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Power = 100W; Voltage = 240V; Time = 5 hours Total energy used = Power x time = 100 x 5 = 500wh converting to kwh, we have \(\frac{500}{1000} = 0.5kWh \times N5 = N2.50\)

E = 3.0V; \(\frac{R}{l} = 10\Omega m^{-1}\) l = 30cm = 0.3m; R = \(10\Omega m^{-1} \times 0.3m = 3\Omega\) \(\frac{E}{V} = \frac{l_{1}}{l_{2}} \implies \frac{3}{V} = \frac{1}{0.3}\) = \(\frac{3\times 0.3}{1} = 0.9V\)

No expanded working has been attached for this question yet. The verified answer is Option B.

No expanded working has been attached for this question yet. The verified answer is Option D.

No expanded working has been attached for this question yet. The verified answer is Option A.

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Let R represent the horizontal distance covered at time t. Using h = ut + \(\frac{1}{2}gt^2\) but U\(_v\) = 0 h = \(\frac{1}{2} gt^{2}\) 20 = \(\frac{1}{2} \times 10 \times t^2\) 20 = 5t\(^2\) t\(^2\) = \(\frac{20}{5}\) = 4 t = 2s Range(R) = U\(_x\) x t = 15 x 2 = 30m or R = u x \(\sqrt{\frac{2h}{g}}\) = 15 x \(\sqrt{\frac{2 \times 20}{10}}\) = 15 x \(\sqrt{4}\) = 15 x 2 = 30m

(i) Osmosis (ii) Brownian motion (iii) Diffusion (iv) Capilarity (v) Viscosity

Using hookes law \(\frac{F_1}{e_1} = \frac{F_2}{e_2}\) but e\(_1\) = l\(_1\) = l\(_o\) \(\frac{F_1}{l_1 - l_o}\) = \(\frac{F_2}{e_2}\) \(\frac{4}{14 - \_o}\) = \(\frac{6}{4}\); 16 = 6(14 - \(_o\)) 14 = 84 - 6\(l_o\); 16 - 84 = -6\(l_o\) \(l_o = \frac{-68}{-6}\) = 11.33cm

(a)(i) Purity of liquid (ii) Temperature of liquid (iii) Nature / type of liquid. (b) Mosquito larvae can be made to sink by the addition of any of; (i) Detergent / Soap (ii) Alcohol (iii) Heat (iv) Camphor .

Advantages of flourescent tubes over filament bulbs are; (i) They produce less heat (ii) They last longer (iii) They are brighter (iv) They consume less electrical energy

The advantages of P – n; junction diode over diode valves are: (i) It is not bulky (ii) It does not need time to warm up (iii) It is cheaper to manufacture in large quantities. (iv) It needs low voltage to operate.

(a) From a displacement - time graph, the following can be deduced (i) Uniform velocity (ii) Instantaneous velocity (iii) Initial displacement (b) The force would remain the same Gram = \(F_2 = \frac{Gmm}{r^2}\) F\(_2\) = \(\frac{G2m \times 2m}{(2r)^2}\) =\(\frac{G4m^2}{4r^2}\) = \(\frac{Gm^2}{r^2}\) = \(\frac{Gmm}{r^2}\) \(F_1 = F_2\) (c)Factors that affect maximum height of a bullet include (i) Initial velocity / speed (ii) Acceleration due to gravity (iii) Angle of projection (iv) Air resistance (d)Practical examples of mechanical resonance include: (i) Shattering of glass due to high pitch note. (ii) Throwing of 2 paper rider from a vibrating stretched string (iii) Collapse of a bridge by matching soldiers (iv) Car bodies rattle (at very high speed) (e) acceleration = g sin \(\theta\) – \(\mu\) cos \(\theta\) a =10 x sin 30 – 0.3 x 10 x cos30 =10 x 0.5 –0.3 x 10 x 0.8660 = 5 – 2.598 = 2.4m/s\(^2\) Distance covered, s = \(\frac{4}{\sin30}\) = \(\frac{4}{0.5}\) = 8m Using S = Ut + \(\frac{1}{2}\)at\(^2\) 8 =\(\frac{1}{2}\) x 2.4 x r\(^2\) 8 = 1.2\(t^2\) \(r^2\) = \(\frac{8}{1.2}\) = 2.58s

(a) A body is said to be in stable equillibrium if it tends to return to its original position after being slightly displaced. (b) i.e. V.R = Number of pulleys in the system V.R = 3 (c) \(P_1 = 4.5 \times 10^5\)Pa; \(_2 = 4.6 \times 10^5\)Pa T\(_1\) = 20\(^{o}\) + 273 = 293k T\(_2\) = ? \(\frac{P_1}{T_1} = \frac{P_2}{T_2}; \frac{4.5 \times 10^5}{293} = \frac{4.6 \times 10^5}{T_2}\) \(T_2 = \frac{4.6 \times 10^5 \times 293}{4.5 \times 10^5}\) \(T_2\) = 299.5k or 26.5\(^o\)C (d) \(\alpha\) Resistance thermometer \(\frac{a}{b} = \frac{c}{d}; \frac{R \theta - 5.67}{7.75 - 5.67} = \frac{R - 273}{373 - 273}\) \(\frac{R \theta - 5.67}{1,88} = \frac{R - 273}{100}\) \(\beta\) Gas thermometer \(\frac{a}{b} = \frac{c}{d}; \frac{G\theta - (7.13) x 10^4}{(9.74 - 7.13) \times 10^4} = \frac{G - 273}{373 - 273}\) \(\frac{G\theta - 7.13}{2.16} = \frac{G - 273}{100}\)

(a) A Wavefront is a surface over which particles vibrate in the same phase. (b)(i) Glass prisms are used practically for (I) changing direction of light I dispersion (II) viewing distant objects (Binoculars) (III) determing refractive index of glass (IV) viewing objects over obstacles (periscope) (ii) The factors include: (I) Angle of incidence (II) refractive index of the prism (III) refracting angle of the prism (iii) (c) (ii) (\(alpha\)) Magnifiation = \(\frac{F_o}{F_e}\); 5 = \(\frac{100}{Fe}\) fe = 20cm (\(\beta\)) length of telescope = \(F_o + F_e = 100 + 20\) = 120cm