A cylindrical rod has temperature ' $\mathrm{T}_1$ ' and ' $\mathrm{T}_2$ ' at its ends. The rate of flow of…

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
  1. $\mathrm{Q}_2=\frac{\mathrm{Q}_1}{2}$
  2. $\mathrm{Q}_2=\frac{\mathrm{Q}_1}{4}$
  3. $\mathrm{Q}_2=4 \mathrm{Q}_1$
  4. $\mathrm{Q}_2=2 \mathrm{Q}_1$

Solution

$\begin{aligned} & \mathrm{Q}_1=\frac{\mathrm{kA}\left(\mathrm{T}_1-\mathrm{T}_2\right)}{\mathrm{L}_1}=\frac{\mathrm{k} \pi \mathrm{r}_1^2\left(\mathrm{~T}_1-\mathrm{T}_2\right)}{\mathrm{L}_1} \\ & \mathrm{Q}_2=\frac{\mathrm{k} \pi \mathrm{r}_2^2\left(\mathrm{~T}_1-\mathrm{T}_2\right)}{\mathrm{L}_2} \\ & \frac{\mathrm{Q}_2}{\mathrm{Q}_1}=\frac{\mathrm{r}_2^2}{\mathrm{r}_1^2} \cdot \frac{\mathrm{L}_1}{\mathrm{~L}_2}=(2)^2 \cdot \frac{1}{2}=2 \\ & \therefore \mathrm{Q}_2=2 \mathrm{Q}_1\end{aligned}$

Asked in: MHT CET 2021 (24 Sep Shift 1)

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