Question

Rate of flow of heat through a cylindrical rod is '\(H_1\)'. 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 and if the rate of flow of heat becomes '\(H_2\)' then \(H_2 =\)

• A. \(\frac{\text{H}_1}{2}\)
• B. \(4\text{H}_1\)
• C. \(2\text{H}_1\)
• D. \(\frac{\text{H}_1}{4}\)

Hint:

H is proportional to A / L for a fixed temperature difference.

Answer 1

The Correct Option is C

Step 1: Ratio method:
$\dfrac{H_2}{H_1} = \dfrac{A_2}{A_1}\cdot\dfrac{L_1}{L_2}$ at the same $\Delta T$ and $k$.

Step 2: Substitute:
$\dfrac{A_2}{A_1} = 2^2 = 4$ and $\dfrac{L_1}{L_2} = \dfrac12$.

Step 3: Result:
$\dfrac{H_2}{H_1} = 4\times\dfrac12 = 2$, so $H_2 = 2H_1$, option (C).

Final Answer:
The heat flow becomes 2 H1. \[ \boxed{\text{(C) }2H_1} \]