms\\(^{-2}\\)]\n\n\n\n\n(a) Give two examples each of:\n(i) rotational motion;\n(ii) linear motion.\n(b) Describe a laboratory experiment to determine the density of an irregularly shaped solid.\n(c) State Newton\'s second law of motion\n(d) Explain the term inertia.\n(e)\n\nThe diagram above illustrates a body of mass 5.0 kg being pulled by a horizontal force F. If the body accelerates at 2.0 ms\\(^{-2}\\) and experiences a frictional force of 5 N, calculate the:\n(i) net force on it;\n(ii) magnitude of F;\n(iii) coefficient of kinetic friction. [ g = 10 ms\\(^{-2}\\)]', NULL, '\n\nExplanation\n(a)(i) Rotational Motion; Examples:\n(1) Motion of a wind mill.\n(2) Motion of a turntable or disc.\n(3) Rotation of the earth about its axis.\n(4) Movement of the blade of an electric fan, etc.\n\n(ii) Linear Motion; Examples are:\n(1) An athlete running on a straight track.\n(2) A car moving on a straight road.\n(3)A ball rolling on a level ground, etc.\n\n(b) In the laboratory, the mass(m) of an irregular shaped solid can be determined using a chemical or beam balance using a graduated measuring cylinder fill it partially with water and the initial volume V\\(_2\\) recorded, then completely immersed the irregular solid shape in the water and the final volume V recorded. Volume of the solid = V\\(_2\\) - V\\(_{1}\\)\nBut density = \\(\\frac{Mass}{ Volume}\\)\n= \\(\\frac{M}{V_2 - V_1}\\)\n(c) Newton\'s second law of motion states that the time rate of change of momentum of a body is directly propotional to the applied force acting on it and takes place in the direction of the force.\n(d) Inertia is the reluctance of a body to move if it is at rest or to stop if it is already in motion. The more the mass of a body, the greater is its inertia.\n(e)(i) Net force = F - F\\(_1\\) = Ma\n5.0 x 2.0 = 10N\n(ii) F = Ma + F\\(_1\\) = 10 + 5 = 15.0N\n(iii) ', '', 'Explanation\n(a)(i) Rotational Motion; Examples:\n(1) Motion of a wind mill.\n(2) Motion of a turntable or disc.\n(3) Rotation of the earth about its axis.\n(4) Movement of the blade of an electric fan, etc.\n\n(ii) Linear Motion; Examples are:\n(1) An athlete running on a straight track.\n(2) A car moving on a straight road.\n(3)A ball rolling on a level ground, etc.\n\n(b) In the laboratory, the mass(m) of an irregular shaped solid can be determined using a chemical or beam balance using a graduated measuring cylinder fill it partially with water and the initial volume V\\(_2\\) recorded, then completely immersed the irregular solid shape in the water and the final volume V recorded. Volume of the solid = V\\(_2\\) - V\\(_{1}\\)\nBut density = \\(\\frac{Mass}{ Volume}\\)\n= \\(\\frac{M}{V_2 - V_1}\\)\n(c) Newton\'s second law of motion states that the time rate of change of momentum of a body is directly propotional to the applied force acting on it and takes place in the direction of the force.\n(d) Inertia is the reluctance of a body to move if it is at rest or to stop if it is already in motion. The more the mass of a body, the greater is its inertia.\n(e)(i) Net force = F - F\\(_1\\) = Ma\n5.0 x 2.0 = 10N\n(ii) F = Ma + F\\(_1\\) = 10 + 5 = 15.0N\n(iii) ', '(a)(i) Rotational Motion; Examples:\n(1) Motion of a wind mill.\n(2) Motion of a turntable or disc.\n(3) Rotation of the earth about its axis.\n(4) Movement of the blade of an electric fan, etc.\n\n(ii) Linear Motion; Examples are:\n(1) An athlete running on a straight track.\n(2) A car moving on a straight road.\n(3)A ball rolling on a level ground, etc.\n~(b) In the laboratory, the mass(m) of an irregular shaped solid can be determined using a chemical or beam balance using a graduated measuring cylinder fill it partially with water and the initial volume V\\(_2\\) recorded, then completely immersed the irregular solid shape in the water and the final volume V recorded. Volume of the solid = V\\(_2\\) - V\\(_{1}\\)\nBut density = \\(\\frac{Mass}{ Volume}\\)\n= \\(\\frac{M}{V_2 - V_1}\\)\n(c) Newton\'s second law of motion states that the time rate of change of momentum of a body is directly propotional to the applied force acting on it and takes place in the direction of the force.\n(d) Inertia is the reluctance of a body to move if it is at rest or to stop if it is already in motion. The more the mass of a body, the greater is its inertia.\n(e)(i) Net force = F - F\\(_1\\) = Ma\n5.0 x 2.0 = 10N\n(ii) F = Ma + F\\(_1\\) = 10 + 5 = 15.0N~(c) Newton\'s second law of motion states that the time rate of change of momentum of a body is directly propotional to the applied force acting on it and takes place in the direction of the force.\n(d) Inertia is the reluctance of a body to move if it is at rest or to stop if it is already in motion. The more the mass of a body, the greater is its inertia.\n(e)(i) Net force = F - F\\(_1\\) = Ma\n5.0 x 2.0 = 10N\n(ii) F = Ma + F\\(_1\\) = 10 + 5 = 15.0N~(d) Inertia is the reluctance of a body to move if it is at rest or to stop if it is already in motion. The more the mass of a body, the greater is its inertia.\n(e)(i) Net force = F - F\\(_1\\) = Ma\n5.0 x 2.0 = 10N\n(ii) F = Ma + F\\(_1\\) = 10 + 5 = 15.0N~(e)(i) Net force = F - F\\(_1\\) = Ma\n5.0 x 2.0 = 10N\n(ii) F = Ma + F\\(_1\\) = 10 + 5 = 15.0N', 'WAEC-PK-1', 'WAEC-PHY', 1),
(57, '2015', '\n\nQuestion\n\na) Define heat capacity and state its unit.\n(b) List two effects of heat on a substance.\n(c) Explain how a tightly fitted glass stopper could be removed from a reagent bottle.\n(d) A quantity of pepper soup of mass 800 g poured into a plastic container with a tight-fitting lid has a temperature of 30°C. The container is then placed in a microwave oven, rated 1200 W and operated for 3 minutes.\n(i) Calculate the final temperature attained by the soup. (Assuming no heat losses).\n(ii) Explain why containers with tight-fitting lids are not suitable for use in microwave cooking.\n(iii) When the soup is brought out and allowed to cool, a dent is observed on the container. Explain. [Take specific heat capacity of the soup = 4000 Jkg\\(^{-1}\\) K\\(^{-1}\\)]\n\n', '', 'a) Define heat capacity and state its unit.\n(b) List two effects of heat on a substance.\n(c) Explain how a tightly fitted glass stopper could be removed from a reagent bottle.\n(d) A quantity of pepper soup of mass 800 g poured into a plastic container with a tight-fitting lid has a temperature of 30°C. The container is then placed in a microwave oven, rated 1200 W and operated for 3 minutes.\n(i) Calculate the final temperature attained by the soup. (Assuming no heat losses).\n(ii) Explain why containers with tight-fitting lids are not suitable for use in microwave cooking.\n(iii) When the soup is brought out and allowed to cool, a dent is observed on the container. Explain. [Take specific heat capacity of the soup = 4000 Jkg\\(^{-1}\\) K\\(^{-1}\\)]', '~(b) List two effects of heat on a substance.\n(c) Explain how a tightly fitted glass stopper could be removed from a reagent bottle.\n(d) A quantity of pepper soup of mass 800 g poured into a plastic container with a tight-fitting lid has a temperature of 30°C. The container is then placed in a microwave oven, rated 1200 W and operated for 3 minutes.\n(i) Calculate the final temperature attained by the soup. (Assuming no heat losses).\n(ii) Explain why containers with tight-fitting lids are not suitable for use in microwave cooking.\n(iii) When the soup is brought out and allowed to cool, a dent is observed on the container. Explain. [Take specific heat capacity of the soup = 4000 Jkg\\(^{-1}\\) K\\(^{-1}\\)]~(c) Explain how a tightly fitted glass stopper could be removed from a reagent bottle.\n(d) A quantity of pepper soup of mass 800 g poured into a plastic container with a tight-fitting lid has a temperature of 30°C. The container is then placed in a microwave oven, rated 1200 W and operated for 3 minutes.\n(i) Calculate the final temperature attained by the soup. (Assuming no heat losses).\n(ii) Explain why containers with tight-fitting lids are not suitable for use in microwave cooking.\n(iii) When the soup is brought out and allowed to cool, a dent is observed on the container. Explain. [Take specific heat capacity of the soup = 4000 Jkg\\(^{-1}\\) K\\(^{-1}\\)]~(d) A quantity of pepper soup of mass 800 g poured into a plastic container with a tight-fitting lid has a temperature of 30°C. The container is then placed in a microwave oven, rated 1200 W and operated for 3 minutes.\n(i) Calculate the final temperature attained by the soup. (Assuming no heat losses).\n(ii) Explain why containers with tight-fitting lids are not suitable for use in microwave cooking.\n(iii) When the soup is brought out and allowed to cool, a dent is observed on the container. Explain. [Take specific heat capacity of the soup = 4000 Jkg\\(^{-1}\\) K\\(^{-1}\\)]', NULL, '\n\nExplanation\n(a) Heat capacity is the quantity of heat required to change the temperature of a body by 1K. The unit is JK\\(^{-1}\\)\n(b) Effects of heat on a substance are:\n(i) Increase in temperature.\n(ii) Change of state.\n(iii) Change in colour .\n(iv) Change in electrical resistance, etc.\n(c) To remove a tightly fitted glass cork from a reagent bottle, warm the bottle neck using warm cloth to make the glass bottle to expand creating space for easy removal of the cork.\n(d)(i) Pt = MC\\(\\Delta \\theta\\)\n\\(\\Delta \\theta = \\frac{pt}{Mc} = \\frac{1200 \\times 3 \\times 60}{0.8 \\times 4000}\\) =67.5°\n= final temp \\(\\theta\\) = 67.5 + 30° = 97.5°C\n\n(ii) Containers with tight-fitted lids are not suitable because with increase in temperature the steam molecules from the container are confined in the container, thereby increasing the rate of collission of the steam with the walls of the container, hence in pressure inside the container. Thus the high pressure can burst the tightly-fitted container and the hot content released.\n(iii) The observed dent on the container is due to the condensation of the steam in the container which-will lead to a decrease in pressure in the greater container making the atmospheric pressure greater than the pressure in the container.\n\n', '', 'Explanation\n(a) Heat capacity is the quantity of heat required to change the temperature of a body by 1K. The unit is JK\\(^{-1}\\)\n(b) Effects of heat on a substance are:\n(i) Increase in temperature.\n(ii) Change of state.\n(iii) Change in colour .\n(iv) Change in electrical resistance, etc.\n(c) To remove a tightly fitted glass cork from a reagent bottle, warm the bottle neck using warm cloth to make the glass bottle to expand creating space for easy removal of the cork.\n(d)(i) Pt = MC\\(\\Delta \\theta\\)\n\\(\\Delta \\theta = \\frac{pt}{Mc} = \\frac{1200 \\times 3 \\times 60}{0.8 \\times 4000}\\) =67.5°\n= final temp \\(\\theta\\) = 67.5 + 30° = 97.5°C\n\n(ii) Containers with tight-fitted lids are not suitable because with increase in temperature the steam molecules from the container are confined in the container, thereby increasing the rate of collission of the steam with the walls of the container, hence in pressure inside the container. Thus the high pressure can burst the tightly-fitted container and the hot content released.\n(iii) The observed dent on the container is due to the condensation of the steam in the container which-will lead to a decrease in pressure in the greater container making the atmospheric pressure greater than the pressure in the container.\n\n', '(a) Heat capacity is the quantity of heat required to change the temperature of a body by 1K. The unit is JK\\(^{-1}\\)~(b) Effects of heat on a substance are:\n(i) Increase in temperature.\n(ii) Change of state.\n(iii) Change in colour .\n(iv) Change in electrical resistance, etc.\n(c) To remove a tightly fitted glass cork from a reagent bottle, warm the bottle neck using warm cloth to make the glass bottle to expand creating space for easy removal of the cork.\n(d)(i) Pt = MC\\(\\Delta \\theta\\)\n\\(\\Delta \\theta = \\frac{pt}{Mc} = \\frac{1200 \\times 3 \\times 60}{0.8 \\times 4000}\\) =67.5°\n= final temp \\(\\theta\\) = 67.5 + 30° = 97.5°C\n\n(ii) Containers with tight-fitted lids are not suitable because with increase in temperature the steam molecules from the container are confined in the container, thereby increasing the rate of collission of the steam with the walls of the container, hence in pressure inside the container. Thus the high pressure can burst the tightly-fitted container and the hot content released.\n(iii) The observed dent on the container is due to the condensation of the steam in the container which-will lead to a decrease in pressure in the greater container making the atmospheric pressure greater than the pressure in the container.\n~(c) To remove a tightly fitted glass cork from a reagent bottle, warm the bottle neck using warm cloth to make the glass bottle to expand creating space for easy removal of the cork.\n(d)(i) Pt = MC\\(\\Delta \\theta\\)\n\\(\\Delta \\theta = \\frac{pt}{Mc} = \\frac{1200 \\times 3 \\times 60}{0.8 \\times 4000}\\) =67.5°\n= final temp \\(\\theta\\) = 67.5 + 30° = 97.5°C\n\n(ii) Containers with tight-fitted lids are not suitable because with increase in temperature the steam molecules from the container are confined in the container, thereby increasing the rate of collission of the steam with the walls of the container, hence in pressure inside the container. Thus the high pressure can burst the tightly-fitted container and the hot content released.\n(iii) The observed dent on the container is due to the condensation of the steam in the container which-will lead to a decrease in pressure in the greater container making the atmospheric pressure greater than the pressure in the container.\n~(d)(i) Pt = MC\\(\\Delta \\theta\\)\n\\(\\Delta \\theta = \\frac{pt}{Mc} = \\frac{1200 \\times 3 \\times 60}{0.8 \\times 4000}\\) =67.5°\n= final temp \\(\\theta\\) = 67.5 + 30° = 97.5°C\n\n(ii) Containers with tight-fitted lids are not suitable because with increase in temperature the steam molecules from the container are confined in the container, thereby increasing the rate of collission of the steam with the walls of the container, hence in pressure inside the container. Thus the high pressure can burst the tightly-fitted container and the hot content released.\n(iii) The observed dent on the container is due to the condensation of the steam in the container which-will lead to a decrease in pressure in the greater container making the atmospheric pressure greater than the pressure in the container.\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(58, '2015', '\n\nQuestion\n\n(a) State the three characteristics of sound and the factor on which each of them depends.\n(b) Explain resonance as applied to sound.\n(c) What role does echo play in the construction of a concert hall?\n(d) The surface of an ear drum (assumed circular) has a radius 2.1 mm. It resonates with an amplitude of 0.8 x 10\\(^{-7}\\) in as a result of impulses received from an external body vibrating at 2400 Hz. If the resulting pressure change on the ear drum is 3.6 x 10\\(^{-5}\\) NM\\(^{-2}\\), calculate the:\n(i) period of oscillation;\n(ii) velocity;\n(iii) acceleration;\n(iv) force. [\\(\\pi\\) = 3.14 ].\n\n', '', '(a) State the three characteristics of sound and the factor on which each of them depends.\n(b) Explain resonance as applied to sound.\n(c) What role does echo play in the construction of a concert hall?