Q115 / 150 · MHT CET 2024 (15 May Shift 2) · Physics

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

  • A
    4Q
  • Correct answer
    B
    2 Q
  • C
    $\frac{\mathrm{Q}}{2}$
  • D
    $\quad \frac{\mathrm{Q}}{4}$

Solution

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

Hence, the correct answer is option: 'B'.

Relevant Questions