MHT CET · Physics · Thermal Properties of Matter

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
B 2 Q
C \(\frac{\mathrm{Q}}{2}\)
D \(\quad \frac{\mathrm{Q}}{4}\)

Verified Solution

Answer & Solution

Correct Answer

(B) 2 Q

Step-by-step Solution

Detailed explanation

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

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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