Physics 2023 Jamb Examination

Equilibrium is categorized based on how an object reacts to a small displacement:
1. Stable Equilibrium (A): The ball is at the bottom of a curve. If displaced, its center of gravity rises, and it tends to return to its original position.
2. Unstable Equilibrium (B): The ball is balanced on a peak. If displaced, its center of gravity lowers, and it moves further away from the original position.
3. Neutral Equilibrium (C): The ball is on a flat plane. If displaced, the height of its center of gravity remains constant, and it stays in its new position.

When sunlight enters a raindrop, it undergoes refraction as it slows down. This causes the light to split into its component colors, a process known as dispersion. The light then hits the back of the raindrop and undergoes total internal reflection, finally refracting again as it exits the drop. While 'reflection' is part of the process, specifically total internal reflection, dispersion and refraction are the primary drivers of the visual effect.

The rate of thermal conduction is calculated using the formula:\[ \frac{Q}{t} = \frac{kA(\theta_2 - \theta_1)}{L} \]Given:
\( k = 0.6071 \), \( A = 620 \), \( \Delta\theta = 25 - 15 = 10 \), \( L = 1.5 \)\[ \text{Rate} = \frac{0.6071 \times 620 \times 10}{1.5} \] \[ \text{Rate} = \frac{3764.02}{1.5} \approx 2509 \, \text{W} \approx 2.5 \, \text{kW} \]

Stars, including our Sun, generate energy through nuclear fusion. This is a process where light atomic nuclei (typically Hydrogen) are squeezed together under extreme pressure and temperature to form a heavier nucleus (Helium), releasing a massive amount of energy in the process.

According to the Work-Energy Theorem, net work done is equal to the change in kinetic energy:\[ W = \Delta K.E. = \frac{1}{2}m(v^2 - u^2) \] \[ W = \frac{1}{2}(1200)(15^2 - 10^2) \] \[ W = 600(225 - 100) = 600 \times 125 = 75,000 \, \text{J} \] \[ W = 7.5 \times 10^4 \, \text{J} \]

Young's modulus \( E = \frac{\text{Stress}}{\text{Strain}} \)
1. Force (\( F \)) = \( mg = 1.5 \times 10 = 15 \, \text{N} \)
2. Area (\( A \)) = \( \pi r^2 = \pi (0.0002)^2 \approx 1.2566 \times 10^{-7} \, \text{m}^2 \)
3. Stress = \( F/A = 15 / 1.2566 \times 10^{-7} \approx 1.193 \times 10^8 \, \text{Pa} \)
4. Strain = \( 0.5\% = 0.005 \)
5. \( E = (1.193 \times 10^8) / 0.005 \approx 2.387 \times 10^{10} \, \text{Nm}^{-2} \).

Tension provides the centripetal force: \( T = m\omega^2r \)
1. Convert rev/min to rad/s (\( \omega \)):\[ \omega = 120 \times \frac{2\pi}{60} = 4\pi \, \text{rad/s} \approx 12.566 \, \text{rad/s} \]2. Calculate tension:\[ T = 2 \times (4\pi)^2 \times 1.2 = 2.4 \times 16\pi^2 \approx 2.4 \times 157.9 \approx 379 \, \text{N} \]

1. Velocity Ratio (V.R.) = \( \frac{\text{Distance moved by Effort}}{\text{Distance moved by Load}} = \frac{100}{10} = 10 \) (Note: Effort arm is 110 - 10 = 100m).
2. M.A. = \( \text{Efficiency} \times V.R. = 0.80 \times 10 = 8 \)
3. M.A. = \( \text{Load} / \text{Effort} \)
4. \( 8 = (200 \times 10) / E \Rightarrow E = 2000 / 8 = 250 \, \text{N} \).

Using Boyle's Law (\( P_1V_1 = P_2V_2 \)):
1. \( P_1 = 10.34 + 12 = 22.34 \, \text{m of water} \)
2. \( P_2 = 10.34 \, \text{m of water} \)
3. \( V \propto r^3 \), so \( P_1r_1^3 = P_2r_2^3 \)\[ 22.34 \times (4.5)^3 = 10.34 \times r_2^3 \] \[ r_2^3 = \frac{22.34}{10.34} \times 4.5^3 \approx 2.16 \times 91.125 \approx 196.8 \] \[ r_2 = \sqrt[3]{196.8} \approx 5.82 \, \text{cm} \]

1. Capacitance (\( C \)) = \( \varepsilon_0 A / d \)\[ C = (8.85 \times 10^{-12} \times 0.8) / 0.02 = 3.54 \times 10^{-10} \, \text{F} \]2. Charge (\( Q \)) = \( CV \)\[ Q = (3.54 \times 10^{-10}) \times 120 = 4.248 \times 10^{-8} \, \text{C} = 42.48 \, \text{nC} \]