Physics 2025 Jamb Examination

This problem is solved using the principle of conservation of energy (Heat lost = Heat gained). "Boiled water" is at \(100^\circ\text{C}\). Let the mass of boiled water be \(m_b\). \[ \text{Heat lost by boiled water} = m_b \times c \times (100 - 60) \] \[ \text{Heat gained by cool water} = 8 \times c \times (60 - 25) \] Since the specific heat capacity (\(c\)) is the same for both, they cancel out: \[ m_b(40) = 8(35) \] \[ 40m_b = 280 \] \[ m_b = \frac{280}{40} = 7\, \text{kg} \]

Thermal conductivity is a measure of a material's ability to conduct heat. Generally, gases are the poorest conductors (best insulators) because their molecules are far apart, making energy transfer through collisions inefficient. Air is an excellent insulator with a very low thermal conductivity compared to liquids like water or solids like glass and ash.

Total heat lost involves two stages: condensation of steam at \(100^\circ\text{C}\) and cooling of the resulting water to \(15^\circ\text{C}\). \[ Q = mL_v + mc_w\Delta\theta \] Given \(Q = 5400\, \text{kJ} = 5.4 \times 10^6\, \text{J}\): \[ 5,400,000 = m(2.26 \times 10^6) + m(4200 \times (100 - 15)) \] \[ 5,400,000 = m(2,260,000 + 357,000) \] \[ 5,400,000 = m(2,617,000) \] \[ m = \frac{5,400,000}{2,617,000} \approx 2.063 \, \text{kg} \]

In semiconductors, electricity is conducted by two types of charge carriers. Doping creates an abundance of either free electrons (in n-type) or 'holes' (in p-type). A hole is effectively the absence of an electron in the valence band, which behaves like a positive charge carrier. Unlike in metals (where only electrons flow) or electrolytes (where ions flow), semiconductors rely on the movement of both electrons and holes.

The relationship between the refractive index (\(n\)) of a medium and its critical angle (\(C\)) is given by the formula: \[ n = \frac{1}{\sin C} \] Given \(C = 45^\circ\): \[ n = \frac{1}{\sin 45^\circ} \] Since \(\sin 45^\circ = \frac{1}{\sqrt{2}}\): \[ n = \frac{1}{1/\sqrt{2}} = \sqrt{2} \]

To solve this, we use the principle of moments about the pivot (the knife edge at 10 cm). The uniform metre rule has its weight (100 N) acting at its center of gravity, which is the 50 cm mark. 1. Distance of weight from pivot = 50 cm - 10 cm = 40 cm. 2. Distance of tension (T) from pivot = 85 cm - 10 cm = 75 cm. Sum of clockwise moments = Sum of anticlockwise moments: \[ 100 \times 40 = T \times 75 \] \[ 4000 = 75T \] \[ T = \frac{4000}{75} \approx 53.33 \text{ N} \]

Food contains energy stored in the chemical bonds of its molecules (carbohydrates, fats, and proteins). When we digest food, these bonds are broken through metabolic processes, releasing the energy our bodies need to function. This stored energy is a specific form of potential energy known as chemical energy.

The energy (E) stored in an inductor is given by the formula: \[ E = \frac{1}{2}LI^2 \] Where \(E = 2.5 \text{ J}\) and \(L = 3 \text{ H}\). \[ 2.5 = \frac{1}{2} \times 3 \times I^2 \] \[ 5.0 = 3I^2 \] \[ I^2 = \frac{5}{3} \approx 1.666... \] \[ I = \sqrt{1.666} \approx 1.29 \text{ A} \]

Efficiency is defined as the ratio of useful work output to total work input, expressed as a percentage. \[ \text{Efficiency} = \frac{\text{Work Output}}{\text{Work Input}} \times 100 \] \[ 80 = \frac{\text{Work Output}}{200} \times 100 \] \[ 0.8 = \frac{\text{Work Output}}{200} \] \[ \text{Work Output} = 0.8 \times 200 = 160 \text{ J} \]

According to the Law of Floatation, the fraction of the volume immersed is the ratio of the density of the floating object to the density of the liquid. First, find the density of the wood: \( \text{Relative Density} \times \text{Density of water} = 0.4 \times 1000 = 400 \text{ kg m}^{-3} \). \[ \text{Fraction immersed} = \frac{\text{Density of object}}{\text{Density of liquid}} \] \[ \text{Fraction} = \frac{400}{1600} = \frac{1}{4} = 0.25 \]