Question Library

Net force in the horizontal (x) direction: \(F_x = 8 \, \text{N} - 4 \, \text{N} = 4 \, \text{N} \quad \text{(to the right)}\) Net force in the vertical (y) direction: \(F_y = 15 \, \text{N} - 12 \, \text{N} = 3 \, \text{N} \quad \text{(3 N upward)}\) Magnitude of the resultant force: \(R = \sqrt{F_x^2 + F_y^2} = \sqrt{(4)^2 + (3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \, \text{N}\)

At constant pressure, volume ∝ absolute temperature (Charles’ law). \(\frac{V_1}{T_1} = \frac{V_2}{T_2}\) \(\frac{200}{273} = \frac{V_2}{546}\) v\(_2\) = \(\frac{200 \times 546}{273}\) = 400cm\(^3\)

The pressure exerted by the water at the bottom of the reservoir is given by: \(P = \frac{F}{A}\) Given: \( F = 5000 \, \text{N} \) (weight of water), \( L = 10 \, \text{m} \) (length), \( B = 5 \, \text{m} \) (breadth) Calculate the area \( A \): \(A = L \times B = 10 \, \text{m} \times 5 \, \text{m} = 50 \, \text{m}^2\) Calculate the pressure \( P \): \(P = \frac{F}{A} = \frac{5000 \, \text{N}}{50 \, \text{m}^2} = 100 \, \text{Pa}\)

Resistance thermometers respond faster because they have small sensor mass and use direct electrical detection. Liquid-in-glass and gas thermometers are slower due to thermal expansion and larger thermal inertia, often taking minutes to equilibrate.

Using Charles' law: \(\frac{V_1}{T_1} = \frac{V_2}{T_2} \quad \text{(T in Kelvin, constant P)}\) Calculate \( T_1 \): \(T_1 = 57 + 273 = 330 \, \text{K}\) Since \( V_2 = \frac{1}{5} V_1 \) (volume shrinks to one-fifth), we find \( T_2 \): \(T_2 = T_1 \times \frac{V_2}{V_1} = 330 \times \frac{1}{5} = 66 \, \text{K}\) In ^\circC, this is: \(T_2 = 66 - 273 = -207 \, ^\circC\)

Answer: D Given a rod of length \( L = 0.2 \, \text{m} \) with a temperature gradient of \( 416 \, \text{K/m} \) and one end at \( T_1 = 233^\circ \text{C} \): \(\Delta T = \text{gradient} \times L = 416 \, \text{K/m} \times 0.2 \, \text{m} = 83.2 \, \text{K}\) The temperature at the other end \( T_2 \) is: \(T_2 = T_1 + \Delta T = 233^\circ \text{C} + 83.2 \, \text{C} = 316.2^\circ \text{C}\) Thus, the temperature at the other end is \( 316.2^\circ \text{C}\).

Clouds are collections of condensed water droplets or ice crystals suspended in the air, but high above the ground (typically hundreds of meters to kilometers up). Fog is the same phenomenon (condensed suspended droplets, often with dust), but at ground level, reducing visibility. The question specifies "in the air" near the surface (implied by dust particles and suspension context), so fog is correct, not cloud.

Weight on Earth: \(W_e = 2.5 \times 10^3 \, \text{N}, \quad g_e = 10 \, \text{m/s}^2\) Mass: \(m = \frac{W_e}{g_e} = \frac{2500}{10} = 250 \, \text{kg}\) Moon's gravity is approximately \( \frac{1}{6} \) of Earth's: \(g_m = \frac{10}{6} \, \text{m/s}^2\) Weight on the Moon: \(W_m = m \times g_m = 250 \times \frac{10}{6} = \frac{2500}{6} \approx 416.67 \, \text{N}\) Thus, the weight on the Moon is: 416.67N

The gravitational force \( F \) between two masses is given by Newton's law of universal gravitation: \(F = G \frac{m_1 m_2}{r^2}\) where \( G \) is the gravitational constant, \( m_1 \) and \( m_2 \) are the masses (P and Q), and \( r \) is the distance between their centers. Originally, \( F = 10 \, \text{N} \). If both masses are doubled (\( m_1' = 2P \), \( m_2' = 2Q \)) and \( r \) remains constant, the new force is: \(F' = G \frac{(2P)(2Q)}{r^2} = G \frac{4PQ}{r^2} = 4 \left( G \frac{PQ}{r^2} \right) = 4 \times 10 = 40 \, \text{N}\)

B. Increases When a positively charged rod is brought near the cap of a positively charged gold-leaf electroscope (as indicated by the diverging leaves with + signs in the diagram), electrostatic induction occurs. The rod attracts negative charges toward the cap and repels additional positive charges toward the leaves. This increases the positive charge on the leaves, enhancing their mutual repulsion and thus increasing the divergence.

The neutral point around a magnet is where the magnetic field strength is zero, meaning no magnetic effects (like deflection of a compass needle) are observed, often due to cancellation by Earth's magnetic field.

Soft iron has high magnetic permeability, so it easily attracts and redirects magnetic field lines, shielding areas from external magnetic fields. Aluminum, brass, and copper do not have this property and are ineffective for magnetic shielding.

This process disrupts the alignment of magnetic domains, effectively demagnetizing the bar.

Answer: C An electric bell works using an electromagnet.

The capacitance \( C \) of a parallel-plate capacitor is calculated using the formula: \(C = \epsilon_0 \frac{A}{d}\) where: \(\epsilon_0 = 8.85 \times 10^{-12} \, \text{F/m},\) \(A = \pi r^2 \quad \text{(area of one plate)},\) and \( d \) is the separation between plates. \(r = 6 \, \text{cm} = 0.06 \, \text{m},\) \(d = 0.12 \, \text{cm} = 0.0012 \, \text{m}\) Calculate the area: \(A = \pi r^2 = 3.142 \times (0.06)^2 = 3.142 \times 0.0036 = 0.0113112 \, \text{m}^2\) Calculate the capacitance: \(C = 8.85 \times 10^{-12} \times \frac{0.0113112}{0.0012} = 8.85 \times 10^{-12} \times 9426 \approx 8.34 \times 10^{-11} \, \text{F}\)