JEE Physics Thermal Properties Of Matter MCQs Set 06

Answer: (b) 30 g

Explanation:
1. Heat given by water = \(200 \times 1 \times 15 = 3000\) cal.
2. Each gram of ice absorbs \(0.5 \times 20 = 10\) cal to reach \(0^\circ\)C, 80 cal to melt, and \(1 \times 10 = 10\) cal to reach \(10^\circ\)C, a total of 100 cal.
3. \(m = 3000/100 = 30\) g.

Teacher's Note:
1. In calorimetry write the heat gained by every stage of the cold body and equate to heat lost by the hot body.

Answer: (c) B and D cannot be in thermal equilibrium

Explanation:
1. Suppose B and D were in thermal equilibrium. Then A with B, B with D and D with C would all be in equilibrium.
2. By the zeroth law, A would then be in equilibrium with C, which contradicts the given data.
3. So B and D cannot be in thermal equilibrium.

Teacher's Note:
1. The zeroth law is transitive and can be chained through any number of bodies.

Answer: (c) material of the body

Explanation:
1. The net rate is \(\epsilon \sigma A (T^4 - T_0^4)\).
2. It depends on the temperatures and on the emissivity, which is a property of the surface.
3. It does not depend on the material of the body as such.

Teacher's Note:
1. Surface condition and not bulk material decides radiative exchange.

Answer: (b) \((\epsilon_P/\epsilon_Q)^{1/4}\,\theta_P\)

Explanation:
1. Power \(= \epsilon \sigma A T^4\), so \(\epsilon_P \theta_P^4 = \epsilon_Q \theta_Q^4\).
2. \(\theta_Q^4 = (\epsilon_P/\epsilon_Q)\,\theta_P^4\).
3. \(\theta_Q = (\epsilon_P/\epsilon_Q)^{1/4}\theta_P\).

Teacher's Note:
1. Equate powers and take the fourth root.

Answer: (b) 25/3

Explanation:
1. Use Newton's law with mean temperature: \(\dfrac{50-40}{5} = k(45 - 20) = 25k\), so \(k = 0.08\) per minute.
2. For \(40 \to 30\): \(\dfrac{10}{t} = k(35 - 20) = 15k = 1.2\).
3. \(t = 10/1.2 = 25/3\) minutes.

Teacher's Note:
1. The mean-temperature form of Newton's law is accurate for small temperature drops and is the standard exam method.

Answer: (a) \(k_1 > k_2\)

Explanation:
1. In steady state the heat current \(kA\,(dT/dx)\) is the same in both rods.
2. So the rod with the steeper temperature gradient has the smaller conductivity.
3. The second rod is steeper, so \(k_2 < k_1\).

Teacher's Note:
1. Good conductors have flat temperature profiles, poor conductors have steep ones.

Answer: (c) 1/32

Explanation:
1. The maximum spectral energy density of a black body is proportional to \(T^5\).
2. \(T_1 = 1000\) K and \(T_2 = 2000\) K, so \(E_m/E_{2m} = (1000/2000)^5 = 1/32\).

Teacher's Note:
1. Wien's displacement law gives \(\lambda_m T = \text{const}\), and the peak intensity scales as \(T^5\).

Answer: (d) \(15^\circ\)C

Explanation:
1. In steady state the heat flow per unit area through both plates is the same: \(3k(\theta - 0)/d = k(60 - \theta)/d\).
2. \(3\theta = 60 - \theta\), so \(\theta = 15^\circ\)C.

Teacher's Note:
1. The poorer conductor carries the larger temperature drop.

Answer: (b) \(l_1/l_2 = \alpha_2/\alpha_1\)

Explanation:
1. The difference of lengths stays constant if both lengths expand by the same amount.
2. \(l_1\alpha_1\Delta T = l_2\alpha_2\Delta T\), so \(l_1\alpha_1 = l_2\alpha_2\).
3. Hence \(l_1/l_2 = \alpha_2/\alpha_1\).

Teacher's Note:
1. Equal absolute expansion keeps the difference constant, which means \(l\alpha\) must be equal.

Answer: (d) 5 s slow

Explanation:
1. The period \(T \propto \sqrt{l}\), so \(\Delta T/T = \tfrac{1}{2}\alpha \Delta\theta = \tfrac{1}{2} \times 10^{-6} \times 10 = 5 \times 10^{-6}\).
2. The period increases, so the clock runs slow.
3. Time lost in \(10^6\) s \(= 5 \times 10^{-6} \times 10^6 = 5\) s.

Teacher's Note:
1. Heating lengthens the pendulum, so the clock loses time.