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(a) Strain is the ratio of the extension to original length of an object. (b) Strain = \(\frac{e}{l_o} = \frac{2l_o - l_o}{l_o}\) = 1

S/N d(cm) I(A) I\(^{1}\)(A\(^{-1}\)) Log I\(^{-1}\) 1 25 0.65 1.54 0.19 2 35 0.55 1.82 0.26 3 50 0.44 2.50 0.40 4 65 0.30 3.33 0.52 5 80 0.23 4.35 0.64 vii. Scale: 2cm to 10 units on x-axis 2cm to 0.1 unit on y-axis viii. Slopes(s) = \(\frac{x_{2} - x_{1}}{y_{2} - y_{1}}\) = \(\frac{80 -25}{0.64 - 0.20}\) = \(\frac{55}{0.44}\) = 125 (ix) Precautions (a) Key opened in between reading or keys opened when readings were not taken (b) Tight connections ensured (c) Avoided parallax error when taking reading on ammeter or metre rule (d) Repeated readings shown (e) Avoided zero error on ammeter or metre rule (f) Clean terminals ensured. (b)i. 1\(^{-1}\) = \(\frac{1}{1.5}\) = 0.67 Log(1\(^{-1}\)) =-0.174 ii. Factors affecting the resistance of a wire - Length of wire - Cross-sectional area/ thickness/ diameter/ radius of wire - Resistivity or nature of wire - Temperature.

\(\lambda\) max = \(\frac{2.9 \times 10^{-3}}{330}\) = 8.8 x 10\(^{-6}\)m v = f\(\lambda\) or f = \(\frac{v}{\lambda}\) where v = velocity of light = 3.0 x10\(^{-8}\) m/s \(\lambda\) = 8.8 x 10\(^{-6}\) f = \(\frac{v}{\lambda}\) = \(\frac{3.0 \times 10^8}{8.8 \times 10^{-6}}\) = 3.4 x 10\(^{13}\) Hz

Table of value/observation \(a_0\) = 2.00cm S/N x(cm) a(cm) v(cm) y=\(\frac{a}{a_{0}}\) p=\(\frac{1+y^{2}}{y}\) T=x+v(cm) 1 25.0 2.60 49.0 1.30 2.07 74.00 2 30.0 2.40 46.0 1.20 2.03 76.00 3 35.0 2.10 45.0 1.05 2.01 80.00 4 40.0 1.80 43.0 0.90 2.01 83.00 5 45.0 1.60 40.0 0.80 2.05 85.00 viii. viii. Slopes(s) = \(\frac{x_{2} - x_{1}}{y_{2} - y_{1}}\) = \(\frac{76 - 70}{2.073 - 2.040}\) = \(\frac{6}{0.033}\) = 181.8 x. Intercept (c) = 81 xi. Evaluation k = \(\frac{c}{2}\) = \(\frac{81}{2}\) = 40.5 xii. Precautions; (a) Conxial arrangement of optical instruments (b) Avoided parallax error in reading metre rule (c) Lens kept upright (d) Repeated readings shown on the table (e) Surface of lens cleaned (f) Avoided zero error on metre rule (b)i. The distance between the optical centre and the principal focus of the lens is 20cm or if the parallel beam of light is incident on the lens, it is brought to a focus at a point 20cm from the lens. ii. M = 5cm, F = 20c, x = u M = \(\frac{v}{u}\) 5 = \(\frac{v}{u}\) v = 5u Using the formula \(\frac{1}{v} + \frac{1}{u} = \frac{1}{f}\) \(\frac{1}{5u} + \frac{1}{u} = \frac{1}{20}\) \(\frac{1+5}{5u} = \frac{1}{20}\) 5u = 20(1 + 5) 5u = 20 x 6 5u = 120 u = \(\frac{120}{5}\) = 24 x = 24cm

\(\sqrt{108} + \sqrt{125} - \sqrt{75}\) = \(\sqrt{3 \times 36} + \sqrt{5 \times 25} - \sqrt{3 \times 25}\) = \(6 \sqrt{3} + 5 \sqrt{5} - 5 \sqrt{3}\) = \(\sqrt{3} + 5\sqrt{5}\)

\([64^{\frac{1}{2}} + 125^{\frac{1}{3}}]^2\) = \([\sqrt{64} + \sqrt[3] {125}]^2\) \([8 + 5]^2\) = \([13]^2\) = 169