\n(d) The surface of an ear drum (assumed circular) has a radius 2.1 mm. It resonates with an amplitude of 0.8 x 10\\(^{-7}\\) in as a result of impulses received from an external body vibrating at 2400 Hz. If the resulting pressure change on the ear drum is 3.6 x 10\\(^{-5}\\) NM\\(^{-2}\\), calculate the:\n(i) period of oscillation;\n(ii) velocity;\n(iii) acceleration;\n(iv) force. [\\(\\pi\\) = 3.14 ].', '(a) State the three characteristics of sound and the factor on which each of them depends.~(b) Explain resonance as applied to sound.\n(c) What role does echo play in the construction of a concert hall?\n(d) The surface of an ear drum (assumed circular) has a radius 2.1 mm. It resonates with an amplitude of 0.8 x 10\\(^{-7}\\) in as a result of impulses received from an external body vibrating at 2400 Hz. If the resulting pressure change on the ear drum is 3.6 x 10\\(^{-5}\\) NM\\(^{-2}\\), calculate the:\n(i) period of oscillation;\n(ii) velocity;\n(iii) acceleration;\n(iv) force. [\\(\\pi\\) = 3.14 ].~(c) What role does echo play in the construction of a concert hall?\n(d) The surface of an ear drum (assumed circular) has a radius 2.1 mm. It resonates with an amplitude of 0.8 x 10\\(^{-7}\\) in as a result of impulses received from an external body vibrating at 2400 Hz. If the resulting pressure change on the ear drum is 3.6 x 10\\(^{-5}\\) NM\\(^{-2}\\), calculate the:\n(i) period of oscillation;\n(ii) velocity;\n(iii) acceleration;\n(iv) force. [\\(\\pi\\) = 3.14 ].~(d) The surface of an ear drum (assumed circular) has a radius 2.1 mm. It resonates with an amplitude of 0.8 x 10\\(^{-7}\\) in as a result of impulses received from an external body vibrating at 2400 Hz. If the resulting pressure change on the ear drum is 3.6 x 10\\(^{-5}\\) NM\\(^{-2}\\), calculate the:\n(i) period of oscillation;\n(ii) velocity;\n(iii) acceleration;\n(iv) force. [\\(\\pi\\) = 3.14 ].', NULL, '\n\nExplanation\n(a) Characteristics of sound and their corresponding Factors\n\n\n\n\nCharacteristics\n\n\nfactors\n\n\n\n\n- Pitch\n- Timber or quality\n- Loudness or intensity\n\n\n- Frequency\n- Harmonics or Overtone\n- Amplitude\n\n\n\n\n\n(b) Resonance is a phenomenon which occurs when an external vibrating source of sound such as a sounding tunning fork set another system such as molecules of column into vibration; at a stage the system vibrates with maximum amplitude producing, a loud sound.\n\n(c) Echo enables the construction engineers to line the interior wall with sound absorbing materials to prevent echo.\n(d)(i) Period T = \\(\\frac{1}{f} = \\frac{1}{2400} = 4.2 \\times 10^{-7}\\)\n= 1.2 x 10\\(^{-3}\\)ms\\(^{-1}\\)\n(ii) Velocity V = \\(\\omega r = 2 \\pi fr\\)\n= 2 x 3.142 x 2400 x 0.8 x 10\\(^{-7}\\)\n= 1.2 x 10\\(^{-3}\\)ms\\(^{-1}\\)\n(iii) Acceleration a = \\(\\omega^2r\\)\n= (\\(2 \\pi f)^2r = (2 \\times 3.14 \\times 2400)^2 \\times 0.8 \\times 10^{-7}\\)\n= 18.2 ms\\(^{-2}\\)\n(iv) p = \\(\\frac{F}{A}\\)\nforce F = PA\n= 3.6 x 10\\(^{-5} \\times 3.14 \\times (2.1 \\times 10^{-3})^2\\)\n= 5.0 x 10\\(^{-11}\\) N\n\n\n\n', '', 'Explanation\n(a) Characteristics of sound and their corresponding Factors\n\n\n\n\nCharacteristics\n\n\nfactors\n\n\n\n\n- Pitch\n- Timber or quality\n- Loudness or intensity\n\n\n- Frequency\n- Harmonics or Overtone\n- Amplitude\n\n\n\n\n\n(b) Resonance is a phenomenon which occurs when an external vibrating source of sound such as a sounding tunning fork set another system such as molecules of column into vibration; at a stage the system vibrates with maximum amplitude producing, a loud sound.\n\n(c) Echo enables the construction engineers to line the interior wall with sound absorbing materials to prevent echo.\n(d)(i) Period T = \\(\\frac{1}{f} = \\frac{1}{2400} = 4.2 \\times 10^{-7}\\)\n= 1.2 x 10\\(^{-3}\\)ms\\(^{-1}\\)\n(ii) Velocity V = \\(\\omega r = 2 \\pi fr\\)\n= 2 x 3.142 x 2400 x 0.8 x 10\\(^{-7}\\)\n= 1.2 x 10\\(^{-3}\\)ms\\(^{-1}\\)\n(iii) Acceleration a = \\(\\omega^2r\\)\n= (\\(2 \\pi f)^2r = (2 \\times 3.14 \\times 2400)^2 \\times 0.8 \\times 10^{-7}\\)\n= 18.2 ms\\(^{-2}\\)\n(iv) p = \\(\\frac{F}{A}\\)\nforce F = PA\n= 3.6 x 10\\(^{-5} \\times 3.14 \\times (2.1 \\times 10^{-3})^2\\)\n= 5.0 x 10\\(^{-11}\\) N\n\n\n\n', '(a) Characteristics of sound and their corresponding Factors\n\n\n\n\nCharacteristics\n\n\nfactors\n\n\n\n\n- Pitch\n- Timber or quality\n- Loudness or intensity\n\n\n- Frequency\n- Harmonics or Overtone\n- Amplitude\n\n\n\n\n~(b) Resonance is a phenomenon which occurs when an external vibrating source of sound such as a sounding tunning fork set another system such as molecules of column into vibration; at a stage the system vibrates with maximum amplitude producing, a loud sound.\n\n(c) Echo enables the construction engineers to line the interior wall with sound absorbing materials to prevent echo.\n(d)(i) Period T = \\(\\frac{1}{f} = \\frac{1}{2400} = 4.2 \\times 10^{-7}\\)\n= 1.2 x 10\\(^{-3}\\)ms\\(^{-1}\\)\n(ii) Velocity V = \\(\\omega r = 2 \\pi fr\\)\n= 2 x 3.142 x 2400 x 0.8 x 10\\(^{-7}\\)\n= 1.2 x 10\\(^{-3}\\)ms\\(^{-1}\\)\n(iii) Acceleration a = \\(\\omega^2r\\)\n= (\\(2 \\pi f)^2r = (2 \\times 3.14 \\times 2400)^2 \\times 0.8 \\times 10^{-7}\\)\n= 18.2 ms\\(^{-2}\\)\n(iv) p = \\(\\frac{F}{A}\\)\nforce F = PA\n= 3.6 x 10\\(^{-5} \\times 3.14 \\times (2.1 \\times 10^{-3})^2\\)\n= 5.0 x 10\\(^{-11}\\) N\n\n\n~(c) Echo enables the construction engineers to line the interior wall with sound absorbing materials to prevent echo.\n(d)(i) Period T = \\(\\frac{1}{f} = \\frac{1}{2400} = 4.2 \\times 10^{-7}\\)\n= 1.2 x 10\\(^{-3}\\)ms\\(^{-1}\\)\n(ii) Velocity V = \\(\\omega r = 2 \\pi fr\\)\n= 2 x 3.142 x 2400 x 0.8 x 10\\(^{-7}\\)\n= 1.2 x 10\\(^{-3}\\)ms\\(^{-1}\\)\n(iii) Acceleration a = \\(\\omega^2r\\)\n= (\\(2 \\pi f)^2r = (2 \\times 3.14 \\times 2400)^2 \\times 0.8 \\times 10^{-7}\\)\n= 18.2 ms\\(^{-2}\\)\n(iv) p = \\(\\frac{F}{A}\\)\nforce F = PA\n= 3.6 x 10\\(^{-5} \\times 3.14 \\times (2.1 \\times 10^{-3})^2\\)\n= 5.0 x 10\\(^{-11}\\) N\n\n\n~(d)(i) Period T = \\(\\frac{1}{f} = \\frac{1}{2400} = 4.2 \\times 10^{-7}\\)\n= 1.2 x 10\\(^{-3}\\)ms\\(^{-1}\\)\n(ii) Velocity V = \\(\\omega r = 2 \\pi fr\\)\n= 2 x 3.142 x 2400 x 0.8 x 10\\(^{-7}\\)\n= 1.2 x 10\\(^{-3}\\)ms\\(^{-1}\\)\n(iii) Acceleration a = \\(\\omega^2r\\)\n= (\\(2 \\pi f)^2r = (2 \\times 3.14 \\times 2400)^2 \\times 0.8 \\times 10^{-7}\\)\n= 18.2 ms\\(^{-2}\\)\n(iv) p = \\(\\frac{F}{A}\\)\nforce F = PA\n= 3.6 x 10\\(^{-5} \\times 3.14 \\times (2.1 \\times 10^{-3})^2\\)\n= 5.0 x 10\\(^{-11}\\) N\n\n\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(59, '2015', '\n\nQuestion\n\n(a) Define electromotive force.\n(b) State:\n(i) the principle of operation of a potentiometer,\n(ii) two advantages that a potentiometer has over a voltmeter in measuring potential difference.\n(c)(i) Sketch and label a diagram of a gold-leaf electroscope.\n(ii) Give one use of a gold-leaf electroscope.\n(d)(i) Explain the action of a magnetic relay.\n(ii) List two factors which determine the magnitude of an induced emf in a coil.\n(iii) A current of 5 A passes through a straight wire in a uniform magnetic field of flux density 2.0 x10\\(^{-3}\\) T. Calculate the force per unit length exerted on the wire when it is inclined at 30° to the field.\n\n', '', '(a) Define electromotive force.\n(b) State:\n(i) the principle of operation of a potentiometer,\n(ii) two advantages that a potentiometer has over a voltmeter in measuring potential difference.\n(c)(i) Sketch and label a diagram of a gold-leaf electroscope.\n(ii) Give one use of a gold-leaf electroscope.\n(d)(i) Explain the action of a magnetic relay.\n(ii) List two factors which determine the magnitude of an induced emf in a coil.\n(iii) A current of 5 A passes through a straight wire in a uniform magnetic field of flux density 2.0 x10\\(^{-3}\\) T. Calculate the force per unit length exerted on the wire when it is inclined at 30° to the field.', '(a) Define electromotive force.~(b) State:\n(i) the principle of operation of a potentiometer,\n(ii) two advantages that a potentiometer has over a voltmeter in measuring potential difference.\n(c)(i) Sketch and label a diagram of a gold-leaf electroscope.\n(ii) Give one use of a gold-leaf electroscope.\n(d)(i) Explain the action of a magnetic relay.\n(ii) List two factors which determine the magnitude of an induced emf in a coil.\n(iii) A current of 5 A passes through a straight wire in a uniform magnetic field of flux density 2.0 x10\\(^{-3}\\) T. Calculate the force per unit length exerted on the wire when it is inclined at 30° to the field.~(c)(i) Sketch and label a diagram of a gold-leaf electroscope.\n(ii) Give one use of a gold-leaf electroscope.\n(d)(i) Explain the action of a magnetic relay.\n(ii) List two factors which determine the magnitude of an induced emf in a coil.\n(iii) A current of 5 A passes through a straight wire in a uniform magnetic field of flux density 2.0 x10\\(^{-3}\\) T. Calculate the force per unit length exerted on the wire when it is inclined at 30° to the field.~(d)(i) Explain the action of a magnetic relay.\n(ii) List two factors which determine the magnitude of an induced emf in a coil.\n(iii) A current of 5 A passes through a straight wire in a uniform magnetic field of flux density 2.0 x10\\(^{-3}\\) T. Calculate the force per unit length exerted on the wire when it is inclined at 30° to the field.', NULL, '\n\nExplanation\n(a) Elecromotive force is the energy needed to take a unit positive charge round a circuit in which a cell is connected. Or The energy per unit charge transformed in the cell. Or e.m.f is the p.d across the terminals of the cell or battery or generator in an open circuit.\n\n(b)(i) The principle of operation of a potentiometer is based on the fact that the p.d across the wire is directly proportional to the length of the wire.V \\(\\alpha\\) L\nwhere V = p.d across the wire.\nL = length of the wire.\n(ii) Advantages of potentiometer over voltmeter in measuring p.d are:\n(1) It is more sensitive.\n(2) It is more accurate.\n(3) It eliminates error due to internal resistance of the cell.\n\n(c)(i) Labelled sketch of a Gold leaf electroscope:\n\n(ii) Uses of Gold leaf electroscope are:\n(1) To detetct the nature of electrostatic charge.\n(2) To detect electrostatic charge.\n(3) To determine the magnitude of the induced charges.\n\n(d)(i) The mode of action of magnetic relay relies on the principle of electromagnetic induction. A small current passing through one circuit controls another circuit containing devices which require larger current.\n(ii) Factors affecting the magnitude of induced e.m.f in a coil are:\n(1) the number of turns in the coil.\n(2) the magnetic field strength.\n(3) the cross-sectional area of the coil.\n(4) the speed of rotation of the coil.\n(iii) F = IBLsin\\(\\theta\\)\nforce per unit length\n\\(\\frac{F}{L}\\) = IBsin\\(\\theta\\)\n= 5 x 2.0 x 10\\(^{-3}\\) x sin30\n= 5.0 x 10\\(^{-3}\\)Nm\\(^{-1}\\)\n\n\n', '', 'Explanation\n(a) Elecromotive force is the energy needed to take a unit positive charge round a circuit in which a cell is connected. Or The energy per unit charge transformed in the cell. Or e.m.f is the p.d across the terminals of the cell or battery or generator in an open circuit.\n\n(b)(i) The principle of operation of a potentiometer is based on the fact that the p.d across the wire is directly proportional to the length of the wire.V \\(\\alpha\\) L\nwhere V = p.d across the wire.\nL = length of the wire.\n(ii) Advantages of potentiometer over voltmeter in measuring p.d are:\n(1) It is more sensitive.\n(2) It is more accurate.\n(3) It eliminates error due to internal resistance of the cell.\n\n(c)(i) Labelled sketch of a Gold leaf electroscope:\n\n(ii) Uses of Gold leaf electroscope are:\n(1) To detetct the nature of electrostatic charge.\n(2) To detect electrostatic charge.\n(3) To determine the magnitude of the induced charges.\n\n(d)(i) The mode of action of magnetic relay relies on the principle of electromagnetic induction. A small current passing through one circuit controls another circuit containing devices which require larger current.\n(ii) Factors affecting the magnitude of induced e.m.f in a coil are:\n(1) the number of turns in the coil.\n(2) the magnetic field strength.\n(3) the cross-sectional area of the coil.\n(4) the speed of rotation of the coil.\n(iii) F = IBLsin\\(\\theta\\)\nforce per unit length\n\\(\\frac{F}{L}\\) = IBsin\\(\\theta\\)\n= 5 x 2.0 x 10\\(^{-3}\\) x sin30\n= 5.0 x 10\\(^{-3}\\)Nm\\(^{-1}\\)\n\n\n', '(a) Elecromotive force is the energy needed to take a unit positive charge round a circuit in which a cell is connected. Or The energy per unit charge transformed in the cell. Or e.m.f is the p.d across the terminals of the cell or battery or generator in an open circuit.\n~(b)(i) The principle of operation of a potentiometer is based on the fact that the p.d across the wire is directly proportional to the length of the wire.V \\(\\alpha\\) L\nwhere V = p.d across the wire.\nL = length of the wire.\n(ii) Advantages of potentiometer over voltmeter in measuring p.d are:\n(1) It is more sensitive.\n(2) It is more accurate.\n(3) It eliminates error due to internal resistance of the cell.\n\n(c)(i) Labelled sketch of a Gold leaf electroscope:\n\n(ii) Uses of Gold leaf electroscope are:\n(1) To detetct the nature of electrostatic charge.\n(2) To detect electrostatic charge.\n(3) To determine the magnitude of the induced charges.\n\n(d)(i) The mode of action of magnetic relay relies on the principle of electromagnetic induction. A small current passing through one circuit controls another circuit containing devices which require larger current.\n(ii) Factors affecting the magnitude of induced e.m.f in a coil are:\n(1) the number of turns in the coil.\n(2) the magnetic field strength.\n(3) the cross-sectional area of the coil.\n(4) the speed of rotation of the coil.\n(iii) F = IBLsin\\(\\theta\\)\nforce per unit length\n\\(\\frac{F}{L}\\) = IBsin\\(\\theta\\)\n= 5 x 2.0 x 10\\(^{-3}\\) x sin30\n= 5.0 x 10\\(^{-3}\\)Nm\\(^{-1}\\)\n\n~(c)(i) Labelled sketch of a Gold leaf electroscope:\n\n(ii) Uses of Gold leaf electroscope are:\n(1) To detetct the nature of electrostatic charge.\n(2) To detect electrostatic charge.\n(3) To determine the magnitude of the induced charges.\n\n(d)(i) The mode of action of magnetic relay relies on the principle of electromagnetic induction. A small current passing through one circuit controls another circuit containing devices which require larger current.\n(ii) Factors affecting the magnitude of induced e.m.f in a coil are:\n(1) the number of turns in the coil.\n(2) the magnetic field strength.\n(3) the cross-sectional area of the coil.\n(4) the speed of rotation of the coil.\n(iii) F = IBLsin\\(\\theta\\)\nforce per unit length\n\\(\\frac{F}{L}\\) = IBsin\\(\\theta\\)\n= 5 x 2.0 x 10\\(^{-3}\\) x sin30\n= 5.0 x 10\\(^{-3}\\)Nm\\(^{-1}\\)\n\n~(d)(i) The mode of action of magnetic relay relies on the principle of electromagnetic induction. A small current passing through one circuit controls another circuit containing devices which require larger current.\n(ii) Factors affecting the magnitude of induced e.m.f in a coil are:\n(1) the number of turns in the coil.\n(2) the magnetic field strength.\n(3) the cross-sectional area of the coil.\n(4) the speed of rotation of the coil.\n(iii) F = IBLsin\\(\\theta\\)\nforce per unit length\n\\(\\frac{F}{L}\\) = IBsin\\(\\theta\\)\n= 5 x 2.0 x 10\\(^{-3}\\) x sin30\n= 5.0 x 10\\(^{-3}\\)Nm\\(^{-1}\\)\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
