Medium MCQ +4 / -1 PYQ · JEE Mains 2022

Read the following statements :

A. When small temperature difference between a liquid and its surrounding is doubled, the rate of loss of heat of the liquid becomes twice.

B. Two bodies $P$ and $Q$ having equal surface areas are maintained at temperature $10^{\circ} \mathrm{C}$ and $20^{\circ} \mathrm{C}$. The thermal radiation emitted in a given time by $\mathrm{P}$ and $\mathrm{Q}$ are in the ratio $1: 1.15$.

C. A Carnot Engine working between $100 \mathrm{~K}$ and $400 \mathrm{~K}$ has an efficiency of $75 \%$.

D. When small temperature difference between a liquid and its surrounding is quadrupled, the rate of loss of heat of the liquid becomes twice.

Choose the correct answer from the options given below :

  1. A A, B, C only Correct answer
  2. B A, B only
  3. C A, C only
  4. D B, C, D only

Solution

<p>From Newton's cooling law ${{dQ} \over {dt}} = - k(T - {T_s})$ the statement A is correct.</p> <p>For B</p> <p>$U = \sigma eA{T^4}$</p> <p>So, $${{{U_1}} \over {{U_2}}} = {\left( {{{283} \over {293}}} \right)^4} \simeq {1 \over {1.15}}$$</p> <p>Statement B is correct</p> <p>For C</p> <p>$\eta = 1 - {{{T_1}} \over {{T_2}}} = 1 - {{100} \over {400}} = {3 \over 4}$</p> <p>So, efficiency is 75% C is correct</p> <p>For D</p> <p>From Newton's law of cooling ${{dQ} \over {dt}} = - k(T - {T_s})$</p> <p>The statement is wrong.</p>

About this question

Subject: Physics · Chapter: Thermodynamics · Topic: Zeroth and First Law

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