Question

The ends of a metallic rod are at temperatures \( T_1 \) and \( T_2 \), and the rate of flow of heat through it is \( Q \, {J s}^{-1} \). If all the dimensions of the rod are halved, keeping the end temperatures constant, the new rate of flow of heat will be:

• A. \( 2Q \)
• B. \( \frac{Q}{8} \)
• C. \( \frac{Q}{4} \)
• D. \( \frac{Q}{2} \)
• E. \( Q \)

Hint:

When the dimensions of the rod are halved, the heat transfer rate decreases as the length decreases and the area decreases proportionally.

Answer 1

The Correct Option is D

Step 1: The rate of heat transfer \( Q \) through a metallic rod is given by the formula: \[ Q = \frac{kA(T_1 - T_2)}{L} \] where: 
- \( k \) is the thermal conductivity of the material,
- \( A \) is the cross-sectional area,
- \( (T_1 - T_2) \) is the temperature difference across the ends of the rod,
- \( L \) is the length of the rod.
Step 2: If all dimensions of the rod are halved, the new length \( L' = \frac{L}{2} \) and the new cross-sectional area \( A' = \frac{A}{4} \). 
Step 3: The new rate of heat transfer \( Q' \) is: \[ Q' = \frac{kA'(T_1 - T_2)}{L'} \] Substitute the values for \( A' \) and \( L' \): \[ Q' = \frac{k \left(\frac{A}{4}\right)(T_1 - T_2)}{\frac{L}{2}} = \frac{kA(T_1 - T_2)}{2L} = \frac{Q}{2} \] Thus, the new rate of heat transfer is \( \frac{Q}{2} \).