(60, '2015', '\n\nQuestion\n\n(a) Write Einstein\'s photoelectric equation and identh: each component of the equation.\n(b) For a photocell, star; one factor each that is responsible fo: the:\n(i) emission (ii) rate of emission;\n(iii) energy of photoelectrons.\n(c)(i) Two nuclear equations are given below:\n\\(^{222}_{p}RN\\) \\(\\to\\) \\(^{218}_{84}PO + ^q_2He\\)...................A\n\\(^{214}_{83}RN\\) \\(\\to\\) \\(^{214}_{84}PO + ^m_nX\\)...................B\n\nDetermine the values of: (\\(\\alpha\\)) p and q in equation A; (\\(\\beta\\)) in and n in equation B and identify X.\n(ii) Give a reason why it is important to dispose o radioactive waste safely.\n(d)(i) A certain atom emits ultra violet photon of wavelength 2.4 x10\\(^{-7}\\)m. Calculate the energy of the photon:\n----------------- - 6.0 x 10\\(^{-19}\\)J\n---------------- - 8.2 x 10\\(^{-19}\\)J\n---------------- - 8.8 x 10\\(^{-19}\\)J\n---------------- - 16.7 x 10\\(^{-19}\\)J\n(ii) The figure above illustrates the energy levels o the atom. Copy the figure in your answer booklet anc indicate on it, the energy level transitions which cause the emission of the photon in (d)(i) above. [h= 6.6 x 10\\(^{-34}\\) Js; c = 3.0 x 108 ms\\(^{-1}\\)].\n\n\n\n\n(a) Write Einstein\'s photoelectric equation and identh: each component of the equation.\n(b) For a photocell, star; one factor each that is responsible fo: the:\n(i) emission (ii) rate of emission;\n(iii) energy of photoelectrons.\n(c)(i) Two nuclear equations are given below:\n\\(^{222}_{p}RN\\) \\(\\to\\) \\(^{218}_{84}PO + ^q_2He\\)...................A\n\\(^{214}_{83}RN\\) \\(\\to\\) \\(^{214}_{84}PO + ^m_nX\\)...................B\n\nDetermine the values of: (\\(\\alpha\\)) p and q in equation A; (\\(\\beta\\)) in and n in equation B and identify X.\n(ii) Give a reason why it is important to dispose o radioactive waste safely.\n(d)(i) A certain atom emits ultra violet photon of wavelength 2.4 x10\\(^{-7}\\)m. Calculate the energy of the photon:\n----------------- - 6.0 x 10\\(^{-19}\\)J\n---------------- - 8.2 x 10\\(^{-19}\\)J\n---------------- - 8.8 x 10\\(^{-19}\\)J\n---------------- - 16.7 x 10\\(^{-19}\\)J\n(ii) The figure above illustrates the energy levels o the atom. Copy the figure in your answer booklet anc indicate on it, the energy level transitions which cause the emission of the photon in (d)(i) above. [h= 6.6 x 10\\(^{-34}\\) Js; c = 3.0 x 108 ms\\(^{-1}\\)].\n\n', '', '(a) Write Einstein\'s photoelectric equation and identh: each component of the equation.\n(b) For a photocell, star; one factor each that is responsible fo: the:\n(i) emission (ii) rate of emission;\n(iii) energy of photoelectrons.\n(c)(i) Two nuclear equations are given below:\n\\(^{222}_{p}RN\\) \\(\\to\\) \\(^{218}_{84}PO + ^q_2He\\)...................A\n\\(^{214}_{83}RN\\) \\(\\to\\) \\(^{214}_{84}PO + ^m_nX\\)...................B\n\nDetermine the values of: (\\(\\alpha\\)) p and q in equation A; (\\(\\beta\\)) in and n in equation B and identify X.\n(ii) Give a reason why it is important to dispose o radioactive waste safely.\n(d)(i) A certain atom emits ultra violet photon of wavelength 2.4 x10\\(^{-7}\\)m. Calculate the energy of the photon:\n----------------- - 6.0 x 10\\(^{-19}\\)J\n---------------- - 8.2 x 10\\(^{-19}\\)J\n---------------- - 8.8 x 10\\(^{-19}\\)J\n---------------- - 16.7 x 10\\(^{-19}\\)J\n(ii) The figure above illustrates the energy levels o the atom. Copy the figure in your answer booklet anc indicate on it, the energy level transitions which cause the emission of the photon in (d)(i) above. [h= 6.6 x 10\\(^{-34}\\) Js; c = 3.0 x 108 ms\\(^{-1}\\)].\n\n\n\n\n(a) Write Einstein\'s photoelectric equation and identh: each component of the equation.\n(b) For a photocell, star; one factor each that is responsible fo: the:\n(i) emission (ii) rate of emission;\n(iii) energy of photoelectrons.\n(c)(i) Two nuclear equations are given below:\n\\(^{222}_{p}RN\\) \\(\\to\\) \\(^{218}_{84}PO + ^q_2He\\)...................A\n\\(^{214}_{83}RN\\) \\(\\to\\) \\(^{214}_{84}PO + ^m_nX\\)...................B\n\nDetermine the values of: (\\(\\alpha\\)) p and q in equation A; (\\(\\beta\\)) in and n in equation B and identify X.\n(ii) Give a reason why it is important to dispose o radioactive waste safely.\n(d)(i) A certain atom emits ultra violet photon of wavelength 2.4 x10\\(^{-7}\\)m. Calculate the energy of the photon:\n----------------- - 6.0 x 10\\(^{-19}\\)J\n---------------- - 8.2 x 10\\(^{-19}\\)J\n---------------- - 8.8 x 10\\(^{-19}\\)J\n---------------- - 16.7 x 10\\(^{-19}\\)J\n(ii) The figure above illustrates the energy levels o the atom. Copy the figure in your answer booklet anc indicate on it, the energy level transitions which cause the emission of the photon in (d)(i) above. [h= 6.6 x 10\\(^{-34}\\) Js; c = 3.0 x 108 ms\\(^{-1}\\)].', '(a) Write Einstein\'s photoelectric equation and identh: each component of the equation.\n(b) For a photocell, star; one factor each that is responsible fo: the:\n(i) emission (ii) rate of emission;\n(iii) energy of photoelectrons.\n(c)(i) Two nuclear equations are given below:\n\\(^{222}_{p}RN\\) \\(\\to\\) \\(^{218}_{84}PO + ^q_2He\\)...................A\n\\(^{214}_{83}RN\\) \\(\\to\\) \\(^{214}_{84}PO + ^m_nX\\)...................B\n\nDetermine the values of: (\\(\\alpha\\)) p and q in equation A; (\\(\\beta\\)) in and n in equation B and identify X.\n(ii) Give a reason why it is important to dispose o radioactive waste safely.\n(d)(i) A certain atom emits ultra violet photon of wavelength 2.4 x10\\(^{-7}\\)m. Calculate the energy of the photon:\n----------------- - 6.0 x 10\\(^{-19}\\)J\n---------------- - 8.2 x 10\\(^{-19}\\)J\n---------------- - 8.8 x 10\\(^{-19}\\)J\n---------------- - 16.7 x 10\\(^{-19}\\)J\n(ii) The figure above illustrates the energy levels o the atom. Copy the figure in your answer booklet anc indicate on it, the energy level transitions which cause the emission of the photon in (d)(i) above. [h= 6.6 x 10\\(^{-34}\\) Js; c = 3.0 x 108 ms\\(^{-1}\\)].\n\n\n\n\n(a) Write Einstein\'s photoelectric equation and identh: each component of the equation.~(b) For a photocell, star; one factor each that is responsible fo: the:\n(i) emission (ii) rate of emission;\n(iii) energy of photoelectrons.\n(c)(i) Two nuclear equations are given below:\n\\(^{222}_{p}RN\\) \\(\\to\\) \\(^{218}_{84}PO + ^q_2He\\)...................A\n\\(^{214}_{83}RN\\) \\(\\to\\) \\(^{214}_{84}PO + ^m_nX\\)...................B\n\nDetermine the values of: (\\(\\alpha\\)) p and q in equation A; (\\(\\beta\\)) in and n in equation B and identify X.\n(ii) Give a reason why it is important to dispose o radioactive waste safely.\n(d)(i) A certain atom emits ultra violet photon of wavelength 2.4 x10\\(^{-7}\\)m. Calculate the energy of the photon:\n----------------- - 6.0 x 10\\(^{-19}\\)J\n---------------- - 8.2 x 10\\(^{-19}\\)J\n---------------- - 8.8 x 10\\(^{-19}\\)J\n---------------- - 16.7 x 10\\(^{-19}\\)J\n(ii) The figure above illustrates the energy levels o the atom. Copy the figure in your answer booklet anc indicate on it, the energy level transitions which cause the emission of the photon in (d)(i) above. [h= 6.6 x 10\\(^{-34}\\) Js; c = 3.0 x 108 ms\\(^{-1}\\)].\n\n\n\n\n(a) Write Einstein\'s photoelectric equation and identh: each component of the equation.\n(b) For a photocell, star; one factor each that is responsible fo: the:\n(i) emission (ii) rate of emission;\n(iii) energy of photoelectrons.\n(c)(i) Two nuclear equations are given below:\n\\(^{222}_{p}RN\\) \\(\\to\\) \\(^{218}_{84}PO + ^q_2He\\)...................A\n\\(^{214}_{83}RN\\) \\(\\to\\) \\(^{214}_{84}PO + ^m_nX\\)...................B\n\nDetermine the values of: (\\(\\alpha\\)) p and q in equation A; (\\(\\beta\\)) in and n in equation B and identify X.\n(ii) Give a reason why it is important to dispose o radioactive waste safely.\n(d)(i) A certain atom emits ultra violet photon of wavelength 2.4 x10\\(^{-7}\\)m. Calculate the energy of the photon:\n----------------- - 6.0 x 10\\(^{-19}\\)J\n---------------- - 8.2 x 10\\(^{-19}\\)J\n---------------- - 8.8 x 10\\(^{-19}\\)J\n---------------- - 16.7 x 10\\(^{-19}\\)J\n(ii) The figure above illustrates the energy levels o the atom. Copy the figure in your answer booklet anc indicate on it, the energy level transitions which cause the emission of the photon in (d)(i) above. [h= 6.6 x 10\\(^{-34}\\) Js; c = 3.0 x 108 ms\\(^{-1}\\)].~(c)(i) Two nuclear equations are given below:\n\\(^{222}_{p}RN\\) \\(\\to\\) \\(^{218}_{84}PO + ^q_2He\\)...................A\n\\(^{214}_{83}RN\\) \\(\\to\\) \\(^{214}_{84}PO + ^m_nX\\)...................B\n\nDetermine the values of: (\\(\\alpha\\)) p and q in equation A; (\\(\\beta\\)) in and n in equation B and identify X.\n(ii) Give a reason why it is important to dispose o radioactive waste safely.\n(d)(i) A certain atom emits ultra violet photon of wavelength 2.4 x10\\(^{-7}\\)m. Calculate the energy of the photon:\n----------------- - 6.0 x 10\\(^{-19}\\)J\n---------------- - 8.2 x 10\\(^{-19}\\)J\n---------------- - 8.8 x 10\\(^{-19}\\)J\n---------------- - 16.7 x 10\\(^{-19}\\)J\n(ii) The figure above illustrates the energy levels o the atom. Copy the figure in your answer booklet anc indicate on it, the energy level transitions which cause the emission of the photon in (d)(i) above. [h= 6.6 x 10\\(^{-34}\\) Js; c = 3.0 x 108 ms\\(^{-1}\\)].\n\n\n\n\n(a) Write Einstein\'s photoelectric equation and identh: each component of the equation.\n(b) For a photocell, star; one factor each that is responsible fo: the:\n(i) emission (ii) rate of emission;\n(iii) energy of photoelectrons.\n(c)(i) Two nuclear equations are given below:\n\\(^{222}_{p}RN\\) \\(\\to\\) \\(^{218}_{84}PO + ^q_2He\\)...................A\n\\(^{214}_{83}RN\\) \\(\\to\\) \\(^{214}_{84}PO + ^m_nX\\)...................B\n\nDetermine the values of: (\\(\\alpha\\)) p and q in equation A; (\\(\\beta\\)) in and n in equation B and identify X.\n(ii) Give a reason why it is important to dispose o radioactive waste safely.\n(d)(i) A certain atom emits ultra violet photon of wavelength 2.4 x10\\(^{-7}\\)m. Calculate the energy of the photon:\n----------------- - 6.0 x 10\\(^{-19}\\)J\n---------------- - 8.2 x 10\\(^{-19}\\)J\n---------------- - 8.8 x 10\\(^{-19}\\)J\n---------------- - 16.7 x 10\\(^{-19}\\)J\n(ii) The figure above illustrates the energy levels o the atom. Copy the figure in your answer booklet anc indicate on it, the energy level transitions which cause the emission of the photon in (d)(i) above. [h= 6.6 x 10\\(^{-34}\\) Js; c = 3.0 x 108 ms\\(^{-1}\\)].~(d)(i) A certain atom emits ultra violet photon of wavelength 2.4 x10\\(^{-7}\\)m. Calculate the energy of the photon:\n----------------- - 6.0 x 10\\(^{-19}\\)J\n---------------- - 8.2 x 10\\(^{-19}\\)J\n---------------- - 8.8 x 10\\(^{-19}\\)J\n---------------- - 16.7 x 10\\(^{-19}\\)J\n(ii) The figure above illustrates the energy levels o the atom. Copy the figure in your answer booklet anc indicate on it, the energy level transitions which cause the emission of the photon in (d)(i) above. [h= 6.6 x 10\\(^{-34}\\) Js; c = 3.0 x 108 ms\\(^{-1}\\)].\n\n\n\n\n(a) Write Einstein\'s photoelectric equation and identh: each component of the equation.\n(b) For a photocell, star; one factor each that is responsible fo: the:\n(i) emission (ii) rate of emission;\n(iii) energy of photoelectrons.\n(c)(i) Two nuclear equations are given below:\n\\(^{222}_{p}RN\\) \\(\\to\\) \\(^{218}_{84}PO + ^q_2He\\)...................A\n\\(^{214}_{83}RN\\) \\(\\to\\) \\(^{214}_{84}PO + ^m_nX\\)...................B\n\nDetermine the values of: (\\(\\alpha\\)) p and q in equation A; (\\(\\beta\\)) in and n in equation B and identify X.\n(ii) Give a reason why it is important to dispose o radioactive waste safely.\n(d)(i) A certain atom emits ultra violet photon of wavelength 2.4 x10\\(^{-7}\\)m. Calculate the energy of the photon:\n----------------- - 6.0 x 10\\(^{-19}\\)J\n---------------- - 8.2 x 10\\(^{-19}\\)J\n---------------- - 8.8 x 10\\(^{-19}\\)J\n---------------- - 16.7 x 10\\(^{-19}\\)J\n(ii) The figure above illustrates the energy levels o the atom. Copy the figure in your answer booklet anc indicate on it, the energy level transitions which cause the emission of the photon in (d)(i) above. [h= 6.6 x 10\\(^{-34}\\) Js; c = 3.0 x 108 ms\\(^{-1}\\)].', NULL, '\n\nExplanation\n(a) Einstein\'s photoelectric equation is\nE = W + K Emax, where\nE = Energy of incident radiation\nW = Work function\nK.E\\(_{\\text{max}}\\) = Maximum kinetic energy of photoelectron.\n\n(b) For photocell factors responsible for:\n(i) Emission \\(\\to\\) frequency or wavelength of incident radiation.\n(ii) Rate of emission \\(\\to\\) intensity of incident radiation or number of photons.\n(iii) Energy of photoelectrons ---> frequency or wavelength of incident radiation.\n\n(c)(i) (\\(\\alpha\\)) 218 + q = 222\nq = 4\n(\\(\\alpha\\)) P = 84 + 2 = 86,\nm = 214 — 214 = 0\nn = 83 — 84 = —1\nx is an electron or beta particle.\n\nii) Needs to safely dispose radioactive waste:\n(1) it emits harmful radiations.\n(2) it is toxic\n\n(d)(i) E = \\(\\frac{hc}{\\lambda}\\)\n= \\(\\frac{6.6 \\times 10^{-34} \\times 3.0 \\times 10^{8}}{2.4 \\times 10^{8}}\\)\n= 8.25 x 10\\(^{-19}\\)J\n(ii)\nE\\(_4\\) ----------------- - 6.0 x 10\\(^{-19}\\)J\nE\\(_3\\) ---------------- - 8.2 x 10\\(^{-19}\\)J\nE\\(_2\\) ---------------- - 8.8 x 10\\(^{-19}\\)J\nE\\(_1\\) ---------------- - 16.7 x 10\\(^{-19}\\)J\nThe energy level transitions which cause the emission of the photon in (d)(i) is. E\\(_3\\)\n\n', '', 'Explanation\n(a) Einstein\'s photoelectric equation is\nE = W + K Emax, where\nE = Energy of incident radiation\nW = Work function\nK.E\\(_{\\text{max}}\\) = Maximum kinetic energy of photoelectron.\n\n(b) For photocell factors responsible for:\n(i) Emission \\(\\to\\) frequency or wavelength of incident radiation.\n(ii) Rate of emission \\(\\to\\) intensity of incident radiation or number of photons.\n(iii) Energy of photoelectrons ---> frequency or wavelength of incident radiation.\n\n(c)(i) (\\(\\alpha\\)) 218 + q = 222\nq = 4\n(\\(\\alpha\\)) P = 84 + 2 = 86,\nm = 214 — 214 = 0\nn = 83 — 84 = —1\nx is an electron or beta particle.\n\nii) Needs to safely dispose radioactive waste:\n(1) it emits harmful radiations.\n(2) it is toxic\n\n(d)(i) E = \\(\\frac{hc}{\\lambda}\\)\n= \\(\\frac{6.6 \\times 10^{-34} \\times 3.0 \\times 10^{8}}{2.4 \\times 10^{8}}\\)\n= 8.25 x 10\\(^{-19}\\)J\n(ii)\nE\\(_4\\) ----------------- - 6.0 x 10\\(^{-19}\\)J\nE\\(_3\\) ---------------- - 8.2 x 10\\(^{-19}\\)J\nE\\(_2\\) ---------------- - 8.8 x 10\\(^{-19}\\)J\nE\\(_1\\) ---------------- - 16.7 x 10\\(^{-19}\\)J\nThe energy level transitions which cause the emission of the photon in (d)(i) is. E\\(_3\\)\n\n', '(a) Einstein\'s photoelectric equation is\nE = W + K Emax, where\nE = Energy of incident radiation\nW = Work function\nK.E\\(_{\\text{max}}\\) = Maximum kinetic energy of photoelectron.\n~(b) For photocell factors responsible for:\n(i) Emission \\(\\to\\) frequency or wavelength of incident radiation.\n(ii) Rate of emission \\(\\to\\) intensity of incident radiation or number of photons.\n(iii) Energy of photoelectrons ---> frequency or wavelength of incident radiation.\n\n(c)(i) (\\(\\alpha\\)) 218 + q = 222\nq = 4\n(\\(\\alpha\\)) P = 84 + 2 = 86,\nm = 214 — 214 = 0\nn = 83 — 84 = —1\nx is an electron or beta particle.