Question

Rods \( A \) and \( B \) have their lengths in the ratio \( 1 : 2 \). Their thermal conductivities are \( K_1 \) and \( K_2 \) respectively. The temperatures at the ends of each rod are \( T_1 \) and \( T_2 \). If the rate of flow of heat through the rods is equal, the ratio of area of cross section of \( A \) to that of \( B \) is

• A. \( \frac{2K_2}{K_1} \)
• B. \( \frac{K_2}{4K_1} \)
• C. \( \frac{K_2}{K_1} \)
• D. \( \frac{K_2}{2K_1} \)

Hint:

For equal heat flow through rods, use \( \frac{KA}{L} \) comparison when temperature difference is same.

Answer 1

The Correct Option is D

Step 1: Use heat conduction formula.
Rate of heat flow through a rod is:
\[ H = \frac{KA\Delta T}{L} \]

Step 2: Write heat flow for rod \( A \).
\[ H_A = \frac{K_1 A_A (T_1 - T_2)}{L_A} \]

Step 3: Write heat flow for rod \( B \).
\[ H_B = \frac{K_2 A_B (T_1 - T_2)}{L_B} \]

Step 4: Use condition of equal heat flow.
\[ H_A = H_B \]
\[ \frac{K_1 A_A (T_1 - T_2)}{L_A} = \frac{K_2 A_B (T_1 - T_2)}{L_B} \]

Step 5: Cancel common temperature difference.
\[ \frac{K_1 A_A}{L_A} = \frac{K_2 A_B}{L_B} \]

Step 6: Rearrange for area ratio.
\[ \frac{A_A}{A_B} = \frac{K_2 L_A}{K_1 L_B} \]

Step 7: Substitute length ratio.
Given:
\[ L_A : L_B = 1 : 2 \]
\[ \frac{A_A}{A_B} = \frac{K_2 \times 1}{K_1 \times 2} \]
\[ \boxed{\frac{A_A}{A_B} = \frac{K_2}{2K_1}} \]