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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

(a)(i) A dielectric is an insulating material (solid, liquid or gas) separating the two plates of a capacitor. (ii) Using C = \(\frac{\xi_1.\xi_oA}{d} = \frac{900 \times 8.85 \times 10^{-12} \times 9.6 \times 10^{-3}}{2.25 \times 10^{-4}}\) = 3.4 x 10\(^{-7}\) (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. (ii) \(\frac{1}{R_1} = \frac{1}{400} + \frac{1}{400}\) \(R_1 = 200\Omega\) \(\frac{1}{R_2} = \frac{1}{800} + \frac{1}{800}\) \(R_1 = 400\Omega\) RT= 200 + 400 = 600\(\Omega\) I = \(\frac{V_T}{R_T} = \frac{6}{600}\) = 0.01A Voltmeter reading = Pd across R\(_2\) = 0.01A Voltmeter reading = Pd across R\(_2\) = 0.01 x 400 = 4v (c) \(\frac{1}{R_p} = \frac{1}{2} + \frac{1}{2} = \frac{1 + 1}{2} = \frac{2}{2}\) Note: This catadation is for the two resistors in parallel. R\(_T = 2 + 1 + 2 = 5\Omega\)

(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. (ii) (\(\alpha\)) graphite; (\(\beta\)) Boron rods; (\(\gamma\) )Liquid sodium Material Function Graphite Slows down high energetic neutrons in the reaction chamber. Boron rods To absorb excess neutrons so as to control fission rate Liquid sodium Extraction / conduction of excess heat from the reaction chamber. (b)(i) E\(_n\), n = \(\alpha\) ---------------------------------- 0.eV E\(_4\), n = 5 ---------------------------------------------------- - 0.54eV E\(_3\), n = 4 ---------------------------------------------------- - 0.85eV E\(_2\), n = 3 ---------------------------------------------------- 1.51eV E\(_1\), n = 2 ------------------------------------------------------ 3.39eV (ii) Using \(\lambda\) = \(\frac{hc}{\Delta E}\) = \(\frac{6.6 \times 10^{-34} \times 3.0 \times 10^8}{(-1.51 - ( - 13.60)) \times 1.6 \times 10^{-19}}\) = 1.02 x 10\(^{-7}\)m (c) \(\lambda\) = \(\frac{0.693}{ T^{\frac{1}{2}}}\) = \(\frac{0.693}{5572}\) = 0.0001244 years\(^{-1}\) N = Noe\(^{- \lambda t}\) e\(^{0.0001244t}\) = 1.2632 0.0001244t = 0.234 t = 1878 years

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

S/N I/A V/V LogV/V LogI/A 1.0 0.10 0.30 -0.552 -1.000 2.0 0.15 0.50 -0.301 -0.823 3.0 0.20 0.70 -0.154 -0.698 4.0 0.25 0.80 -0.096 0.602 5.0 0.30 0.90 -0.045 0.522[ Slopes S = \(\frac{\bigtriangleup \text{logI}}{\bigtriangleup \text{logV}}\) = \(\frac{\text{logl}_{2}} {\text{logV}_{2}}\) - \(\frac{\text{logl}_{1}}{\text{logV}_{2}}\) S = \(\frac{-1.00 - (-0.70)}{-0.15 - (0.50)}\) = \(\frac{0.30}{0.10}\) S= 3 PRECUATIONS 1. avoided parallax error when reading ammeter and voltmeter 2. opened the key when readings are not being taken to ensure that the cells do not run down. 3. ensured that the jockey is not dragged along the potentiometer wire. (b)i. As x increases, the brightness of the bulb increases. As x increases, voltage and current through the bulb also increases, hence brightness increases ii. Electrical devices whose action does not obey ohm's law. They are: Diode, Capacitor, Transistor, Thermistor and Thyristor. SEE GRAPH ABOVE

S/N L/CM t/s T=t/o/S LogT/S LogL/cm 1 80.0 17.90 1.79 0.252 1.903 2 70.0 17.30 1.73 0.238 1.845 3 60.0 15.34 1.53 0.185 1.775 4 50.0 13.90 1.39 0.145 1.698 5 40.0 13.06 1.31 0.155 1.602 Slope (s) = \(\frac{\bigtriangleup \text{logT}_{2}}{\bigtriangleup\text{logL}_{2}}\) - \(\frac{\bigtriangleup \text{logT}_{1}}{\bigtriangleup\text{logL}_{1}}\) = \(\frac{0.15 - 0.05}{0.164 - 0.148}\) = \(\frac{0.100}{0.016}\) = 6.25 PRECAUTIONS 1. avoided zero error in the reciding of the stop watch 2. avoided Parallax in reading the metre rule and the stop watch 3. ensured that the support of the bifilar pendulum is rigid 4. ensured smooth and regular oscillations in a horizontal plane. b. i. Simple harmonic motion is defined as the motion of an object whose acceleration is proportional to the distance from a fixed point and is always directed towards that point see graph above