Physics 2025 Jamb Examination

Using the formula \(Q = mc\Delta\theta\): \[ c = \frac{Q}{m\Delta\theta} \] \[ c = \frac{1000}{0.025 \times 15} \] \[ c = \frac{1000}{0.375} = 2666.67 \, \text{Jkg}^{-1}\text{K}^{-1} \]

Assuming the speed of sound given in the question (60 m/s) is to be used (though the physical speed is usually ~340 m/s), distance is calculated as speed multiplied by time. \[ \text{Distance} = \text{Speed} \times \text{Time} = 60 \, \text{m/s} \times 4 \, \text{s} = 240 \, \text{m} \]

A gas thermometer is the most sensitive because gases have a high volume expansion coefficient and low heat capacity. This allows them to respond to very small changes in temperature more readily than liquid-in-glass or resistance thermometers.

Magnification (M) is given by \(M = \frac{\text{image distance (v)}}{\text{object distance (u)}}\). Given \(M = 1/5\) and \(v = 25 \, \text{cm} = 0.25 \, \text{m}\): \[ \frac{1}{5} = \frac{0.25}{u} \] \[ u = 0.25 \times 5 = 1.25 \, \text{m} \]

We use the de Broglie wavelength formula: \(\lambda = \frac{h}{mv}\). First, convert wavelength to metres: \(4.0 \times 10^{-7} \, \text{mm} = 4.0 \times 10^{-10} \, \text{m}\). \[ m = \frac{h}{\lambda v} \] \[ m = \frac{6.63 \times 10^{-34}}{(4.0 \times 10^{-10}) \times (2.7 \times 10^8)} \] \[ m = \frac{6.63 \times 10^{-34}}{1.08 \times 10^{-1}} \approx 6.14 \times 10^{-33} \, \text{kg} \] (Note: Based on typical question variations, double-check the wavelength power for specific option matching).

The electric field intensity (E) at a distance (r) from a point charge (q) is given by the formula: \[ E = \frac{kq}{r^2} \] Where \(k = 9.0 \times 10^9\, \text{Nm}^2\text{C}^{-2}\). Rearranging to solve for \(r^2\): \[ r^2 = \frac{kq}{E} \] \[ r^2 = \frac{(9.0 \times 10^9) \times (1.2 \times 10^{-7})}{4.8 \times 10^{-4}} \] \[ r^2 = \frac{1080}{4.8 \times 10^{-4}} = 225 \times 10^4 = 2,250,000 \] \[ r = \sqrt{2,250,000} = 1500\, \text{m} \] Since \(1000\, \text{m} = 1\, \text{km}\), the distance is 1.5 km.

In an oscillating simple pendulum, energy is continuously transformed between potential energy (PE) and kinetic energy (KE). At the maximum displacements (extremes P and S), the bob momentarily stops, so its KE is zero and its PE is at a maximum. As it moves toward the equilibrium position (Q), PE is converted into KE. At point Q, the bob is at its lowest vertical height (minimum PE) and traveling at its maximum speed. Therefore, the maximum kinetic energy is obtained at the mean position, Q.

Dust, smoke, and salt particles in the atmosphere act as 'condensation nuclei'. When water vapor in the air cools, it condenses around these suspended particles. When this happens at high altitudes, it forms clouds. When it occurs near the ground, it is called fog or mist. However, the term for the general collection of these condensed particles suspended in the atmosphere is a cloud.

Impulse (I) is equal to the change in momentum: \[ I = m(v - u) \] First, find the final velocity (v) using: \[ v = u + at \] \[ v = 4 + (5 \times 2) = 14\, \text{m/s} \] Now substitute into the impulse equation: \[ 60 = m(14 - 4) \] \[ 60 = 10m \] \[ m = 6\, \text{kg} \]

A 'bore' or hole in a material expands at the same rate as the material itself (as if the hole were filled with that material). Since the bore is given in units of volume (cm\(^{3}\)), we use the cubic expansivity (\(\gamma\)), where \(\gamma = 3\alpha\). \[ \gamma = 3 \times (24 \times 10^{-6}) = 72 \times 10^{-6}\, \text{K}^{-1} \] Temperature change \(\Delta\theta = 340 - 34 = 306\, ^\circ\text{C} \). New Volume \(V_2 = V_1(1 + \gamma\Delta\theta)\): \[ V_2 = 3.48(1 + (72 \times 10^{-6} \times 306)) \] \[ V_2 = 3.48(1 + 0.022032) \] \[ V_2 = 3.48 \times 1.022032 \approx 3.5566\, \text{cm}^3 \] Rounding gives 3.56 cm\(^{3}\).