Physics 2018 Jamb Examination

Power is defined as the rate of doing work. In linear motion, work is the product of force and displacement. Therefore, power can be expressed as:\[ P = \frac{\text{Work}}{\text{Time}} = \frac{F \times s}{t} \]Since velocity \( V \) is displacement \( s \) divided by time \( t \), we can substitute to get:\[ P = F \times V \]To find the force provided by the engine, we rearrange the formula to isolate \( F \):\[ F = \frac{P}{V} \]

Pressure is defined as Force per unit Area. To find its dimensions, we break it down into fundamental units of Mass (M), Length (L), and Time (T). \[ \text{Pressure} = \frac{\text{Force}}{\text{Area}} \]Force has dimensions of mass \( \times \) acceleration:\[ F = M \times LT^{-2} = MLT^{-2} \]Area has dimensions of length squared:\[ A = L^{2} \]Therefore, Pressure is:\[ P = \frac{MLT^{-2}}{L^{2}} = ML^{1-2}T^{-2} = ML^{-1}T^{-2} \]

Viscosity is the internal friction or resistance of a fluid to flow. For most liquids, viscosity decreases as temperature increases. When palm oil is heated, the kinetic energy of its molecules increases, which helps to overcome the attractive intermolecular forces that keep the liquid 'thick'. Consequently, the oil becomes thinner (less viscous) and flows more readily out of the bottle.

Impulse is defined as the change in momentum (\( mv - mu \)). When an object hits a wall 'normally' (perpendicularly) and rebounds, its velocity changes direction. In a perfectly elastic or standard physics problem context without a specified rebound speed, we assume it rebounds with the same speed in the opposite direction for maximum impulse:\[ \text{Impulse} = m(v - u) \]Let initial velocity \( u = 2 \; \text{ms}^{-1} \) and final velocity \( v = -2 \; \text{ms}^{-1} \):\[ \text{Impulse} = 5.0(-2 - 2) = 5.0(-4) = -20 \; \text{kgms}^{-1} \]The magnitude is 20.0 \( \text{kgms}^{-1} \). If the question implies the ball stops or just refers to the momentum of the strike, the value would differ, but for a standard rebound, 20 is the physical result.

Efficiency is generally defined as the ratio of useful output energy to total input energy.

In the context of energy flow through stages (where E1 might be potential energy, E2 kinetic energy, and E3 electrical energy), the efficiency of a specific stage or the whole process is the final desired energy divided by the initial energy source. Usually, for a hydroelectric system, if E2 is the output and E1 is the input, efficiency is E2/E1. Given the options, B or A are common representations depending on which labels represent input and output.

First, find the upthrust in water (\( U_w \)):\[ U_w = \text{Weight in air} - \text{Weight in water} = 30 - 21 = 9\text{N} \]Upthrust is proportional to the density of the liquid. The upthrust in the new liquid (\( U_l \)) is:\[ U_l = \text{Relative Density} \times U_w = 1.4 \times 9 = 12.6\text{N} \]The weight in the liquid is the weight in air minus the new upthrust:\[ \text{Weight in liquid} = 30 - 12.6 = 17.4\text{N} \]

For measuring very small dimensions accurately (in the range of millimeters), the micrometer screw gauge is the most suitable tool. It has a much higher precision (usually 0.01 mm) compared to a vernier calliper (0.1 mm) or a metre rule (1 mm), making it ideal for narrow pipes or wires.

Energy loss in transmission lines occurs primarily due to the resistance of the wires, which converts electrical energy into heat. This is known as Joule heating, calculated as \( P_{loss} = I^{2}R \). By transmitting at high voltage, the current (\( I \)) is reduced for the same amount of power. Since power loss depends on the square of the current, reducing the current significantly decreases the energy lost as heat.

We use the lens formula:\[ \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \]Where \( f = 20 \; \text{cm} \) and the object distance \( u = 100 \; \text{cm} \). We need to find the image distance \( v \) (distance from lens to film):\[ \frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{20} - \frac{1}{100} \]\[ \frac{1}{v} = \frac{5 - 1}{100} = \frac{4}{100} \]\[ v = \frac{100}{4} = 25 \; \text{cm} \]

In a mirror, magnification \( M = |v/u| \). For a convex mirror, the image is virtual and behind the mirror, so we use the sign convention \( v = -6 \; \text{cm} \).
From magnification:\[ 0.6 = \frac{-(-6)}{u} \implies 0.6 = \frac{6}{u} \implies u = 10 \; \text{cm} \]Now use the mirror formula:\[ \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \]\[ \frac{1}{f} = \frac{1}{10} + \frac{1}{-6} = \frac{3 - 5}{30} = \frac{-2}{30} \]\[ f = -15 \; \text{cm} \]The negative sign confirms it is a diverging (convex) mirror.