\n\nii) Needs to safely dispose radioactive waste:\n(1) it emits harmful radiations.\n(2) it is toxic\n\n(d)(i) E = \\(\\frac{hc}{\\lambda}\\)\n= \\(\\frac{6.6 \\times 10^{-34} \\times 3.0 \\times 10^{8}}{2.4 \\times 10^{8}}\\)\n= 8.25 x 10\\(^{-19}\\)J\n(ii)\nE\\(_4\\) ----------------- - 6.0 x 10\\(^{-19}\\)J\nE\\(_3\\) ---------------- - 8.2 x 10\\(^{-19}\\)J\nE\\(_2\\) ---------------- - 8.8 x 10\\(^{-19}\\)J\nE\\(_1\\) ---------------- - 16.7 x 10\\(^{-19}\\)J\nThe energy level transitions which cause the emission of the photon in (d)(i) is. E\\(_3\\)\n~(c)(i) (\\(\\alpha\\)) 218 + q = 222\nq = 4\n(\\(\\alpha\\)) P = 84 + 2 = 86,\nm = 214 — 214 = 0\nn = 83 — 84 = —1\nx is an electron or beta particle.\n\nii) Needs to safely dispose radioactive waste:\n(1) it emits harmful radiations.\n(2) it is toxic\n\n(d)(i) E = \\(\\frac{hc}{\\lambda}\\)\n= \\(\\frac{6.6 \\times 10^{-34} \\times 3.0 \\times 10^{8}}{2.4 \\times 10^{8}}\\)\n= 8.25 x 10\\(^{-19}\\)J\n(ii)\nE\\(_4\\) ----------------- - 6.0 x 10\\(^{-19}\\)J\nE\\(_3\\) ---------------- - 8.2 x 10\\(^{-19}\\)J\nE\\(_2\\) ---------------- - 8.8 x 10\\(^{-19}\\)J\nE\\(_1\\) ---------------- - 16.7 x 10\\(^{-19}\\)J\nThe energy level transitions which cause the emission of the photon in (d)(i) is. E\\(_3\\)\n~(d)(i) E = \\(\\frac{hc}{\\lambda}\\)\n= \\(\\frac{6.6 \\times 10^{-34} \\times 3.0 \\times 10^{8}}{2.4 \\times 10^{8}}\\)\n= 8.25 x 10\\(^{-19}\\)J\n(ii)\nE\\(_4\\) ----------------- - 6.0 x 10\\(^{-19}\\)J\nE\\(_3\\) ---------------- - 8.2 x 10\\(^{-19}\\)J\nE\\(_2\\) ---------------- - 8.8 x 10\\(^{-19}\\)J\nE\\(_1\\) ---------------- - 16.7 x 10\\(^{-19}\\)J\nThe energy level transitions which cause the emission of the photon in (d)(i) is. E\\(_3\\)\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(61, '2016', '\n\nQuestion\n\nA projectile is released with a speed u at an angle \\(\\theta\\)to the horizontal. With the aid of a diagram, show that the time of flight is equal to \\(\\frac{2uSin\\theta}{g}\\), where g is the acceleration of free fall.\n\n', 'A projectile is released with a speed u at an angle \\(\\theta\\)to the horizontal. With the aid of a diagram, show that the time of flight is equal to \\(\\frac{2uSin\\theta}{g}\\), where g is the acceleration of free fall.', '', '', NULL, '\n\nExplanation\nVertical component velocity Vy = Usin\\(\\theta\\)\nTime to reach maximium height = t\nV = USin\\(\\theta\\) — gt\nBut V at maximum height = O\n.-. O = Usin\\(\\theta\\) - gt\nHence, t = \\(\\frac{U sin \\theta}{g}\\)\nBut time for a projectile to reach maximum he is equal to time to return to projection plane\nTime of flight T = \\(\\frac{2Usin \\theta}{g}\\)\n\n\n', 'Vertical component velocity Vy = Usin\\(\\theta\\)\nTime to reach maximium height = t\nV = USin\\(\\theta\\) — gt\nBut V at maximum height = O\n.-. O = Usin\\(\\theta\\) - gt\nHence, t = \\(\\frac{U sin \\theta}{g}\\)\nBut time for a projectile to reach maximum he is equal to time to return to projection plane\nTime of flight T = \\(\\frac{2Usin \\theta}{g}\\)\n\n\n', '', '', 'WAEC-PK-1', 'WAEC-PHY', 1),
(62, '2016', '\n\nQuestion\n\nThe horizontal component of the initial speed of a particle projected at 30° to the horizontal is 50 ms\\(^{-1}\\). If the acceleration cf free fall due to gravity is 10ms\\(^{-2}\\), determine its: (a) initial speed; (b) speed at maximum height reached.\n\n', 'The horizontal component of the initial speed of a particle projected at 30° to the horizontal is 50 ms\\(^{-1}\\). If the acceleration cf free fall due to gravity is 10ms\\(^{-2}\\), determine its: (a) initial speed; (b) speed at maximum height reached.', '', '', NULL, '\n\nExplanation\n(a) If initial velocity = U Then horizontal component\n\\(U_x = Ucos\\theta\\)\nUcos\\(\\theta\\) = 50\nU = \\(\\frac{50}{cos \\theta} = \\frac{50}{cos 30^o}\\) = 57.7m/s\n\n(b) But at maximum height V = O\nV\\(_y\\) = O\nV\\(_x\\) = U\\(_x\\) = 50m/s\n\n\n', '', 'Explanation\n(a) If initial velocity = U Then horizontal component\n\\(U_x = Ucos\\theta\\)\nUcos\\(\\theta\\) = 50\nU = \\(\\frac{50}{cos \\theta} = \\frac{50}{cos 30^o}\\) = 57.7m/s\n\n(b) But at maximum height V = O\nV\\(_y\\) = O\nV\\(_x\\) = U\\(_x\\) = 50m/s\n\n\n', '(a) If initial velocity = U Then horizontal component\n\\(U_x = Ucos\\theta\\)\nUcos\\(\\theta\\) = 50\nU = \\(\\frac{50}{cos \\theta} = \\frac{50}{cos 30^o}\\) = 57.7m/s\n~(b) But at maximum height V = O\nV\\(_y\\) = O\nV\\(_x\\) = U\\(_x\\) = 50m/s\n\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(63, '2016', '\n\nQuestion\n\nIn an electrolysis experiment, the ammeter records a steady current of 1 A. The mass of copper deposited in 30 minutes is 0.66 g. Calculate the error in the ammeter reading. [Electrochemical equivalent of copper = 0.00033 g C\\(^{-1}\\)]\n\n', 'In an electrolysis experiment, the ammeter records a steady current of 1 A. The mass of copper deposited in 30 minutes is 0.66 g. Calculate the error in the ammeter reading. [Electrochemical equivalent of copper = 0.00033 g C\\(^{-1}\\)]', '', '', NULL, '\n\nExplanation\nM = zit\nI = \\(\\frac{M}{Zt}\\) = \\(\\frac{0.66}{3 \\times 60 \\times 33 \\times 10^{-5}}\\)\n= 1.11A\nError = 1.11 - 1\n= 0.11A\n\n', 'M = zit\nI = \\(\\frac{M}{Zt}\\) = \\(\\frac{0.66}{3 \\times 60 \\times 33 \\times 10^{-5}}\\)\n= 1.11A\nError = 1.11 - 1\n= 0.11A\n\n', '', '', 'WAEC-PK-1', 'WAEC-PHY', 1),
(64, '2016', '\n\nQuestion\n\nA mass of 11.0 kg is suspended from a rigid support by an aluminum wire of length 2.0 m, diameter 2.0 mm and Young\'s modulus 7.0 x 10\\(^{11}\\) Nm\\(^{-2}\\). Determine the extension produced. [g = 10 ms\\(^{-2}\\); \\(\\pi\\) = 3.142]\n\n', 'A mass of 11.0 kg is suspended from a rigid support by an aluminum wire of length 2.0 m, diameter 2.0 mm and Young\'s modulus 7.0 x 10\\(^{11}\\) Nm\\(^{-2}\\). Determine the extension produced. [g = 10 ms\\(^{-2}\\); \\(\\pi\\) = 3.142]', '', '', NULL, '\n\nExplanation\nE = \\(\\frac{Fl}{Ae}\\)\nE = 7.0 x 10\\(^{-11}\\) Nm\\(^{-2}\\);\nF = mass x acceleration due to gravity\n= 11 x 10 = 110N\nl = 2.0m\nA = \\(\\pi\\) r\\(^2\\)\nd = 2.0 mm \\(\\implies\\) r = 1.0 mm\nA = 3.142 x \\((\\frac{10}{1000})^2\\)\ne = \\(\\frac{Fl}{EA} = \\frac{110 \\times 20}{7.0 \\times 10^{11} \\times 3.142(\\frac{10}{1000})^2}\\)\n= 1.0 x 10\\(^{-5}\\)m\n\n', 'E = \\(\\frac{Fl}{Ae}\\)\nE = 7.0 x 10\\(^{-11}\\) Nm\\(^{-2}\\);\nF = mass x acceleration due to gravity\n= 11 x 10 = 110N\nl = 2.0m\nA = \\(\\pi\\) r\\(^2\\)\nd = 2.0 mm \\(\\implies\\) r = 1.0 mm\nA = 3.142 x \\((\\frac{10}{1000})^2\\)\ne = \\(\\frac{Fl}{EA} = \\frac{110 \\times 20}{7.0 \\times 10^{11} \\times 3.142(\\frac{10}{1000})^2}\\)\n= 1.0 x 10\\(^{-5}\\)m\n\n', '', '', 'WAEC-PK-1', 'WAEC-PHY', 1),
(65, '2016', '\n\nQuestion\n\n(a) What is a polarizer?\n(b) With the aid of a diagram, explain how a polarizer can be used to polarize a beam of unpolarized light.\n\n', '', '(a) What is a polarizer?\n(b) With the aid of a diagram, explain how a polarizer can be used to polarize a beam of unpolarized light.', '(a) What is a polarizer?~(b) With the aid of a diagram, explain how a polarizer can be used to polarize a beam of unpolarized light.', NULL, '\n\nExplanation\nPolarizer is an optical device which allows vibrations of light to pass throagh in only one phone perpendicular to the direalioniof propagation\n(b)\n\nUnpolarized Polarizer Polarized l light ight Unpolarized light vibrates in all directions and a polarized is used to block all vibrations, and transmits only light vibrations polarized in one plane perpendicular to the direction of propagation.\n\n', '', 'Explanation\nPolarizer is an optical device which allows vibrations of light to pass throagh in only one phone perpendicular to the direalioniof propagation\n(b)\n\nUnpolarized Polarizer Polarized l light ight Unpolarized light vibrates in all directions and a polarized is used to block all vibrations, and transmits only light vibrations polarized in one plane perpendicular to the direction of propagation.\n\n', '~(b)\n\nUnpolarized Polarizer Polarized l light ight Unpolarized light vibrates in all directions and a polarized is used to block all vibrations, and transmits only light vibrations polarized in one plane perpendicular to the direction of propagation.\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(66, '2016', '\n\nQuestion\n\nName one use of \'LASER\' in each of the following areas:\n(a) communication;\n(b) medicine;\n(c) security\n\n', '', 'Name one use of \'LASER\' in each of the following areas:\n(a) communication;\n(b) medicine;\n(c) security', '(a) communication;~(b) medicine;\n(c) security~(c) security', NULL, '\n\nExplanation\nUses of LASER:\n(a) Communication for holograms production, CD, VCD, DVD playing, data transfer, fiber-optic cable and space communication.\n(b) Medicine - for welding retina of the eye, cervical cancer surgery, boring holes in the skull, cauterizing blood vessels, testing biological samples, monitoring glucose level in diabetic patient, dentistry, etc.\n(c) Security - for guidance system of missiles, aircrafts and satellites, generation of isotopes for nuclear weapons or reactors or enriched uranium production.\n\n', '', 'Explanation\nUses of LASER:\n(a) Communication for holograms production, CD, VCD, DVD playing, data transfer, fiber-optic cable and space communication.\n(b) Medicine - for welding retina of the eye, cervical cancer surgery, boring holes in the skull, cauterizing blood vessels, testing biological samples, monitoring glucose level in diabetic patient, dentistry, etc.\n(c) Security - for guidance system of missiles, aircrafts and satellites, generation of isotopes for nuclear weapons or reactors or enriched uranium production.\n\n', '(a) Communication for holograms production, CD, VCD, DVD playing, data transfer, fiber-optic cable and space communication.~(b) Medicine - for welding retina of the eye, cervical cancer surgery, boring holes in the skull, cauterizing blood vessels, testing biological samples, monitoring glucose level in diabetic patient, dentistry, etc.\n(c) Security - for guidance system of missiles, aircrafts and satellites, generation of isotopes for nuclear weapons or reactors or enriched uranium production.\n~(c) Security - for guidance system of missiles, aircrafts and satellites, generation of isotopes for nuclear weapons or reactors or enriched uranium production.\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(67, '2016', '\n\nQuestion\n\nThe accelerating potential in a cathode ray oscilloscope is 2.5 kV. Calculate the maximum speed of the accelerated electrons. [ e = 1.6 x 10\\(^{-19}\\) C; Me = 9.1 x 10\\(^{-31}\\) kg]\n\n', 'The accelerating potential in a cathode ray oscilloscope is 2.5 kV. Calculate the maximum speed of the accelerated electrons. [ e = 1.6 x 10\\(^{-19}\\) C; Me = 9.1 x 10\\(^{-31}\\) kg]', '', '', NULL, '\n\nExplanation\n\\(\\frac{1}{2}\\) MeV\\(^2\\) = eV\n\\(V^2 = \\frac{2eV}{Me} = \\frac{2 \\times 1.6 \\times 10^{-19} \\times 2500}{9.1 \\times 10^{-31}}\\)\n= 8.79 x 10\\(^{-14}\\)\nV = \\(\\sqrt{8.79 \\times 10^{14}}\\)\n= 2.965 x 10\\(^7\\) m/s\n\n\n', '\\(\\frac{1}{2}\\) MeV\\(^2\\) = eV\n\\(V^2 = \\frac{2eV}{Me} = \\frac{2 \\times 1.6 \\times 10^{-19} \\times 2500}{9.1 \\times 10^{-31}}\\)\n= 8.79 x 10\\(^{-14}\\)\nV = \\(\\sqrt{8.79 \\times 10^{14}}\\)\n= 2.965 x 10\\(^7\\) m/s\n\n\n', '', '', 'WAEC-PK-1', 'WAEC-PHY', 1),
(68, '2016', '\n\nQuestion\n\nWrite down the name of:\n(a) two particles used in explaining the wave nature of matter;\n(b) one device whose invention is based on the wave nature of matter.\n\n', '', 'Write down the name of:\n(a) two particles used in explaining the wave nature of matter;\n(b) one device whose invention is based on the wave nature of matter.', '(a) two particles used in explaining the wave nature of matter;~(b) one device whose invention is based on the wave nature of matter.', NULL, '\n\nExplanation\n(a) Particles used in explaining the wave nature of matter are: hydrogen atom or proton, helium atom, electrons, neutron and helium nuclei or apha particles.\n(b) Inventions based on thewave nature of water devices are:\n- Electron microscope and scanning tunneling microscope (STM).\n\n', '', 'Explanation\n(a) Particles used in explaining the wave nature of matter are: hydrogen atom or proton, helium atom, electrons, neutron and helium nuclei or apha particles.\n(b) Inventions based on thewave nature of water devices are:\n- Electron microscope and scanning tunneling microscope (STM).\n\n', '(a) Particles used in explaining the wave nature of matter are: hydrogen atom or proton, helium atom, electrons, neutron and helium nuclei or apha particles.~(b) Inventions based on thewave nature of water devices are:\n- Electron microscope and scanning tunneling microscope (STM).\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(69, '2016', '\n\nQuestion\n\n(a) what is Brownian motion?\n(b) State the two inferences that can be drawn from Brownian motion experiment.\n\n', '', '(a) what is Brownian motion?\n(b) State the two inferences that can be drawn from Brownian motion experiment.', '(a) what is Brownian motion?~(b) State the two inferences that can be drawn from Brownian motion experiment.', NULL, '\n\nExplanation\n(a) Brownian motiorbigtcontinuous random movement of particles sus erided in a fluid.\n(b) Inferences that can be drawn from Brownian experiment include;\n- existence of molecules.\n- kinetic motion of molecules.\n\n', '', 'Explanation\n(a) Brownian motiorbigtcontinuous random movement of particles sus erided in a fluid.\n(b) Inferences that can be drawn from Brownian experiment include;\n- existence of molecules.\n- kinetic motion of molecules.\n\n', '(a) Brownian motiorbigtcontinuous random movement of particles sus erided in a fluid.~(b) Inferences that can be drawn from Brownian experiment include;\n- existence of molecules.\n- kinetic motion of molecules.\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(70, '2016', '\n\nQuestion\n\n(a) State the triangle law of vector addition.\n(b) Name the four physical quantities that are associated with the equationq of linear motion.\n(c) Using the same set of axes, sketch and label two graphs to illustrate the variation of potential energy and kinetic energy with time for a body in simple harmonic motion.\n(d)\n\nA light spiral spring of force constant K lies on a horizontal frictionless surface and has one end fixed to a vertical wall. A block P of mass 2.0 kg placed against the free end of the spring is pushed a distance 5 cm towards the wall with 10J of energy as illustrated in the diagram above. The block is released and after 0.25s, it collides inelastically with a stationary block Q of mass 4.0 kg. Calculate the:\n(i) value of k;\n(ii) force used to compress the spring;\n(iii) acceleration of the block p after release;\n(iv) common speed after collision of the blocks.\n\n\n', '', '(a) State the triangle law of vector addition.\n(b) Name the four physical quantities that are associated with the equationq of linear motion.\n(c) Using the same set of axes, sketch and label two graphs to illustrate the variation of potential energy and kinetic energy with time for a body in simple harmonic motion.