Question

The two ends of a rod of length ' \( x \) ' and uniform cross-sectional area ' \( A \) ' are kept at temperatures ' \( T_1 \) ' and ' \( T_2 \) ' respectively (\( T_1>T_2 \)). If the rate of heat transfer is ' \( Q/t \) ', through the rod in steady state, then the coefficient of thermal conductivity ' \( K \) ' is

• A. \( \frac{A Q}{tx(T_1-T_2)} \)
• B. \( \frac{x Q}{t A(T_1-T_2)} \)
• C. \( \frac{x A Q}{t(T_1-T_2)} \)
• D. \( \frac{Q}{tx A(T_1-T_2)} \)

Hint:

Fourier's Law of Heat Conduction: $Q = \frac{KA\Delta T t}{L}$.

Answer 1

The Correct Option is B

Step 1: Fundamental Formula
The rate of heat flow is $\frac{Q}{t} = \frac{KA(T_1 - T_2)}{x}$.
Step 2: Rearrange for K
$K = \frac{(Q/t) \cdot x}{A(T_1 - T_2)}$.
Step 3: Formatting
$K = \frac{xQ}{tA(T_1 - T_2)}$.
Final Answer: (B)