Q. The two ends of a rod of length $L$ and a uniform cross-sectional area $A$ are kept at two temperature $T_{1}$ and $T_{2}$ $(T_{1} > T_{2})$. The rate of heat transfer, $\frac{dQ}{dt}$, through the rod in a steady state is given by
A) $\frac{dQ}{dt} = \frac{kL(T_{1} - T_{2})}{A}$
B) $\frac{dQ}{dt} = \frac{k(T_{1} - T_{2})}{LA}$
C) $\frac{dQ}{dt} = kLA(T_{1} - T_{2})$
D) $\frac{dQ}{dt} = \frac{kA(T_{1} - T_{2})}{L}$
Explanation
The correct answer is Option D: $\frac{dQ}{dt} = \frac{kA(T_{1} - T_{2})}{L}$.
A detailed explanation for this question is being prepared.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.