\n(d)\n\nA light spiral spring of force constant K lies on a horizontal frictionless surface and has one end fixed to a vertical wall. A block P of mass 2.0 kg placed against the free end of the spring is pushed a distance 5 cm towards the wall with 10J of energy as illustrated in the diagram above. The block is released and after 0.25s, it collides inelastically with a stationary block Q of mass 4.0 kg. Calculate the:\n(i) value of k;\n(ii) force used to compress the spring;\n(iii) acceleration of the block p after release;\n(iv) common speed after collision of the blocks.\n', '(a) State the triangle law of vector addition.~(b) Name the four physical quantities that are associated with the equationq of linear motion.\n(c) Using the same set of axes, sketch and label two graphs to illustrate the variation of potential energy and kinetic energy with time for a body in simple harmonic motion.\n(d)\n\nA light spiral spring of force constant K lies on a horizontal frictionless surface and has one end fixed to a vertical wall. A block P of mass 2.0 kg placed against the free end of the spring is pushed a distance 5 cm towards the wall with 10J of energy as illustrated in the diagram above. The block is released and after 0.25s, it collides inelastically with a stationary block Q of mass 4.0 kg. Calculate the:\n(i) value of k;\n(ii) force used to compress the spring;\n(iii) acceleration of the block p after release;\n(iv) common speed after collision of the blocks.\n~(c) Using the same set of axes, sketch and label two graphs to illustrate the variation of potential energy and kinetic energy with time for a body in simple harmonic motion.\n(d)\n\nA light spiral spring of force constant K lies on a horizontal frictionless surface and has one end fixed to a vertical wall. A block P of mass 2.0 kg placed against the free end of the spring is pushed a distance 5 cm towards the wall with 10J of energy as illustrated in the diagram above. The block is released and after 0.25s, it collides inelastically with a stationary block Q of mass 4.0 kg. Calculate the:\n(i) value of k;\n(ii) force used to compress the spring;\n(iii) acceleration of the block p after release;\n(iv) common speed after collision of the blocks.\n~(d)\n\nA light spiral spring of force constant K lies on a horizontal frictionless surface and has one end fixed to a vertical wall. A block P of mass 2.0 kg placed against the free end of the spring is pushed a distance 5 cm towards the wall with 10J of energy as illustrated in the diagram above. The block is released and after 0.25s, it collides inelastically with a stationary block Q of mass 4.0 kg. Calculate the:\n(i) value of k;\n(ii) force used to compress the spring;\n(iii) acceleration of the block p after release;\n(iv) common speed after collision of the blocks.\n', NULL, '\n\nExplanation\n(a) Triangle law of vector addition states that if three forces are in equilibrium, they can be represented in magnitude and direction by the three sides of a triangle taken in order.\n(b) Physical quantities associated with equations of motion are:\n- time\n- speed/ velocity\n- distance or displacement and acceleration.\n(c)\n\n\n(d)\nW = \\(\\frac{1}{2}Ke^2\\)\nK = \\(\\frac{2W}{e^2} = \\frac{2 \\times 10}{\\frac{5}{100}} = 8000 NM^{-1}\\)\n(ii) F = Ke = 800 x \\(\\frac{5}{100}\\)\n= 400N\n(iii) F = Ma\na = \\(\\frac{F}{M} = \\frac{400}{2}\\) = 200 ms\\(^{-2}\\)\n(iv) Initial velocity U\\(_p\\) of the block\nP = 200 x 0.25\n= 50 ms\\(^{-1}\\)\nbut \\(M_pU_p + M_qU_q = (M_p + M_Q)\\)\n2 x 50 + 4 x 0 = V(2 + 4)\nV = \\(\\frac{100}{6} = 16.7 ms^{-1}\\)\n\n\n\n', '', 'Explanation\n(a) Triangle law of vector addition states that if three forces are in equilibrium, they can be represented in magnitude and direction by the three sides of a triangle taken in order.\n(b) Physical quantities associated with equations of motion are:\n- time\n- speed/ velocity\n- distance or displacement and acceleration.\n(c)\n\n\n(d)\nW = \\(\\frac{1}{2}Ke^2\\)\nK = \\(\\frac{2W}{e^2} = \\frac{2 \\times 10}{\\frac{5}{100}} = 8000 NM^{-1}\\)\n(ii) F = Ke = 800 x \\(\\frac{5}{100}\\)\n= 400N\n(iii) F = Ma\na = \\(\\frac{F}{M} = \\frac{400}{2}\\) = 200 ms\\(^{-2}\\)\n(iv) Initial velocity U\\(_p\\) of the block\nP = 200 x 0.25\n= 50 ms\\(^{-1}\\)\nbut \\(M_pU_p + M_qU_q = (M_p + M_Q)\\)\n2 x 50 + 4 x 0 = V(2 + 4)\nV = \\(\\frac{100}{6} = 16.7 ms^{-1}\\)\n\n\n\n', '(a) Triangle law of vector addition states that if three forces are in equilibrium, they can be represented in magnitude and direction by the three sides of a triangle taken in order.~(b) Physical quantities associated with equations of motion are:\n- time\n- speed/ velocity\n- distance or displacement and acceleration.\n(c)\n\n\n(d)\nW = \\(\\frac{1}{2}Ke^2\\)\nK = \\(\\frac{2W}{e^2} = \\frac{2 \\times 10}{\\frac{5}{100}} = 8000 NM^{-1}\\)\n(ii) F = Ke = 800 x \\(\\frac{5}{100}\\)\n= 400N\n(iii) F = Ma\na = \\(\\frac{F}{M} = \\frac{400}{2}\\) = 200 ms\\(^{-2}\\)\n(iv) Initial velocity U\\(_p\\) of the block\nP = 200 x 0.25\n= 50 ms\\(^{-1}\\)\nbut \\(M_pU_p + M_qU_q = (M_p + M_Q)\\)\n2 x 50 + 4 x 0 = V(2 + 4)\nV = \\(\\frac{100}{6} = 16.7 ms^{-1}\\)\n\n\n~(c)\n\n\n(d)\nW = \\(\\frac{1}{2}Ke^2\\)\nK = \\(\\frac{2W}{e^2} = \\frac{2 \\times 10}{\\frac{5}{100}} = 8000 NM^{-1}\\)\n(ii) F = Ke = 800 x \\(\\frac{5}{100}\\)\n= 400N\n(iii) F = Ma\na = \\(\\frac{F}{M} = \\frac{400}{2}\\) = 200 ms\\(^{-2}\\)\n(iv) Initial velocity U\\(_p\\) of the block\nP = 200 x 0.25\n= 50 ms\\(^{-1}\\)\nbut \\(M_pU_p + M_qU_q = (M_p + M_Q)\\)\n2 x 50 + 4 x 0 = V(2 + 4)\nV = \\(\\frac{100}{6} = 16.7 ms^{-1}\\)\n\n\n~(d)\nW = \\(\\frac{1}{2}Ke^2\\)\nK = \\(\\frac{2W}{e^2} = \\frac{2 \\times 10}{\\frac{5}{100}} = 8000 NM^{-1}\\)\n(ii) F = Ke = 800 x \\(\\frac{5}{100}\\)\n= 400N\n(iii) F = Ma\na = \\(\\frac{F}{M} = \\frac{400}{2}\\) = 200 ms\\(^{-2}\\)\n(iv) Initial velocity U\\(_p\\) of the block\nP = 200 x 0.25\n= 50 ms\\(^{-1}\\)\nbut \\(M_pU_p + M_qU_q = (M_p + M_Q)\\)\n2 x 50 + 4 x 0 = V(2 + 4)\nV = \\(\\frac{100}{6} = 16.7 ms^{-1}\\)\n\n\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(71, '2016', '\n\nQuestion\n\n(a) On which day would sound wave travel faster: on a hot or cold day? Explain.\n(b) Why are megaphones shaped like funnels?\n(c) A ray of light is incident on a surface of a ectangular glass prism of refractive index 1.5 illustrated in the diagram below.\n\n(i) Copy the diagram a label the angles of: (\\(\\alpha\\)) Incidence (x); (\\(\\beta\\)) Reflection (y); (\\(\\gamma\\)) refraction (z); with t glass letters indicated.\n(ii) Calculate the angle refraction to the nearest whole number.\n(d) A sonomesr wire vibrates in simple harmoi motion with a maximum amplitude of 1.0 cm. Calculate the frequency of vibration of the wire, giv that the magnitade of the maximum acceleration of the wire is 980ms\\(^{-2}\\). [\\(\\pi \\frac{22}{7}\\)]\n\n', '', '(a) On which day would sound wave travel faster: on a hot or cold day? Explain.\n(b) Why are megaphones shaped like funnels?\n(c) A ray of light is incident on a surface of a ectangular glass prism of refractive index 1.5 illustrated in the diagram below.\n\n(i) Copy the diagram a label the angles of: (\\(\\alpha\\)) Incidence (x); (\\(\\beta\\)) Reflection (y); (\\(\\gamma\\)) refraction (z); with t glass letters indicated.\n(ii) Calculate the angle refraction to the nearest whole number.\n(d) A sonomesr wire vibrates in simple harmoi motion with a maximum amplitude of 1.0 cm. Calculate the frequency of vibration of the wire, giv that the magnitade of the maximum acceleration of the wire is 980ms\\(^{-2}\\). [\\(\\pi \\frac{22}{7}\\)]', '(a) On which day would sound wave travel faster: on a hot or cold day? Explain.~(b) Why are megaphones shaped like funnels?\n(c) A ray of light is incident on a surface of a ectangular glass prism of refractive index 1.5 illustrated in the diagram below.\n\n(i) Copy the diagram a label the angles of: (\\(\\alpha\\)) Incidence (x); (\\(\\beta\\)) Reflection (y); (\\(\\gamma\\)) refraction (z); with t glass letters indicated.\n(ii) Calculate the angle refraction to the nearest whole number.\n(d) A sonomesr wire vibrates in simple harmoi motion with a maximum amplitude of 1.0 cm. Calculate the frequency of vibration of the wire, giv that the magnitade of the maximum acceleration of the wire is 980ms\\(^{-2}\\). [\\(\\pi \\frac{22}{7}\\)]~(c) A ray of light is incident on a surface of a ectangular glass prism of refractive index 1.5 illustrated in the diagram below.\n\n(i) Copy the diagram a label the angles of: (\\(\\alpha\\)) Incidence (x); (\\(\\beta\\)) Reflection (y); (\\(\\gamma\\)) refraction (z); with t glass letters indicated.\n(ii) Calculate the angle refraction to the nearest whole number.\n(d) A sonomesr wire vibrates in simple harmoi motion with a maximum amplitude of 1.0 cm. Calculate the frequency of vibration of the wire, giv that the magnitade of the maximum acceleration of the wire is 980ms\\(^{-2}\\). [\\(\\pi \\frac{22}{7}\\)]~(d) A sonomesr wire vibrates in simple harmoi motion with a maximum amplitude of 1.0 cm. Calculate the frequency of vibration of the wire, giv that the magnitade of the maximum acceleration of the wire is 980ms\\(^{-2}\\). [\\(\\pi \\frac{22}{7}\\)]', NULL, '\n\nExplanation\n(a) Sound wave travels faster on a hot day because the speed of sound is proportional to the square root of absolute temperature (V \\(\\alpha \\sqrt{T}\\)), hence the higher the greater the speed of sound wave.\n(b) Megap:iones are shaped like a funnels because at the smaller end of the cone, the sound energy comes into contact will a small mass of air producing greater intensity. As it comes out of the larger open end, the energy sets a larger air mass into vibration, hence the final sound is louder.\n(c) (i)\n\n(ii) n = \\(\\frac{sini}{sinr}\\)\nsinr = \\(\\frac{sini}{n} = \\frac{sin60}{1.5}\\)\n= \\(\\frac{0.8660}{1.5} = 0.5774\\)\nr = sin\\(^{-1}\\)0.5774 = 35\\(^{o}\\)\n(d) a = \\(\\Omega^2r\\) but \\(\\Omega = 2 \\pi f\\)\na = (2\\(\\pi f)^2\\)r\nf = \\(\\sqrt{\\frac{9}{4 \\pi^2 r}} = \\sqrt{\\frac{988}{4(\\frac{22}{7})^2} \\times \\frac{1}{100}}\\)\n= 50Hz\n\n', '', 'Explanation\n(a) Sound wave travels faster on a hot day because the speed of sound is proportional to the square root of absolute temperature (V \\(\\alpha \\sqrt{T}\\)), hence the higher the greater the speed of sound wave.\n(b) Megap:iones are shaped like a funnels because at the smaller end of the cone, the sound energy comes into contact will a small mass of air producing greater intensity. As it comes out of the larger open end, the energy sets a larger air mass into vibration, hence the final sound is louder.\n(c) (i)\n\n(ii) n = \\(\\frac{sini}{sinr}\\)\nsinr = \\(\\frac{sini}{n} = \\frac{sin60}{1.5}\\)\n= \\(\\frac{0.8660}{1.5} = 0.5774\\)\nr = sin\\(^{-1}\\)0.5774 = 35\\(^{o}\\)\n(d) a = \\(\\Omega^2r\\) but \\(\\Omega = 2 \\pi f\\)\na = (2\\(\\pi f)^2\\)r\nf = \\(\\sqrt{\\frac{9}{4 \\pi^2 r}} = \\sqrt{\\frac{988}{4(\\frac{22}{7})^2} \\times \\frac{1}{100}}\\)\n= 50Hz\n\n', '(a) Sound wave travels faster on a hot day because the speed of sound is proportional to the square root of absolute temperature (V \\(\\alpha \\sqrt{T}\\)), hence the higher the greater the speed of sound wave.~(b) Megap:iones are shaped like a funnels because at the smaller end of the cone, the sound energy comes into contact will a small mass of air producing greater intensity. As it comes out of the larger open end, the energy sets a larger air mass into vibration, hence the final sound is louder.\n(c) (i)\n\n(ii) n = \\(\\frac{sini}{sinr}\\)\nsinr = \\(\\frac{sini}{n} = \\frac{sin60}{1.5}\\)\n= \\(\\frac{0.8660}{1.5} = 0.5774\\)\nr = sin\\(^{-1}\\)0.5774 = 35\\(^{o}\\)\n(d) a = \\(\\Omega^2r\\) but \\(\\Omega = 2 \\pi f\\)\na = (2\\(\\pi f)^2\\)r\nf = \\(\\sqrt{\\frac{9}{4 \\pi^2 r}} = \\sqrt{\\frac{988}{4(\\frac{22}{7})^2} \\times \\frac{1}{100}}\\)\n= 50Hz\n~(c) (i)\n\n(ii) n = \\(\\frac{sini}{sinr}\\)\nsinr = \\(\\frac{sini}{n} = \\frac{sin60}{1.5}\\)\n= \\(\\frac{0.8660}{1.5} = 0.5774\\)\nr = sin\\(^{-1}\\)0.5774 = 35\\(^{o}\\)\n(d) a = \\(\\Omega^2r\\) but \\(\\Omega = 2 \\pi f\\)\na = (2\\(\\pi f)^2\\)r\nf = \\(\\sqrt{\\frac{9}{4 \\pi^2 r}} = \\sqrt{\\frac{988}{4(\\frac{22}{7})^2} \\times \\frac{1}{100}}\\)\n= 50Hz\n~(d) a = \\(\\Omega^2r\\) but \\(\\Omega = 2 \\pi f\\)\na = (2\\(\\pi f)^2\\)r\nf = \\(\\sqrt{\\frac{9}{4 \\pi^2 r}} = \\sqrt{\\frac{988}{4(\\frac{22}{7})^2} \\times \\frac{1}{100}}\\)\n= 50Hz\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
(72, '2016', '\n\nQuestion\n\n(a) Explain briefly the purpose of earthing electrical appliance.\n(b) Why does the light frorr bulb connected to a simple cell dim and eventually goes off after a while?\n(c) A coil of incidence 0.007 H, a resistor of resistance 8 \\(\\Omega\\) and a capacitor capacitance 0.001 F are connected in series an a.c. source of frequency \\(\\frac{500}{\\pi}\\)Hz. If the r.m.s voltages across the coil, the resistor and capacitor are 30v, 20v and 70v respectively;\n(i) draw a vector diagram to illustrate the voltage across the components in the circuit.\n(ii) Calculate the: (\\(\\alpha\\)) r.m.s voltage of the source\n(\\(\\beta\\)) r.m.s current in the circuit;\n(\\(\\gamma\\)) power dissipated in the circuit.\niii) write down the sinusoidal equation for the r.m.s voltage, V, in terms of the time, t.\n\n', '', '(a) Explain briefly the purpose of earthing electrical appliance.\n(b) Why does the light frorr bulb connected to a simple cell dim and eventually goes off after a while?\n(c) A coil of incidence 0.007 H, a resistor of resistance 8 \\(\\Omega\\) and a capacitor capacitance 0.001 F are connected in series an a.c. source of frequency \\(\\frac{500}{\\pi}\\)Hz. If the r.m.s voltages across the coil, the resistor and capacitor are 30v, 20v and 70v respectively;\n(i) draw a vector diagram to illustrate the voltage across the components in the circuit.\n(ii) Calculate the: (\\(\\alpha\\)) r.m.s voltage of the source\n(\\(\\beta\\)) r.m.s current in the circuit;\n(\\(\\gamma\\)) power dissipated in the circuit.\niii) write down the sinusoidal equation for the r.m.s voltage, V, in terms of the time, t.', '(a) Explain briefly the purpose of earthing electrical appliance.~(b) Why does the light frorr bulb connected to a simple cell dim and eventually goes off after a while?\n(c) A coil of incidence 0.007 H, a resistor of resistance 8 \\(\\Omega\\) and a capacitor capacitance 0.001 F are connected in series an a.c. source of frequency \\(\\frac{500}{\\pi}\\)Hz. If the r.m.s voltages across the coil, the resistor and capacitor are 30v, 20v and 70v respectively;\n(i) draw a vector diagram to illustrate the voltage across the components in the circuit.\n(ii) Calculate the: (\\(\\alpha\\)) r.m.s voltage of the source\n(\\(\\beta\\)) r.m.s current in the circuit;\n(\\(\\gamma\\)) power dissipated in the circuit.\niii) write down the sinusoidal equation for the r.m.s voltage, V, in terms of the time, t.~(c) A coil of incidence 0.007 H, a resistor of resistance 8 \\(\\Omega\\) and a capacitor capacitance 0.001 F are connected in series an a.c. source of frequency \\(\\frac{500}{\\pi}\\)Hz. If the r.m.s voltages across the coil, the resistor and capacitor are 30v, 20v and 70v respectively;\n(i) draw a vector diagram to illustrate the voltage across the components in the circuit.\n(ii) Calculate the: (\\(\\alpha\\)) r.m.s voltage of the source\n(\\(\\beta\\)) r.m.s current in the circuit;\n(\\(\\gamma\\)) power dissipated in the circuit.