Physics 2025 Jamb Examination

Given:
L = 0.5H, R = 45\(\Omega\), f = 50Hz\[ 
X_L = 2\pi f L = 2 \times \frac{22}{7} \times 50 \times 0.5 = \frac{1100}{7} \approx 157.14\Omega \] Impedance (Z):\[ 
Z = \sqrt{R^2 + X_L^2} \] \[ 
Z = \sqrt{(45)^2 + (157.14)^2} \] \[ 
Z = \sqrt{2025 + 24692.98} \approx \sqrt{26717.98} \] \[ 
Z \approx 163.46\Omega \]

Pressure is defined as Force divided by Area. The weight of the water acts as the force (F = 5000 N). The area (A) is the base of the reservoir: \[A = \text{length} \times \text{breadth} = 10 \times 5 = 50\, \text{m}^2\] \[P = \frac{F}{A} = \frac{5000}{50} = 100\, \text{Pa}\]

A standing wave (or stationary wave) is formed when two waves of the same frequency and amplitude travel in opposite directions and interfere. This typically happens when a wave is reflected back from a boundary and meets the incident wave. The result is a pattern of stationary nodes (zero amplitude) and antinodes (maximum amplitude).

T x 80 + 15 x 10 = W x 50

80T + 150 = 50W - - -- - - - - -(1)

T + 15 = W  - - - - - - - - - - - (2)

80T + 150 = 50(T + 15)

80T + 150 = 50T + 750

30T = 750 - 150

30T = 600

T = 20N

The closest option is 19.2N

The refractive index from air to glass is given by Snell's law:\[ {}^a\eta_g = \frac{\sin i}{\sin r} \] To find the angle of refraction:\[ \sin r = \frac{\sin i}{{}^a\eta_g} \] Substituting the given values:\[ \sin r = \frac{0.5}{1.5} \]\[ \sin r = 0.3333... \]\[ \sin r \approx 0.3 \]

The speed of sound in an ideal gas is given by the formula \(v = \sqrt{\frac{\gamma P}{\rho}}\). According to Boyle's Law, at a constant temperature, if the pressure (P) increases, the density (\(\rho\)) increases proportionally. Because the ratio \(P/\rho\) remains constant at a fixed temperature, the speed of sound remains unchanged by variations in pressure.

A photovoltaic cell (solar cell) is made from semiconductor materials, typically silicon. When photons of light hit the semiconductor, they knock electrons loose from their atoms. If the semiconductor is structured with a p-n junction, these electrons flow through the material to produce an electric current.

For a concave mirror to produce an erect image, the image must be virtual. This only occurs when the object is placed between the focal point (F) and the pole (P). Since the focal length is 20 cm, the object distance must be less than 20 cm.

To find the resultant, we resolve forces into horizontal and vertical components. Net horizontal (\(F_x\)) = \(8\, \text{N} - 4\, \text{N} = 4\, \text{N}\). Net vertical (\(F_y\)) = \(15\, \text{N} - 12\, \text{N} = 3\, \text{N}\). Using Pythagoras: \(R = \sqrt{F_x^2 + F_y^2} = \sqrt{4^2 + 3^2} = \sqrt{25} = 5\, \text{N}\).

The capacitance is \(C = \frac{\epsilon_0 A}{d}\). First, calculate the area \(A = \pi r^2 = 3.142 \times (0.06)^2 \approx 0.01131 \, \text{m}^2\). The separation \(d = 0.0012 \, \text{m}\). \[ C = \frac{8.85 \times 10^{-12} \times 0.01131}{0.0012} \approx 8.34 \times 10^{-11} \, \text{F} \]