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

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,
\(\mathrm{H}_2=\frac{\mathrm{kA}_2 \cdot\left(\mathrm{~T}_2-\mathrm{T}_1\right)}{l_2}\)
\(\begin{array}{ll}\therefore \quad & \frac{\mathrm{H}_2}{\mathrm{H}_1}=\frac{\mathrm{A}_2}{\mathrm{~A}_1} \times \frac{l_1}{l_2} \\ & \text { If } \mathrm{r}_2=2 \mathrm{r}_1, \text { then } \mathrm{A}_2=4 \mathrm{~A}_1 \\ & \text { Also, } l_2=2 l_1 \\ \therefore \quad & \frac{\mathrm{H}_2}{\mathrm{H}_1}=4 \times \frac{1}{2}=2 \\ \therefore \quad & \mathrm{H}_2=2 \mathrm{H}_1\end{array}\)