\niii) write down the sinusoidal equation for the r.m.s voltage, V, in terms of the time, t.', NULL, '\n\nExplanation\n(a) The earth wire is used on electrical appliances as a safety device to prevent electric shock when short circuit occurs in such appliances. The dangerous current is then passed into thq ground through the earth wire because the eartIthas a lower resistance than the body.\n\n(b) Light from a bulb connected to a single cell dim and eventually goes off because of the defects of polarization and local action.\n(c)\n\n(i) \\(V^2\\)\\(_{\\text{rms}}\\) + \\(V_c - L_L)^2\\)\n\\(V^2\\)\\( _{\\text{rms}}\\) = \\(\\sqrt{V^2 + (V_c - V_L)}^2\\)\n= \\(\\sqrt{30^2 + (70 - 30)^2}\\)\n= \\(\\sqrt{900 + 1600}\\)\n= 50v\n(ii) (\\(\\beta\\)) = X\\(_L = 2\\pi fL\\)\n= 2 \\(\\pi \\times \\frac{500}{\\pi} \\times 0.007\\)\n= 7 \\(\\Omega\\)\nBut X\\(_c = \\frac{1}{2 \\pi fC} = \\frac{1}{2 \\pi \\times \\frac{500}{\\pi} \\times 0.001}\\)\n= L\\(\\Omega\\)\nz = \\(\\sqrt{R^2 + (X_L - X_c)^2}\\)\n= \\(\\sqrt{8^2 + (7 - 1)^2} = \\sqrt{64 + 36}\\)\n= \\(\\sqrt{100}\\)\n= 10\\(\\Omega\\)\nI = \\(\\frac{v}{z} = \\frac{50}{10}\\)\n= 5.0 A\n\n(iii) (\\(\\gamma\\)) P = I\\(^2R = 5^2 \\times 8\\) = 200W\n\\(V_o = V_\\text{rms} \\sqrt{2} = 50\\sqrt{2} = 70.7V\\)\nBut V\\(_osin2 \\pi ft\\)\n= 70.7\\(sin 2 \\times 2 \\times \\pi \\times \\frac{500}{\\pi} \\times t\\)\n= 70.7sin 1000t\n\n', '', 'Explanation\n(a) The earth wire is used on electrical appliances as a safety device to prevent electric shock when short circuit occurs in such appliances. The dangerous current is then passed into thq ground through the earth wire because the eartIthas a lower resistance than the body.\n\n(b) Light from a bulb connected to a single cell dim and eventually goes off because of the defects of polarization and local action.\n(c)\n\n(i) \\(V^2\\)\\(_{\\text{rms}}\\) + \\(V_c - L_L)^2\\)\n\\(V^2\\)\\( _{\\text{rms}}\\) = \\(\\sqrt{V^2 + (V_c - V_L)}^2\\)\n= \\(\\sqrt{30^2 + (70 - 30)^2}\\)\n= \\(\\sqrt{900 + 1600}\\)\n= 50v\n(ii) (\\(\\beta\\)) = X\\(_L = 2\\pi fL\\)\n= 2 \\(\\pi \\times \\frac{500}{\\pi} \\times 0.007\\)\n= 7 \\(\\Omega\\)\nBut X\\(_c = \\frac{1}{2 \\pi fC} = \\frac{1}{2 \\pi \\times \\frac{500}{\\pi} \\times 0.001}\\)\n= L\\(\\Omega\\)\nz = \\(\\sqrt{R^2 + (X_L - X_c)^2}\\)\n= \\(\\sqrt{8^2 + (7 - 1)^2} = \\sqrt{64 + 36}\\)\n= \\(\\sqrt{100}\\)\n= 10\\(\\Omega\\)\nI = \\(\\frac{v}{z} = \\frac{50}{10}\\)\n= 5.0 A\n\n(iii) (\\(\\gamma\\)) P = I\\(^2R = 5^2 \\times 8\\) = 200W\n\\(V_o = V_\\text{rms} \\sqrt{2} = 50\\sqrt{2} = 70.7V\\)\nBut V\\(_osin2 \\pi ft\\)\n= 70.7\\(sin 2 \\times 2 \\times \\pi \\times \\frac{500}{\\pi} \\times t\\)\n= 70.7sin 1000t\n\n', '(a) The earth wire is used on electrical appliances as a safety device to prevent electric shock when short circuit occurs in such appliances. The dangerous current is then passed into thq ground through the earth wire because the eartIthas a lower resistance than the body.\n~(b) Light from a bulb connected to a single cell dim and eventually goes off because of the defects of polarization and local action.\n(c)\n\n(i) \\(V^2\\)\\(_{\\text{rms}}\\) + \\(V_c - L_L)^2\\)\n\\(V^2\\)\\( _{\\text{rms}}\\) = \\(\\sqrt{V^2 + (V_c - V_L)}^2\\)\n= \\(\\sqrt{30^2 + (70 - 30)^2}\\)\n= \\(\\sqrt{900 + 1600}\\)\n= 50v\n(ii) (\\(\\beta\\)) = X\\(_L = 2\\pi fL\\)\n= 2 \\(\\pi \\times \\frac{500}{\\pi} \\times 0.007\\)\n= 7 \\(\\Omega\\)\nBut X\\(_c = \\frac{1}{2 \\pi fC} = \\frac{1}{2 \\pi \\times \\frac{500}{\\pi} \\times 0.001}\\)\n= L\\(\\Omega\\)\nz = \\(\\sqrt{R^2 + (X_L - X_c)^2}\\)\n= \\(\\sqrt{8^2 + (7 - 1)^2} = \\sqrt{64 + 36}\\)\n= \\(\\sqrt{100}\\)\n= 10\\(\\Omega\\)\nI = \\(\\frac{v}{z} = \\frac{50}{10}\\)\n= 5.0 A\n\n(iii) (\\(\\gamma\\)) P = I\\(^2R = 5^2 \\times 8\\) = 200W\n\\(V_o = V_\\text{rms} \\sqrt{2} = 50\\sqrt{2} = 70.7V\\)\nBut V\\(_osin2 \\pi ft\\)\n= 70.7\\(sin 2 \\times 2 \\times \\pi \\times \\frac{500}{\\pi} \\times t\\)\n= 70.7sin 1000t\n~(c)\n\n(i) \\(V^2\\)\\(_{\\text{rms}}\\) + \\(V_c - L_L)^2\\)\n\\(V^2\\)\\( _{\\text{rms}}\\) = \\(\\sqrt{V^2 + (V_c - V_L)}^2\\)\n= \\(\\sqrt{30^2 + (70 - 30)^2}\\)\n= \\(\\sqrt{900 + 1600}\\)\n= 50v\n(ii) (\\(\\beta\\)) = X\\(_L = 2\\pi fL\\)\n= 2 \\(\\pi \\times \\frac{500}{\\pi} \\times 0.007\\)\n= 7 \\(\\Omega\\)\nBut X\\(_c = \\frac{1}{2 \\pi fC} = \\frac{1}{2 \\pi \\times \\frac{500}{\\pi} \\times 0.001}\\)\n= L\\(\\Omega\\)\nz = \\(\\sqrt{R^2 + (X_L - X_c)^2}\\)\n= \\(\\sqrt{8^2 + (7 - 1)^2} = \\sqrt{64 + 36}\\)\n= \\(\\sqrt{100}\\)\n= 10\\(\\Omega\\)\nI = \\(\\frac{v}{z} = \\frac{50}{10}\\)\n= 5.0 A\n\n(iii) (\\(\\gamma\\)) P = I\\(^2R = 5^2 \\times 8\\) = 200W\n\\(V_o = V_\\text{rms} \\sqrt{2} = 50\\sqrt{2} = 70.7V\\)\nBut V\\(_osin2 \\pi ft\\)\n= 70.7\\(sin 2 \\times 2 \\times \\pi \\times \\frac{500}{\\pi} \\times t\\)\n= 70.7sin 1000t\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(73, '2016', '\n\nQuestion\n\n(a) Define ionization potential.\n(b)(i) State the three types of emission spectra.\n(ii) Name one source each which produces each of the spectra stated in (b)(i).\n(c) In an x-ray tube, electrons are accelerated the target by a potential difference of 80 A Calculate the:\n(i) speed of the electron;\nii) threshold wavelength of the electron. [h=6.6 x 10\\(^{-34}\\) Js; e = 1.6 x 10\\(^{-19}\\) C; Me = 9.1 x 10\\(^{-31}\\)\nd) An x-ray photon of frequency 4.5 x 10\\(^{-18}\\) strikes an. electron, assumed to be at rest. If t electron absorbs all the photon energy, calculate the speed acquired by the electron. [ h = 6.6 x 10\\(^{-34}\\) Js; Me = 9.1 x 10\\(^{-31}\\) kg ]\n\n\n\n\n(a) Define ionization potential.\n(b)(i) State the three types of emission spectra.\n(ii) Name one source each which produces each of the spectra stated in (b)(i).\n(c) In an x-ray tube, electrons are accelerated the target by a potential difference of 80 A Calculate the:\n(i) speed of the electron;\nii) threshold wavelength of the electron. [h=6.6 x 10\\(^{-34}\\) Js; e = 1.6 x 10\\(^{-19}\\) C; Me = 9.1 x 10\\(^{-31}\\)\nd) An x-ray photon of frequency 4.5 x 10\\(^{-18}\\) strikes an. electron, assumed to be at rest. If t electron absorbs all the photon energy, calculate the speed acquired by the electron. [ h = 6.6 x 10\\(^{-34}\\) Js; Me = 9.1 x 10\\(^{-31}\\) kg ]\n\n', '', '(a) Define ionization potential.\n(b)(i) State the three types of emission spectra.\n(ii) Name one source each which produces each of the spectra stated in (b)(i).\n(c) In an x-ray tube, electrons are accelerated the target by a potential difference of 80 A Calculate the:\n(i) speed of the electron;\nii) threshold wavelength of the electron. [h=6.6 x 10\\(^{-34}\\) Js; e = 1.6 x 10\\(^{-19}\\) C; Me = 9.1 x 10\\(^{-31}\\)\nd) An x-ray photon of frequency 4.5 x 10\\(^{-18}\\) strikes an. electron, assumed to be at rest. If t electron absorbs all the photon energy, calculate the speed acquired by the electron. [ h = 6.6 x 10\\(^{-34}\\) Js; Me = 9.1 x 10\\(^{-31}\\) kg ]\n\n\n\n\n(a) Define ionization potential.\n(b)(i) State the three types of emission spectra.\n(ii) Name one source each which produces each of the spectra stated in (b)(i).\n(c) In an x-ray tube, electrons are accelerated the target by a potential difference of 80 A Calculate the:\n(i) speed of the electron;\nii) threshold wavelength of the electron. [h=6.6 x 10\\(^{-34}\\) Js; e = 1.6 x 10\\(^{-19}\\) C; Me = 9.1 x 10\\(^{-31}\\)\nd) An x-ray photon of frequency 4.5 x 10\\(^{-18}\\) strikes an. electron, assumed to be at rest. If t electron absorbs all the photon energy, calculate the speed acquired by the electron. [ h = 6.6 x 10\\(^{-34}\\) Js; Me = 9.1 x 10\\(^{-31}\\) kg ]', '(a) Define ionization potential.\n(b)(i) State the three types of emission spectra.\n(ii) Name one source each which produces each of the spectra stated in (b)(i).\n(c) In an x-ray tube, electrons are accelerated the target by a potential difference of 80 A Calculate the:\n(i) speed of the electron;\nii) threshold wavelength of the electron. [h=6.6 x 10\\(^{-34}\\) Js; e = 1.6 x 10\\(^{-19}\\) C; Me = 9.1 x 10\\(^{-31}\\)\nd) An x-ray photon of frequency 4.5 x 10\\(^{-18}\\) strikes an. electron, assumed to be at rest. If t electron absorbs all the photon energy, calculate the speed acquired by the electron. [ h = 6.6 x 10\\(^{-34}\\) Js; Me = 9.1 x 10\\(^{-31}\\) kg ]\n\n\n\n\n(a) Define ionization potential.~(b)(i) State the three types of emission spectra.\n(ii) Name one source each which produces each of the spectra stated in (b)(i).\n(c) In an x-ray tube, electrons are accelerated the target by a potential difference of 80 A Calculate the:\n(i) speed of the electron;\nii) threshold wavelength of the electron. [h=6.6 x 10\\(^{-34}\\) Js; e = 1.6 x 10\\(^{-19}\\) C; Me = 9.1 x 10\\(^{-31}\\)\nd) An x-ray photon of frequency 4.5 x 10\\(^{-18}\\) strikes an. electron, assumed to be at rest. If t electron absorbs all the photon energy, calculate the speed acquired by the electron. [ h = 6.6 x 10\\(^{-34}\\) Js; Me = 9.1 x 10\\(^{-31}\\) kg ]\n\n\n\n\n(a) Define ionization potential.\n(b)(i) State the three types of emission spectra.\n(ii) Name one source each which produces each of the spectra stated in (b)(i).\n(c) In an x-ray tube, electrons are accelerated the target by a potential difference of 80 A Calculate the:\n(i) speed of the electron;\nii) threshold wavelength of the electron. [h=6.6 x 10\\(^{-34}\\) Js; e = 1.6 x 10\\(^{-19}\\) C; Me = 9.1 x 10\\(^{-31}\\)\nd) An x-ray photon of frequency 4.5 x 10\\(^{-18}\\) strikes an. electron, assumed to be at rest. If t electron absorbs all the photon energy, calculate the speed acquired by the electron. [ h = 6.6 x 10\\(^{-34}\\) Js; Me = 9.1 x 10\\(^{-31}\\) kg ]~(c) In an x-ray tube, electrons are accelerated the target by a potential difference of 80 A Calculate the:\n(i) speed of the electron;\nii) threshold wavelength of the electron. [h=6.6 x 10\\(^{-34}\\) Js; e = 1.6 x 10\\(^{-19}\\) C; Me = 9.1 x 10\\(^{-31}\\)\nd) An x-ray photon of frequency 4.5 x 10\\(^{-18}\\) strikes an. electron, assumed to be at rest. If t electron absorbs all the photon energy, calculate the speed acquired by the electron. [ h = 6.6 x 10\\(^{-34}\\) Js; Me = 9.1 x 10\\(^{-31}\\) kg ]\n\n\n\n\n(a) Define ionization potential.\n(b)(i) State the three types of emission spectra.\n(ii) Name one source each which produces each of the spectra stated in (b)(i).\n(c) In an x-ray tube, electrons are accelerated the target by a potential difference of 80 A Calculate the:\n(i) speed of the electron;\nii) threshold wavelength of the electron. [h=6.6 x 10\\(^{-34}\\) Js; e = 1.6 x 10\\(^{-19}\\) C; Me = 9.1 x 10\\(^{-31}\\)\nd) An x-ray photon of frequency 4.5 x 10\\(^{-18}\\) strikes an. electron, assumed to be at rest. If t electron absorbs all the photon energy, calculate the speed acquired by the electron. [ h = 6.6 x 10\\(^{-34}\\) Js; Me = 9.1 x 10\\(^{-31}\\) kg ]', NULL, '\n\nExplanation\na) Ionization potential V is the potential energy required to remove an electron from an atom completely to infinity from ground state.\n(b)(i) The three types of emission are\n- band spectrum\n- continuous spectrum and\n- line spectrum.\n(ii) Sources of each emmision spectrum are:\n- Band spectrum\n- from molecules or carbon(IV) oxides in a discharge tube.\n- Continuous spectrum\n- from sun, solids and liquids.\n- Line spectrum\n- from atoms in gases such as hydrogen or neon in a discharge tube.\n(c)(i) eV =\\(\\frac{1}{2} mV^2\\)\nv = \\(\\sqrt{\\frac{2 \\times 1.6 \\times 10^{-19} \\times 80 \\times 10^3}{9.1 \\times 10^{-31}}}\\)\n= \\(\\sqrt{2.81 \\times 10^{16}}\\)\n= 1.68 x 10\\(^8 ms^{-1}\\)\n(ii) \\(\\pi = \\frac{h}{mv} = \\frac{6.6 \\times 10^{-31}}{9.1 \\times 10^{-31} \\times 1.68 \\times 10^8}\\)\n= 4.32 x 10\\(^{-12}\\)m\n\n(d) hf = \\(\\frac{1}{2}mv^2\\)\n6.6 x 10\\(^{-34} \\times 4.5 \\times 10^{18}\\)\n= \\(\\frac{1}{2} \\times 9.1 \\times 10^{-31} \\times v^2\\)\n\\(v^2 = \\frac{2.97 \\times 10^{-15}}{4.55 \\times 10^{-3}}\\)\n\\(v^2 = \\sqrt{6.53 \\times 10^{15}}\\)\n= 8.08 x 10\\(^7 ms^{-1}\\)\n\n\n', '', 'Explanation\na) Ionization potential V is the potential energy required to remove an electron from an atom completely to infinity from ground state.\n(b)(i) The three types of emission are\n- band spectrum\n- continuous spectrum and\n- line spectrum.\n(ii) Sources of each emmision spectrum are:\n- Band spectrum\n- from molecules or carbon(IV) oxides in a discharge tube.\n- Continuous spectrum\n- from sun, solids and liquids.\n- Line spectrum\n- from atoms in gases such as hydrogen or neon in a discharge tube.\n(c)(i) eV =\\(\\frac{1}{2} mV^2\\)\nv = \\(\\sqrt{\\frac{2 \\times 1.6 \\times 10^{-19} \\times 80 \\times 10^3}{9.1 \\times 10^{-31}}}\\)\n= \\(\\sqrt{2.81 \\times 10^{16}}\\)\n= 1.68 x 10\\(^8 ms^{-1}\\)\n(ii) \\(\\pi = \\frac{h}{mv} = \\frac{6.6 \\times 10^{-31}}{9.1 \\times 10^{-31} \\times 1.68 \\times 10^8}\\)\n= 4.32 x 10\\(^{-12}\\)m\n\n(d) hf = \\(\\frac{1}{2}mv^2\\)\n6.6 x 10\\(^{-34} \\times 4.5 \\times 10^{18}\\)\n= \\(\\frac{1}{2} \\times 9.1 \\times 10^{-31} \\times v^2\\)\n\\(v^2 = \\frac{2.97 \\times 10^{-15}}{4.55 \\times 10^{-3}}\\)\n\\(v^2 = \\sqrt{6.53 \\times 10^{15}}\\)\n= 8.08 x 10\\(^7 ms^{-1}\\)\n\n\n', 'a) Ionization potential V is the potential energy required to remove an electron from an atom completely to infinity from ground state.~(b)(i) The three types of emission are\n- band spectrum\n- continuous spectrum and\n- line spectrum.\n(ii) Sources of each emmision spectrum are:\n- Band spectrum\n- from molecules or carbon(IV) oxides in a discharge tube.\n- Continuous spectrum\n- from sun, solids and liquids.\n- Line spectrum\n- from atoms in gases such as hydrogen or neon in a discharge tube.\n(c)(i) eV =\\(\\frac{1}{2} mV^2\\)\nv = \\(\\sqrt{\\frac{2 \\times 1.6 \\times 10^{-19} \\times 80 \\times 10^3}{9.1 \\times 10^{-31}}}\\)\n= \\(\\sqrt{2.81 \\times 10^{16}}\\)\n= 1.68 x 10\\(^8 ms^{-1}\\)\n(ii) \\(\\pi = \\frac{h}{mv} = \\frac{6.6 \\times 10^{-31}}{9.1 \\times 10^{-31} \\times 1.68 \\times 10^8}\\)\n= 4.32 x 10\\(^{-12}\\)m\n\n(d) hf = \\(\\frac{1}{2}mv^2\\)\n6.6 x 10\\(^{-34} \\times 4.5 \\times 10^{18}\\)\n= \\(\\frac{1}{2} \\times 9.1 \\times 10^{-31} \\times v^2\\)\n\\(v^2 = \\frac{2.97 \\times 10^{-15}}{4.55 \\times 10^{-3}}\\)\n\\(v^2 = \\sqrt{6.53 \\times 10^{15}}\\)\n= 8.08 x 10\\(^7 ms^{-1}\\)\n\n~(c)(i) eV =\\(\\frac{1}{2} mV^2\\)\nv = \\(\\sqrt{\\frac{2 \\times 1.6 \\times 10^{-19} \\times 80 \\times 10^3}{9.1 \\times 10^{-31}}}\\)\n= \\(\\sqrt{2.81 \\times 10^{16}}\\)\n= 1.68 x 10\\(^8 ms^{-1}\\)\n(ii) \\(\\pi = \\frac{h}{mv} = \\frac{6.6 \\times 10^{-31}}{9.1 \\times 10^{-31} \\times 1.68 \\times 10^8}\\)\n= 4.32 x 10\\(^{-12}\\)m\n\n(d) hf = \\(\\frac{1}{2}mv^2\\)\n6.6 x 10\\(^{-34} \\times 4.5 \\times 10^{18}\\)\n= \\(\\frac{1}{2} \\times 9.1 \\times 10^{-31} \\times v^2\\)\n\\(v^2 = \\frac{2.97 \\times 10^{-15}}{4.55 \\times 10^{-3}}\\)\n\\(v^2 = \\sqrt{6.53 \\times 10^{15}}\\)\n= 8.08 x 10\\(^7 ms^{-1}\\)\n\n~(d) hf = \\(\\frac{1}{2}mv^2\\)\n6.6 x 10\\(^{-34} \\times 4.5 \\times 10^{18}\\)\n= \\(\\frac{1}{2} \\times 9.1 \\times 10^{-31} \\times v^2\\)\n\\(v^2 = \\frac{2.97 \\times 10^{-15}}{4.55 \\times 10^{-3}}\\)\n\\(v^2 = \\sqrt{6.53 \\times 10^{15}}\\)\n= 8.08 x 10\\(^7 ms^{-1}\\)\n\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(74, '2016', '\n\nQuestion\n\n(a) Define boiling point of a liquid.\n(b) Describe how water in a round bottom flask could be made to boil without heating it. [diagram not necessary]\n(c) State three applications of expansion of metals.