Y \(\alpha \frac{1}{x^2} \rightarrow y = \frac{k}{x^2}\) If x = 3 and y = 100, then, \(\frac{100}{1} = \frac{k}{3^2}\) \(\frac{100}{1} = \frac{k}{9}\) k = 100 x 9 = 900 Substitute 900 for k in y = \(\frac{k}{x^2}\); y = \(\frac{900}{x^2}\) = \(yx^2 = 900\)

\(32_4 = 22_x\) \(3 \times 4^1 + 2 \times 4^o\) = \(2 \times x^1 + 2 \times x^o\) 12 + 2 x 1 = 2x + 2 x 1 14 = 2x + 2 14 - 2 = 2x 12 = 2x x = \(\frac{12}{2}\) x = 6

Population of school = 250 + 150 = 400 60% of 250 = \(\frac{\text{60%}}{\text{100%}}\) x 250 = 150 40% of 150 = \(\frac{\text{40%}}{\text{100%}}\) x 150 = 60 Total number of students who plays football; 150 + 60 = 210 Percentage of school that play football; \(\frac{210}{400}\) x 100% = 52.5%

\(\log_{10}\)(6x - 4) - \(\log_{10}\)2 = 1 \(\log_{10}\)(6x - 4) - \(\log_{10}\)2 = \(\log_{10}\)10 \(\log_{10}\)\(\frac{6x - 4}{2}\) - \(\log_{10}\)10 \(\frac{6x - 4}{2}\) = 10 6x - 4 = 2 x 10 = 20 6x = 20 + 4 6x = 20 x = \(\frac{24}{6}\) x = 4

F = \(\frac{9}{5}\)C + 32 When F = 98.6 98.6 = \(\frac{9}{5}\)C + 32 98.6 - 32 = \(\frac{9}{5}\)C 66.6= \(\frac{9}{5}\)C 66.6 x 5 = 9C C = \(\frac{66.6 \times 5}{9}\) = 37

y + 2x = 4 .....(1) y - 3x = -1 ......(2) Subtract (2) from (1) 2x - (-3x) = 4 - (-1) 2x + 3x = 4 + 1 5x = 5 X = \(\frac{5}{5}\) = 1 Substitute 1 for x in (1); y + 2(1) = 4 y + 2 = 4 y = 4 - 2 = 2 Hence, (x + y) = (1 + 2) = 3

If x : y : z = 3 : 3 : 4, evaluate \(\frac{9x + 3y}{6x - 2y}\) \(\frac{x}{y}\) = \(\frac{2}{3}\) and \(\frac{y}{z}\) = \(\frac{3}{4}\) Thus; x = \(\frac{2}{3}T_1\) and z = \(\frac{3}{5}T_1\) y = \(\frac{3}{7}T_2\) and z = \(\frac{4}{7}T_2\) Using y = y \(\frac{3}{5}T_1\) = \(\frac{3}{7}T_2\); \(\frac{T_1}{T_2}\) = \(\frac{3}{7}\) x \(\frac{5}{3}\) \(\frac{T_1}{T_2}\) = \(\frac{15}{21}\) \(T_1\) = 15 and \(T_2\) = 21 Therefore; x = \(\frac{2}{5}\) x 15 = 6 y = \(\frac{3}{5}\) x 15 = 9 y = \(\frac{3}{7}\) x 21 = 9 (again) z = \(\frac{4}{7}\) x 21 = 12 Hence; \(\frac{9x + 3y}{6z - 2y}\) = \(\frac{9(6) + 3(9)}{6(12) - 2(9)}\) \(\frac{54 + 27}{72 - 18}\) = \(\frac{81}{54}\) = \(\frac{3}{2}\) = 1\(\frac{1}{2}\)

\(\frac{2 - 18m^2}{1 + 3m}\) = \(\frac{2(1 - 9)m^2}{1 + 3m}\) = \(\frac{2(1 + 3m)(1 - 3m)}{1 + 3m}\) = \(2(1 - 3m)\)

Since the curve cuts the x-axis at x = -2 and x = 3, (x + 2)(x - 3) = 0 \(x^2 - 3x + 2x - 6\) = 0 \(x^2 - x - 6\) = 0 Hence, the equation of the curve is y = \(x^2 - x - 6\)