MHT CET · Physics · Thermal Properties of Matter
A cylindrical rod has temperature ' \(\mathrm{T}_1\) ' and ' \(\mathrm{T}_2\) ' at its ends. The rate of flow of heat is ' \(\mathrm{Q}_1\) ' \(\mathrm{cal} \mathrm{s}^{-1}\). If length and radius of the rod are doubled keeping temperature constant, then the rate of flow of heat ' \(Q_2\) ' will be
- A \(\mathrm{Q}_2=\frac{\mathrm{Q}_1}{2}\)
- B \(\mathrm{Q}_2=\frac{\mathrm{Q}_1}{4}\)
- C \(\mathrm{Q}_2=4 \mathrm{Q}_1\)
- D \(\mathrm{Q}_2=2 \mathrm{Q}_1\)
Answer & Solution
Correct Answer
(D) \(\mathrm{Q}_2=2 \mathrm{Q}_1\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \mathrm{Q}_1=\frac{\mathrm{kA}\left(\mathrm{T}_1-\mathrm{T}_2\right)}{\mathrm{L}_1}=\frac{\mathrm{k} \pi \mathrm{r}_1^2\left(\mathrm{~T}_1-\mathrm{T}_2\right)}{\mathrm{L}_1} \\ & \mathrm{Q}_2=\frac{\mathrm{k} \pi \mathrm{r}_2^2\left(\mathrm{~T}_1-\mathrm{T}_2\right)}{\mathrm{L}_2} \\ & \frac{\mathrm{Q}_2}{\mathrm{Q}_1}=\frac{\mathrm{r}_2^2}{\mathrm{r}_1^2} \cdot \frac{\mathrm{L}_1}{\mathrm{~L}_2}=(2)^2 \cdot \frac{1}{2}=2 \\ & \therefore \mathrm{Q}_2=2 \mathrm{Q}_1\end{aligned}\)
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