\n(d) A room with floor measurements 7m x 10 m contains air of mass 250 kg at a temperature of 34°C. The air is cooled until the temperature falls to 24°C. Calculate the: (i) height of the room;\n(ii) quantity of energy extracted to cool the room;\n(iii) which is higher: the calculated value or the actual energy needed to cool the room? Give a reason for your answer. [ Specific heat capacity of air = 1010 Jkg\\(^{-1}\\)K\\(^{-1}\\); density of air = 1.25 kg m\\(^{-3}]\n\n', '', '(a) Define boiling point of a liquid.\n(b) Describe how water in a round bottom flask could be made to boil without heating it. [diagram not necessary]\n(c) State three applications of expansion of metals.\n(d) A room with floor measurements 7m x 10 m contains air of mass 250 kg at a temperature of 34°C. The air is cooled until the temperature falls to 24°C. Calculate the: (i) height of the room;\n(ii) quantity of energy extracted to cool the room;\n(iii) which is higher: the calculated value or the actual energy needed to cool the room? Give a reason for your answer. [ Specific heat capacity of air = 1010 Jkg\\(^{-1}\\)K\\(^{-1}\\); density of air = 1.25 kg m\\(^{-3}]', '(a) Define boiling point of a liquid.~(b) Describe how water in a round bottom flask could be made to boil without heating it. [diagram not necessary]\n(c) State three applications of expansion of metals.\n(d) A room with floor measurements 7m x 10 m contains air of mass 250 kg at a temperature of 34°C. The air is cooled until the temperature falls to 24°C. Calculate the: (i) height of the room;\n(ii) quantity of energy extracted to cool the room;\n(iii) which is higher: the calculated value or the actual energy needed to cool the room? Give a reason for your answer. [ Specific heat capacity of air = 1010 Jkg\\(^{-1}\\)K\\(^{-1}\\); density of air = 1.25 kg m\\(^{-3}]~(c) State three applications of expansion of metals.\n(d) A room with floor measurements 7m x 10 m contains air of mass 250 kg at a temperature of 34°C. The air is cooled until the temperature falls to 24°C. Calculate the: (i) height of the room;\n(ii) quantity of energy extracted to cool the room;\n(iii) which is higher: the calculated value or the actual energy needed to cool the room? Give a reason for your answer. [ Specific heat capacity of air = 1010 Jkg\\(^{-1}\\)K\\(^{-1}\\); density of air = 1.25 kg m\\(^{-3}]~(d) A room with floor measurements 7m x 10 m contains air of mass 250 kg at a temperature of 34°C. The air is cooled until the temperature falls to 24°C. Calculate the: (i) height of the room;\n(ii) quantity of energy extracted to cool the room;\n(iii) which is higher: the calculated value or the actual energy needed to cool the room? Give a reason for your answer. [ Specific heat capacity of air = 1010 Jkg\\(^{-1}\\)K\\(^{-1}\\); density of air = 1.25 kg m\\(^{-3}]', NULL, '\n\nExplanation\n(a) Boiling point of a liquid is the temperature at which saturated vapour pressure of a liquid equals the external atmospheric pressure.\n(b) Fill a round bottom flask partially with water, then pump out the air gradually with a vacuum pump. At a state when the saturation vapour pressure equal atmospheric pressure, the water is observed to be boiling.\n(c) Application of i expansion of metals are in:\n- Riveting two or more metal plates.\n- Fixing of metal rims on metal wheels.\n- Fusing of platinum wire through walls of glass vessels.\n- Automatic fire alarms and\n- Thermostat.\n(d)(i) Volume = \\(\\frac{Mass}{Density} = \\frac{250}{1.25}\\)\n= 200\\(m^2\\)\nBut volume = Ibh\nh = \\(\\frac{v}{Ib} = \\frac{200}{7 \\times 10}\\)\n= 2.86m\n(ii) Q = mc\\(\\theta\\)\nV = 250 x 1010(34 - 24)\n= 252500 or 2.525 x 10\\(^6\\)\n(iii) The actual energy value will be higher in value than the calculated value because --Part of heat is extracted. from the walls in the room and other materials. he room also is not airtight, there are heat leakages in and out of the room.\n\n', '', 'Explanation\n(a) Boiling point of a liquid is the temperature at which saturated vapour pressure of a liquid equals the external atmospheric pressure.\n(b) Fill a round bottom flask partially with water, then pump out the air gradually with a vacuum pump. At a state when the saturation vapour pressure equal atmospheric pressure, the water is observed to be boiling.\n(c) Application of i expansion of metals are in:\n- Riveting two or more metal plates.\n- Fixing of metal rims on metal wheels.\n- Fusing of platinum wire through walls of glass vessels.\n- Automatic fire alarms and\n- Thermostat.\n(d)(i) Volume = \\(\\frac{Mass}{Density} = \\frac{250}{1.25}\\)\n= 200\\(m^2\\)\nBut volume = Ibh\nh = \\(\\frac{v}{Ib} = \\frac{200}{7 \\times 10}\\)\n= 2.86m\n(ii) Q = mc\\(\\theta\\)\nV = 250 x 1010(34 - 24)\n= 252500 or 2.525 x 10\\(^6\\)\n(iii) The actual energy value will be higher in value than the calculated value because --Part of heat is extracted. from the walls in the room and other materials. he room also is not airtight, there are heat leakages in and out of the room.\n\n', '(a) Boiling point of a liquid is the temperature at which saturated vapour pressure of a liquid equals the external atmospheric pressure.~(b) Fill a round bottom flask partially with water, then pump out the air gradually with a vacuum pump. At a state when the saturation vapour pressure equal atmospheric pressure, the water is observed to be boiling.\n(c) Application of i expansion of metals are in:\n- Riveting two or more metal plates.\n- Fixing of metal rims on metal wheels.\n- Fusing of platinum wire through walls of glass vessels.\n- Automatic fire alarms and\n- Thermostat.\n(d)(i) Volume = \\(\\frac{Mass}{Density} = \\frac{250}{1.25}\\)\n= 200\\(m^2\\)\nBut volume = Ibh\nh = \\(\\frac{v}{Ib} = \\frac{200}{7 \\times 10}\\)\n= 2.86m\n(ii) Q = mc\\(\\theta\\)\nV = 250 x 1010(34 - 24)\n= 252500 or 2.525 x 10\\(^6\\)\n(iii) The actual energy value will be higher in value than the calculated value because --Part of heat is extracted. from the walls in the room and other materials. he room also is not airtight, there are heat leakages in and out of the room.\n~(c) Application of i expansion of metals are in:\n- Riveting two or more metal plates.\n- Fixing of metal rims on metal wheels.\n- Fusing of platinum wire through walls of glass vessels.\n- Automatic fire alarms and\n- Thermostat.\n(d)(i) Volume = \\(\\frac{Mass}{Density} = \\frac{250}{1.25}\\)\n= 200\\(m^2\\)\nBut volume = Ibh\nh = \\(\\frac{v}{Ib} = \\frac{200}{7 \\times 10}\\)\n= 2.86m\n(ii) Q = mc\\(\\theta\\)\nV = 250 x 1010(34 - 24)\n= 252500 or 2.525 x 10\\(^6\\)\n(iii) The actual energy value will be higher in value than the calculated value because --Part of heat is extracted. from the walls in the room and other materials. he room also is not airtight, there are heat leakages in and out of the room.\n~(d)(i) Volume = \\(\\frac{Mass}{Density} = \\frac{250}{1.25}\\)\n= 200\\(m^2\\)\nBut volume = Ibh\nh = \\(\\frac{v}{Ib} = \\frac{200}{7 \\times 10}\\)\n= 2.86m\n(ii) Q = mc\\(\\theta\\)\nV = 250 x 1010(34 - 24)\n= 252500 or 2.525 x 10\\(^6\\)\n(iii) The actual energy value will be higher in value than the calculated value because --Part of heat is extracted. from the walls in the room and other materials. he room also is not airtight, there are heat leakages in and out of the room.\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(75, '2016', '\n\nQuestion\n\nName the three basic components P, Q and R that make up a cathode ray tube, as illustrated in the diagram above\n\n\n', 'Name the three basic components P, Q and R that make up a cathode ray tube, as illustrated in the diagram above\n', '', '', NULL, '\n\nExplanation\nThe basic component of the cathode ray tube are:\nP = Electron gun. Q = Deflector places or oscillators.\nR = Screen.\n\n', 'The basic component of the cathode ray tube are:\nP = Electron gun. Q = Deflector places or oscillators.\nR = Screen.\n\n', '', '', 'WAEC-PK-1', 'WAEC-PHY', 1),
(76, '2016', '\n\nQuestion\n\nA ray of light is incident on an air-glass boundary at an angle \\(\\theta\\). If the angle between the partially reflected ray and the refracted ray is 90°, calculate \\(\\theta\\), given that the refractive index of glass is 1.50.\n\n', 'A ray of light is incident on an air-glass boundary at an angle \\(\\theta\\). If the angle between the partially reflected ray and the refracted ray is 90°, calculate \\(\\theta\\), given that the refractive index of glass is 1.50.', '', '', NULL, '\n\nExplanation\nn = \\(\\frac{sin i}{sin r}\\)\n1.5 = \\(\\frac{sin \\theta}{sin(190 - \\theta)} = \\frac{sin \\theta}{cos theta} = \\tan \\theta\\)\nHence, 0 = tan\\(^{-1}\\) 1.5 = 56.31\\(^o\\)\n\n\n', 'n = \\(\\frac{sin i}{sin r}\\)\n1.5 = \\(\\frac{sin \\theta}{sin(190 - \\theta)} = \\frac{sin \\theta}{cos theta} = \\tan \\theta\\)\nHence, 0 = tan\\(^{-1}\\) 1.5 = 56.31\\(^o\\)\n\n\n', '', '', 'WAEC-PK-1', 'WAEC-PHY', 1),
(77, '2016', '\n\nQuestion\n\n(a)\n\n\nIn the diagram illustrated, a body of mass m slides on an inclined plane. Show that the coefficient Mg of friction between the surfaces in contact is tan \\(\\theta\\).\nA spiral spring with a metal extends by 10.5 cm in air. When the metal is fully submerged in water, the spring extends by 6.8 cm. Calculate the relative density of the metal. (Assume Hooke\'s law is obeyed)\n\n\n\n(a)\n\n', '(a)\n\n\nIn the diagram illustrated, a body of mass m slides on an inclined plane. Show that the coefficient Mg of friction between the surfaces in contact is tan \\(\\theta\\).\nA spiral spring with a metal extends by 10.5 cm in air. When the metal is fully submerged in water, the spring extends by 6.8 cm. Calculate the relative density of the metal. (Assume Hooke\'s law is obeyed)\n\n\n\n(a)', '', '', NULL, '\n\nExplanation\nIf F = uniform speed of body\n\\(\\mu = \\frac{F}{R} = \\frac{Mg sin \\theta}{Mg cos \\theta} = tan \\theta\\)\n\n', 'If F = uniform speed of body\n\\(\\mu = \\frac{F}{R} = \\frac{Mg sin \\theta}{Mg cos \\theta} = tan \\theta\\)\n\n', '', '', 'WAEC-PK-1', 'WAEC-PHY', 1),
(78, '2016', '\n\nQuestion\n\n(a) Explain the term electrodes in electric cells.\nb) An electric current passing through an electrolyte for 2 minutes deposited 200 g of a substance. If the electrochemical equivalent of the substance is 8.33 x 10\\(^{-4}\\)g C\\(^{-1}\\), calculate the current passed.\n\n', '(a) Explain the term electrodes in electric cells.\nb) An electric current passing through an electrolyte for 2 minutes deposited 200 g of a substance. If the electrochemical equivalent of the substance is 8.33 x 10\\(^{-4}\\)g C\\(^{-1}\\), calculate the current passed.', '', '', NULL, '\n\nExplanation\n(a) An electrode is a conductor through which electrons flow into and out of the electric cell.\n(b) I = \\(\\frac{M}{zt} = \\frac{200}{8.33 \\times 10^{-4} \\times 60 \\times 2}\\)\n\n\n', '', 'Explanation\n(a) An electrode is a conductor through which electrons flow into and out of the electric cell.\n(b) I = \\(\\frac{M}{zt} = \\frac{200}{8.33 \\times 10^{-4} \\times 60 \\times 2}\\)\n\n\n', '(a) An electrode is a conductor through which electrons flow into and out of the electric cell.~(b) I = \\(\\frac{M}{zt} = \\frac{200}{8.33 \\times 10^{-4} \\times 60 \\times 2}\\)\n\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(79, '2016', '\n\nQuestion\n\nExplain why sound waves cannot be plane Polarized.\n\n', 'Explain why sound waves cannot be plane Polarized.', '', '', NULL, '\n\nExplanation\nSound waves are longitudinal waves which cannot be plane polarized as polarization is a characteristic of transverse waves only.\n\n', 'Sound waves are longitudinal waves which cannot be plane polarized as polarization is a characteristic of transverse waves only.\n\n', '', '', 'WAEC-PK-1', 'WAEC-PHY', 1),
(80, '2016', '\n\nQuestion\n\na) Define surface tension.\n(b) State two methods by which the surface tension of a liquid can be reduced.\n\n', '', 'a) Define surface tension.\n(b) State two methods by which the surface tension of a liquid can be reduced.', '~(b) State two methods by which the surface tension of a liquid can be reduced.', NULL, '\n\nExplanation\n(a) Surface tension is the force per unit length acting in the surface at right angles to one side of a line drawn in the surface of a liquid. It can be defined as the property of a liquid that makes it behave as if its surface is an elastic skin.\n(b) Methods of reducing surface tension are by adding:\n- Soap\n- alcohol\n- oil\n- detergent\n- increasing the temperature.\n\n', '', 'Explanation\n(a) Surface tension is the force per unit length acting in the surface at right angles to one side of a line drawn in the surface of a liquid. It can be defined as the property of a liquid that makes it behave as if its surface is an elastic skin.\n(b) Methods of reducing surface tension are by adding:\n- Soap\n- alcohol\n- oil\n- detergent\n- increasing the temperature.\n\n', '(a) Surface tension is the force per unit length acting in the surface at right angles to one side of a line drawn in the surface of a liquid. It can be defined as the property of a liquid that makes it behave as if its surface is an elastic skin.~(b) Methods of reducing surface tension are by adding:\n- Soap\n- alcohol\n- oil\n- detergent\n- increasing the temperature.\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(81, '2016', '\n\nQuestion\n\nExplain why it is desirable to install an air conditioner near the ceiling of a room and not close to the floor.\n\n', 'Explain why it is desirable to install an air conditioner near the ceiling of a room and not close to the floor.', '', '', NULL, '\n\nExplanation\nAir molecules near the ceiling get cooled, become denser and sink while warmer air moleculus below being lighter rise to the top, get cooled, become denser and sink once more, thereby sets up conventional current.\n\n', 'Air molecules near the ceiling get cooled, become denser and sink while warmer air moleculus below being lighter rise to the top, get cooled, become denser and sink once more, thereby sets up conventional current.\n\n', '', '', 'WAEC-PK-1', 'WAEC-PHY', 1),
(82, '2016', '\n\nQuestion\n\n(a) On what principle does lighting in a fluorescent tube operate?\n(b) State two factors which determine the colour of light produced in a fluorescent tube.\n\n', '', '(a) On what principle does lighting in a fluorescent tube operate?\n(b) State two factors which determine the colour of light produced in a fluorescent tube.', '(a) On what principle does lighting in a fluorescent tube operate?~(b) State two factors which determine the colour of light produced in a fluorescent tube.', NULL, '\n\nExplanation\n(a) Fluorescent tube operate on the principle of electrical discharge through a gas at very low pressure and very high voltage.\n(b) Factors which determine the colour of light in a fluorescent tube are: -voltage across the tube -nature of gas in the tube -nature of powder coating inside the tube.\n\n', '', 'Explanation\n(a) Fluorescent tube operate on the principle of electrical discharge through a gas at very low pressure and very high voltage.\n(b) Factors which determine the colour of light in a fluorescent tube are: -voltage across the tube -nature of gas in the tube -nature of powder coating inside the tube.\n\n', '(a) Fluorescent tube operate on the principle of electrical discharge through a gas at very low pressure and very high voltage.~(b) Factors which determine the colour of light in a fluorescent tube are: -voltage across the tube -nature of gas in the tube -nature of powder coating inside the tube.\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(83, '2016', '\n\nQuestion\n\nThe diagram above illustrates an arrangement of a cathode ray from an electron gun and a bar magnet placed perpendicularly to the direction of the ray. Will the ray bend downward or upward? Explain.\n\n\n', 'The diagram above illustrates an arrangement of a cathode ray from an electron gun and a bar magnet placed perpendicularly to the direction of the ray. Will the ray bend downward or upward? Explain.\n', '', '', NULL, '\n\nExplanation\nThe ray will bend upward. The direction of motion of the charge is due to the force acting on it which is mutually perpendicular to the direction of the magnetic field and the moving charge.\n\n', 'The ray will bend upward. The direction of motion of the charge is due to the force acting on it which is mutually perpendicular to the direction of the magnetic field and the moving charge.\n\n', '', '', 'WAEC-PK-1', 'WAEC-PHY', 1),
