Physics 2025 Jamb Examination

Parallax is the apparent displacement of an object when viewed from two different positions. In optics experiments, specifically when using pins and mirrors, the 'method of no parallax' is used to locate an image. When the object (the real pin) and the image are exactly at the same position, they appear to move together without any relative displacement as the observer's head moves from side to side. If they move together, it indicates that parallax has been eliminated, meaning there is no parallax error.

This is the characteristic pattern produced by a straight conductor carrying electric current. The direction of these circular lines is determined by the Right-Hand Grip Rule, where the thumb points in the direction of the current and the fingers curl in the direction of the magnetic field lines.

Statement I is correct because colour is fundamentally determined by the frequency of the light wave. Statement II is incorrect; in the visible spectrum (ROYGBIV), deviation increases with frequency. Since green has a higher frequency than yellow, green is deviated more than yellow. Statement III is incorrect because rainbows require sunlight and rain droplets acting as prisms, but they don't specifically require 'heavy' rain—just the right optical conditions. Therefore, only statement I is scientifically accurate.

Using Newton's Second Law: \( F_{net} = ma \). The net force is the applied force (\( F_a \)) minus the frictional force (\( F_f \)). \[ F_a - F_f = ma \] \[ 18 - F_f = 40 \times 0.3 \] \[ 18 - F_f = 12 \implies F_f = 6 \, \text{N} \] The frictional force is also defined as \( F_f = \mu R \), where \( R = mg \) (assuming \( g = 10 \, \text{ms}^{-2} \)). \[ 6 = \mu (40 \times 10) \] \[ \mu = \frac{6}{400} = 0.015 \]

Electric power (P) is related to voltage (V) and current (I) by the formula: \[ P = IV \] Rearranging to solve for current: \[ I = \frac{P}{V} \] \[ I = \frac{500}{220} \approx 2.2727 \, \text{A} \]

According to Pascal's Principle, pressure is transmitted equally throughout a fluid. \[ P_1 = P_2 \implies \frac{F_1}{A_1} = \frac{F_2}{A_2} \] Since Area \( A = \pi r^2 \): \[ \frac{F_1}{\pi r_1^2} = \frac{F_2}{\pi r_2^2} \implies \frac{200}{r_1^2} = \frac{3200}{r_2^2} \] \[ \frac{r_1^2}{r_2^2} = \frac{200}{3200} = \frac{1}{16} \] Taking the square root of both sides: \[ \frac{r_1}{r_2} = \sqrt{\frac{1}{16}} = \frac{1}{4} \] So the ratio is 1:4.

Newton's Law of Universal Gravitation states that the force (F) is inversely proportional to the square of the distance (d) between the bodies: \( F \propto \frac{1}{d^2} \). If the distance is doubled (\( 2d \)), the new force (\( F' \)) becomes: \[ F' \propto \frac{1}{(2d)^2} = \frac{1}{4d^2} \] \[ F' = \frac{1}{4} F = \frac{20}{4} = 5 \, \text{N} \]

The cubic (volume) expansivity (\( \gamma \)) is thrice the linear expansivity (\( \alpha \)): \( \gamma = 3\alpha \). \[ \gamma = 3 \times (2.0 \times 10^{-5}) = 6.0 \times 10^{-5} \, \text{K}^{-1} \] Using the formula for volume expansion: \( \Delta V = V_0 \gamma \theta \). \[ 0.0018 = 1 \times (6.0 \times 10^{-5}) \times \theta \] \[ \theta = \frac{0.0018}{6.0 \times 10^{-5}} = \frac{1.8 \times 10^{-3}}{6.0 \times 10^{-5}} = 0.3 \times 10^2 = 30 \, ^\circ\text{C} \]

According to Boyle's Law, for a fixed mass of gas at constant temperature, the pressure (P) of the gas is inversely proportional to its volume (V). This is expressed as \( P \propto \frac{1}{V} \) or \( PV = k \).

Power is the rate of doing work, or Force multiplied by Velocity (\( P = Fv \)). In this case, the force required to lift the mass is its weight: \[ F = mg = 20 \times 10 = 200 \, \text{N} \] The power developed is: \[ P = 200 \, \text{N} \times 5 \, \text{ms}^{-1} = 1000 \, \text{W} \]