Question

Rate of flow of heat through a cylindrical rod is \( H_1 \). The temperature of 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 ' \( H_2 \) '. Then \( H_2 = \)

• A. \( \frac{H_1}{2} \)
• B. \( 2H_1 \)
• C. \( \frac{H_1}{4} \)
• D. \( 4H_1 \)

Hint:

- Heat flow $\propto \frac{A}{L}$ - If dimensions double: $A \rightarrow 4A$, $L \rightarrow 2L$ - Net effect $\Rightarrow$ heat flow doubles

Answer 1

The Correct Option is B

Concept:
Rate of heat conduction through a rod is given by Fourier’s law: \[ H = \frac{kA (T_1 - T_2)}{L} \] where:
• $k$ = thermal conductivity (constant)
• $A$ = cross-sectional area
• $L$ = length of rod

Step 1: Write expression for initial heat flow.
\[ H_1 = \frac{kA (T_1 - T_2)}{L} \]

Step 2: Understand change in dimensions.
If all linear dimensions double:
• Length: $L \rightarrow 2L$
• Radius: $r \rightarrow 2r$ Area of cross-section: \[ A = \pi r^2 \Rightarrow A' = \pi (2r)^2 = 4A \]

Step 3: Write new heat flow expression.
\[ H_2 = \frac{k (4A)(T_1 - T_2)}{2L} \]

Step 4: Simplify.
\[ H_2 = \frac{4}{2} \cdot \frac{kA (T_1 - T_2)}{L} = 2H_1 \]

Step 5: Final relation.
\[ H_2 = 2H_1 \]