Rate of flow of heat through a cylindrical rod is ' $\mathrm{H}_1$ '. The temperature at the ends of the rod…

Rate of flow of heat through a cylindrical rod is ' $\mathrm{H}_1$ '. The temperature at the ends of the rod are ' $T_1$ ' and ' $T_2$ '. If all the dimensions of the rod become double and the temperature difference remains the same, the rate of flow of heat becomes ' $\mathrm{H}_2$ '. Then
  1. $\mathrm{H}_2=4 \mathrm{H}_1$
  2. $\mathrm{H}_2=2 \mathrm{H}_1$
  3. $\mathrm{H}_2=\frac{\mathrm{H}_1}{2}$
  4. $\mathrm{H}_2=\frac{\mathrm{H}_1}{4}$

Solution

Let $l_1$ be the initial length of the rod and $\mathrm{r}_1$ be the radius of the rod. Then. $\mathrm{H}_1=\frac{\mathrm{kA}_1\left(\mathrm{~T}_2-\mathrm{T}_1\right)}{l_1}$
After doubling the dimensions, $\mathrm{H}_2=\frac{\mathrm{kA}_2 \cdot\left(\mathrm{~T}_2-\mathrm{T}_1\right)}{l_2}$ $\begin{array}{ll}\therefore \quad & \frac{\mathrm{H}_2}{\mathrm{H}_1}=\frac{\mathrm{A}_2}{\mathrm{~A}_1} \times \frac{l_1}{l_2} \\ & \text { If } \mathrm{r}_2=2 \mathrm{r}_1, \text { then } \mathrm{A}_2=4 \mathrm{~A}_1 \\ & \text { Also, } l_2=2 l_1 \\ \therefore \quad & \frac{\mathrm{H}_2}{\mathrm{H}_1}=4 \times \frac{1}{2}=2 \\ \therefore \quad & \mathrm{H}_2=2 \mathrm{H}_1\end{array}$

Asked in: MHT CET 2024 (02 May Shift 2)

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