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The quantity of heat needed to melt ice = mL (where m is the mass of ice and L is the specific latent heat of fusion of ice) Q = \(50 \times 336 = 16800J\) but this is 70% of the total heat supplied so that we have, 70% total heat = 16800J \(\frac{70}{100}\) of total heat = 16800J ⇒ Total heat supplied = \(\frac{16800\times100}{70} = 24000J\)

Velocity ratio = \(\frac{Effort distance}{Load distance}\) On an inclined plane, it is the inverse of \(sin\theta\) = \(\frac{1}{sin\theta}\)

When a bubble is deep under water, it has both atmospheric pressure and water pressure (giving it depth), acting on it but as it rises, the water pressure decreases. At this point, the basic pressre acting on the bubble is the atmospheric pressure. Recall that pressure is inversely proportional to volume (Boyles' law), hence, the less the pressure, the more the volume.

The solar cell converts light from the sun to electricity.

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

Latent heat is thermal energy released or absorbed, by a body or a thermodynamic system, during a constant-temperature process.

This is as a result of the Hydrogen bonding in water. Due to the high Kinetic Energy of water, the hydrogen bonds can't hold for too long but from 4^\circC, there's an increase in the hydrogen bonds hence strenghtening it. From the point of production of more hydrogen bonds, the bonding is no longer easily broken. Generally, these and a few factors based on these cause the density of water to be maximum at 4^\circC.

Diffraction refers to the various things that occur when a wave encounters an obstacle. This bending of the wave after it passes a small gap is an example of diffraction.

Electric intensity at a point is defined as the force experienced per unit positive charge at a point placed in the electric field.

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

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

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

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

(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 \(\frac{a}{b} = \frac{c}{d}; \frac{G\theta - (7.13) x 10^4}{(9.74 - 7.13) \times 10^4} = \frac{G - 273}{373 - 273}\) \(\frac{G\theta - 7.13}{2.16} = \frac{G - 273}{100}\)