MHT CET · Physics · Thermal Properties of Matter
A cylindrical rod is having temperatures \(\theta_1\) and \(\theta_2\) at its ends. The rate of heat flow is ' \(\mathrm{Q}\) ' \(\mathrm{J} \mathrm{s}{ }^{-1}\). All the linear dimensions of the rod are doubled by keeping the temperatures constant. What is the new rate of flow of heat?
- A \(\frac{Q}{2}\)
- B \(\frac{Q}{4}\)
- C \(2 Q\)
- D \(\frac{3 Q}{2}\)
Answer & Solution
Correct Answer
(C) \(2 Q\)
Step-by-step Solution
Detailed explanation
\(\mathrm{Q}=\frac{\mathrm{kA} \Delta \theta}{\ell}\)
If the radius of the cylindrical rod is doubled, then its area of cross-section will become four times.
\(
\begin{aligned}
& \therefore \mathrm{A}_2=4 \mathrm{~A}_1 \\
& \text { Also, } \ell_2=2 \ell_1 \\
& \therefore \frac{\mathrm{Q}^{\prime}}{\mathrm{Q}}=\frac{\mathrm{A}_2}{\mathrm{~A}_1} \cdot \frac{\mathrm{d}_1}{\mathrm{~d}_2}=4 \times \frac{1}{2}=2 \\
& \therefore \mathrm{Q}^{\prime}=2 \mathrm{Q}
\end{aligned}
\)
If the radius of the cylindrical rod is doubled, then its area of cross-section will become four times.
\(
\begin{aligned}
& \therefore \mathrm{A}_2=4 \mathrm{~A}_1 \\
& \text { Also, } \ell_2=2 \ell_1 \\
& \therefore \frac{\mathrm{Q}^{\prime}}{\mathrm{Q}}=\frac{\mathrm{A}_2}{\mathrm{~A}_1} \cdot \frac{\mathrm{d}_1}{\mathrm{~d}_2}=4 \times \frac{1}{2}=2 \\
& \therefore \mathrm{Q}^{\prime}=2 \mathrm{Q}
\end{aligned}
\)
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