A cylindrical rod is having temperatures $\theta_1$ and $\theta_2$ at its ends. The rate of heat flow is…

A cylindrical rod is having temperatures $\theta_1$ and $\theta_2$ at its ends. The rate of heat flow is $\mathrm{Q} / \mathrm{S}$. All the linear dimensions of the rod are doubled by keeping the temperature constant. The new rate of flow of heat is
  1. 4Q
  2. 2 Q
  3. $\frac{\mathrm{Q}}{2}$
  4. $\quad \frac{\mathrm{Q}}{4}$

Solution

$\begin{aligned}\left(\frac{Q}{t}\right) & =\frac{K \pi r^2\left(\theta_1-\theta_2\right)}{\Delta x} \propto \frac{r^2}{\Delta x} \\ \therefore \quad \frac{Q}{Q^{\prime}} & =\left(\frac{r_1}{r_2}\right)^2\left(\frac{\Delta x_2}{\Delta x_1}\right)=\left(\frac{1}{2}\right)^2 \times\left(\frac{2}{1}\right)=\frac{1}{2} \\ Q^{\prime} & =2 Q\end{aligned}$ .

Asked in: MHT CET 2024 (15 May Shift 2)

Practice more Kinetic Theory of Gases and Radiation questions on Aicharya