(84, '2016', '\n\nQuestion\n\n(a) Explain wave-particle duality of light.\n(b) Illustrate your answer in (a) with observable phenomena.\n\n', '', '(a) Explain wave-particle duality of light.\n(b) Illustrate your answer in (a) with observable phenomena.', '(a) Explain wave-particle duality of light.~(b) Illustrate your answer in (a) with observable phenomena.', NULL, '\n\nExplanation\n(a) Wave - particle duality of light means that light exists in two forms - wave and particle.\n(b)(i) Wave nature - it has properties such as reflection, refraction, diffraction, polarization and interference.\n(ii) Particle nature such as - Photo-electricity, momentum, radiation of energy from hot bodies and emission and absorption of light.\n\n', '', 'Explanation\n(a) Wave - particle duality of light means that light exists in two forms - wave and particle.\n(b)(i) Wave nature - it has properties such as reflection, refraction, diffraction, polarization and interference.\n(ii) Particle nature such as - Photo-electricity, momentum, radiation of energy from hot bodies and emission and absorption of light.\n\n', '(a) Wave - particle duality of light means that light exists in two forms - wave and particle.~(b)(i) Wave nature - it has properties such as reflection, refraction, diffraction, polarization and interference.\n(ii) Particle nature such as - Photo-electricity, momentum, radiation of energy from hot bodies and emission and absorption of light.\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(85, '2016', '\n\nQuestion\n\n(a) State the principle of conservation of linear momentum.\n(b) Explain the mode of action of a propelled rocket.\n(c) During a training session, two footballers pass a ball repeatedly between themselves. Give two reasons why the to and fro motion of the ball is not simple harmonic.\n(d) A ball is dropped from a height, at the same time as another ball is projected horizontally from the same height.\n(i) Would the balls hit the ground at the same time?\n(ii) Explain your answer in (i).\n(e) A ball of mass 0.10 kg is projected horizontally onto a vertical wall with a speed of 17 ms\\(^{-1}\\). The ball makes contact with the wall for 0.15 s and rebounds horizontally with a speed of 13 ms\\(^{-1}\\).\nCalculate the:\n(i) change in momentum of the ball;\n(ii) average force exerted on the ball during its collision with the wall.\n\n', '', '(a) State the principle of conservation of linear momentum.\n(b) Explain the mode of action of a propelled rocket.\n(c) During a training session, two footballers pass a ball repeatedly between themselves. Give two reasons why the to and fro motion of the ball is not simple harmonic.\n(d) A ball is dropped from a height, at the same time as another ball is projected horizontally from the same height.\n(i) Would the balls hit the ground at the same time?\n(ii) Explain your answer in (i).\n(e) A ball of mass 0.10 kg is projected horizontally onto a vertical wall with a speed of 17 ms\\(^{-1}\\). The ball makes contact with the wall for 0.15 s and rebounds horizontally with a speed of 13 ms\\(^{-1}\\).\nCalculate the:\n(i) change in momentum of the ball;\n(ii) average force exerted on the ball during its collision with the wall.', '(a) State the principle of conservation of linear momentum.~(b) Explain the mode of action of a propelled rocket.\n(c) During a training session, two footballers pass a ball repeatedly between themselves. Give two reasons why the to and fro motion of the ball is not simple harmonic.\n(d) A ball is dropped from a height, at the same time as another ball is projected horizontally from the same height.\n(i) Would the balls hit the ground at the same time?\n(ii) Explain your answer in (i).\n(e) A ball of mass 0.10 kg is projected horizontally onto a vertical wall with a speed of 17 ms\\(^{-1}\\). The ball makes contact with the wall for 0.15 s and rebounds horizontally with a speed of 13 ms\\(^{-1}\\).\nCalculate the:\n(i) change in momentum of the ball;\n(ii) average force exerted on the ball during its collision with the wall.~(c) During a training session, two footballers pass a ball repeatedly between themselves. Give two reasons why the to and fro motion of the ball is not simple harmonic.\n(d) A ball is dropped from a height, at the same time as another ball is projected horizontally from the same height.\n(i) Would the balls hit the ground at the same time?\n(ii) Explain your answer in (i).\n(e) A ball of mass 0.10 kg is projected horizontally onto a vertical wall with a speed of 17 ms\\(^{-1}\\). The ball makes contact with the wall for 0.15 s and rebounds horizontally with a speed of 13 ms\\(^{-1}\\).\nCalculate the:\n(i) change in momentum of the ball;\n(ii) average force exerted on the ball during its collision with the wall.~(d) A ball is dropped from a height, at the same time as another ball is projected horizontally from the same height.\n(i) Would the balls hit the ground at the same time?\n(ii) Explain your answer in (i).\n(e) A ball of mass 0.10 kg is projected horizontally onto a vertical wall with a speed of 17 ms\\(^{-1}\\). The ball makes contact with the wall for 0.15 s and rebounds horizontally with a speed of 13 ms\\(^{-1}\\).\nCalculate the:\n(i) change in momentum of the ball;\n(ii) average force exerted on the ball during its collision with the wall.~(e) A ball of mass 0.10 kg is projected horizontally onto a vertical wall with a speed of 17 ms\\(^{-1}\\). The ball makes contact with the wall for 0.15 s and rebounds horizontally with a speed of 13 ms\\(^{-1}\\).\nCalculate the:\n(i) change in momentum of the ball;\n(ii) average force exerted on the ball during its collision with the wall.', NULL, '\n\nExplanation\n(a) Principle of conservation of linear momentum states that if two or more bodies collide in a closed system, the total momentum after the collision is equal to the total momentum before the collision.\n\n(b) The principle of action of a propelled rocket is that gas burnt within the combustion chamber of the rocket engine is expelled downwards through the exhaust (tail nozzle) at very high speed, hence an equal and opposite momentum is given to the rocket which moves the rocket upward.\n\n(c) The motion is not simple harmonic because:\n-there is no equilibrium position\n-acceleration of the ball is not directly proportional to displacement\n-acceleration of the ball is not directed towards a fixed point.\n\n(d)(i) Both balls will hit the ground at the same time.\n(ii) All free falling objects fall at the same rate because they are pulled towards the earth by gravity and accelerate at a constant rate.\n\n(e)(i) Charge in momentum = m(v-u)\n= 0.10(13 - 17) = 3 kgms\\(^{-1}\\)\n(ii) Average force F = \\(\\frac{\\text{change in momentum}}{\\text{time}}\\) = \\(\\frac{3}{0.15} = 20N\\)\n\n', '', 'Explanation\n(a) Principle of conservation of linear momentum states that if two or more bodies collide in a closed system, the total momentum after the collision is equal to the total momentum before the collision.\n\n(b) The principle of action of a propelled rocket is that gas burnt within the combustion chamber of the rocket engine is expelled downwards through the exhaust (tail nozzle) at very high speed, hence an equal and opposite momentum is given to the rocket which moves the rocket upward.\n\n(c) The motion is not simple harmonic because:\n-there is no equilibrium position\n-acceleration of the ball is not directly proportional to displacement\n-acceleration of the ball is not directed towards a fixed point.\n\n(d)(i) Both balls will hit the ground at the same time.\n(ii) All free falling objects fall at the same rate because they are pulled towards the earth by gravity and accelerate at a constant rate.\n\n(e)(i) Charge in momentum = m(v-u)\n= 0.10(13 - 17) = 3 kgms\\(^{-1}\\)\n(ii) Average force F = \\(\\frac{\\text{change in momentum}}{\\text{time}}\\) = \\(\\frac{3}{0.15} = 20N\\)\n\n', '(a) Principle of conservation of linear momentum states that if two or more bodies collide in a closed system, the total momentum after the collision is equal to the total momentum before the collision.\n~(b) The principle of action of a propelled rocket is that gas burnt within the combustion chamber of the rocket engine is expelled downwards through the exhaust (tail nozzle) at very high speed, hence an equal and opposite momentum is given to the rocket which moves the rocket upward.\n\n(c) The motion is not simple harmonic because:\n-there is no equilibrium position\n-acceleration of the ball is not directly proportional to displacement\n-acceleration of the ball is not directed towards a fixed point.\n\n(d)(i) Both balls will hit the ground at the same time.\n(ii) All free falling objects fall at the same rate because they are pulled towards the earth by gravity and accelerate at a constant rate.\n\n(e)(i) Charge in momentum = m(v-u)\n= 0.10(13 - 17) = 3 kgms\\(^{-1}\\)\n(ii) Average force F = \\(\\frac{\\text{change in momentum}}{\\text{time}}\\) = \\(\\frac{3}{0.15} = 20N\\)\n~(c) The motion is not simple harmonic because:\n-there is no equilibrium position\n-acceleration of the ball is not directly proportional to displacement\n-acceleration of the ball is not directed towards a fixed point.\n\n(d)(i) Both balls will hit the ground at the same time.\n(ii) All free falling objects fall at the same rate because they are pulled towards the earth by gravity and accelerate at a constant rate.\n\n(e)(i) Charge in momentum = m(v-u)\n= 0.10(13 - 17) = 3 kgms\\(^{-1}\\)\n(ii) Average force F = \\(\\frac{\\text{change in momentum}}{\\text{time}}\\) = \\(\\frac{3}{0.15} = 20N\\)\n~(d)(i) Both balls will hit the ground at the same time.\n(ii) All free falling objects fall at the same rate because they are pulled towards the earth by gravity and accelerate at a constant rate.\n\n(e)(i) Charge in momentum = m(v-u)\n= 0.10(13 - 17) = 3 kgms\\(^{-1}\\)\n(ii) Average force F = \\(\\frac{\\text{change in momentum}}{\\text{time}}\\) = \\(\\frac{3}{0.15} = 20N\\)\n~(e)(i) Charge in momentum = m(v-u)\n= 0.10(13 - 17) = 3 kgms\\(^{-1}\\)\n(ii) Average force F = \\(\\frac{\\text{change in momentum}}{\\text{time}}\\) = \\(\\frac{3}{0.15} = 20N\\)\n', 'WAEC-PK-1', 'WAEC-PHY', 1),
(86, '2016', '\n\nQuestion\n\n(a) (i) What is a machine?\n(ii) State two uses of gears.\n(iii) Define the velocity ratio for a pair of gear wheels.\n(iv) How can the mechanical advantage of a gear system be increased?\n\nThe diagram above illustrates the gears system of a bicycle.\n(i) Determine its velocity ratio.\n(ii) If the bicycle has an efficiency of 90%, calculate the effort required to overcome a load of 70N.\n(iii) Why is the calculated effort less than the actual effort required?\n\n', '', '(a) (i) What is a machine?\n(ii) State two uses of gears.\n(iii) Define the velocity ratio for a pair of gear wheels.\n(iv) How can the mechanical advantage of a gear system be increased?\n\nThe diagram above illustrates the gears system of a bicycle.\n(i) Determine its velocity ratio.\n(ii) If the bicycle has an efficiency of 90%, calculate the effort required to overcome a load of 70N.\n(iii) Why is the calculated effort less than the actual effort required?', '', NULL, '\n\nExplanation\n(a)(i) A machine is a device in which an effort applied at one end is used to overcome a load at another point. Machine makes work easier.\n(ii) Uses of gears are: - to change speed to change direction of motion\n- to magnify force.\n(iii) Velocity ratio of a gear wheel is the ratio of number of teeth on driven wheel to number of teeth on driving wheel OR\nVelocity ratio = \\(\\frac{\\text{Time rate of revolution/speed/radius of driven wheel}}{\\text{Time rate of revolution/speed/radius of driving wheel}}\\)\n(iv) To increase the mechanical advantage, the driving gear is made smaller with fewer teeth than the driven gear.\n(b) Number of teeth on driven gear = 12.\nNumber of teeth on driving gear = 18.\n(i) V.R. = \\(\\frac{\\text{no. of teeth on driven gear}}{\\text{no. of teeth on driving gear}}\\)\n= \\(\\frac{12}{8} = \\frac{2}{3}\\)\n(ii) Efficiency - \\(\\frac{M.A}{V.R}\\) x 100% = x 100%\n= \\(\\frac{L/E}{V.R}\\) x 100%\n90 = \\(\\frac{70/E}{2/3}\\) x 100%\n(iii) This is due to: - Friction between the tyres and load.\n- Friction between chain and gear.\n\n', '', 'Explanation\n(a)(i) A machine is a device in which an effort applied at one end is used to overcome a load at another point. Machine makes work easier.\n(ii) Uses of gears are: - to change speed to change direction of motion\n- to magnify force.\n(iii) Velocity ratio of a gear wheel is the ratio of number of teeth on driven wheel to number of teeth on driving wheel OR\nVelocity ratio = \\(\\frac{\\text{Time rate of revolution/speed/radius of driven wheel}}{\\text{Time rate of revolution/speed/radius of driving wheel}}\\)\n(iv) To increase the mechanical advantage, the driving gear is made smaller with fewer teeth than the driven gear.\n(b) Number of teeth on driven gear = 12.\nNumber of teeth on driving gear = 18.\n(i) V.R. = \\(\\frac{\\text{no. of teeth on driven gear}}{\\text{no. of teeth on driving gear}}\\)\n= \\(\\frac{12}{8} = \\frac{2}{3}\\)\n(ii) Efficiency - \\(\\frac{M.A}{V.R}\\) x 100% = x 100%\n= \\(\\frac{L/E}{V.R}\\) x 100%\n90 = \\(\\frac{70/E}{2/3}\\) x 100%\n(iii) This is due to: - Friction between the tyres and load.\n- Friction between chain and gear.\n\n', '(a)(i) A machine is a device in which an effort applied at one end is used to overcome a load at another point. Machine makes work easier.\n(ii) Uses of gears are: - to change speed to change direction of motion\n- to magnify force.\n(iii) Velocity ratio of a gear wheel is the ratio of number of teeth on driven wheel to number of teeth on driving wheel OR\nVelocity ratio = \\(\\frac{\\text{Time rate of revolution/speed/radius of driven wheel}}{\\text{Time rate of revolution/speed/radius of driving wheel}}\\)\n(iv) To increase the mechanical advantage, the driving gear is made smaller with fewer teeth than the driven gear.~(b) Number of teeth on driven gear = 12.\nNumber of teeth on driving gear = 18.\n(i) V.R. = \\(\\frac{\\text{no. of teeth on driven gear}}{\\text{no. of teeth on driving gear}}\\)\n= \\(\\frac{12}{8} = \\frac{2}{3}\\)\n(ii) Efficiency - \\(\\frac{M.A}{V.R}\\) x 100% = x 100%\n= \\(\\frac{L/E}{V.R}\\) x 100%\n90 = \\(\\frac{70/E}{2/3}\\) x 100%\n(iii) This is due to: - Friction between the tyres and load.\n- Friction between chain and gear.\n', 'WAEC-PK-1', 'WAEC-PHY', 1);
