JEE Mains · Physics · STD 11 - Transmission of heat
- JEE Mains 2022
- Hard
- MCQ
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 \(10^{\circ}\,C\) and \(20^{\circ}\,C\). The thermal radiation emitted in a given time by \(P\) and \(Q\) are in the ratio \(1: 1.15\) \(C.\) A carnot Engine working between \(100\,K\) and \(400\,K\) 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 :
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) \(A, B, C \) only
Step-by-step Solution
Detailed explanation
Heat Transfer
A. by Newton's low of colling \(\frac{d \theta}{d t}=\propto \Delta T\)
B. \(H =\frac{d \theta}{d t}=\sigma e AT ^{4} \Rightarrow \frac{ H _{ P }}{ H _{Q}}=\left(\frac{ T _{ P }}{ T _{Q}}\right)^{4}=\left(\frac{283}{293}\right)^{4}\)
\(H _{ p }: H _{ Q }=1(1.03)^{4}=1:(1.03)^{4}=1: 1.15\) \(\Rightarrow B\) is correct
C. \(\eta=1-\frac{100}{400}=\frac{3}{4}=75 \%\)
D. is wrong as \(\frac{d \theta}{d t} \propto \Delta T\)
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