Q. The two ends of a rod of length L and a uniform cross-sectional area A are kept at two temperatures T₁ and T₂ (T₁ > T₂). The rate of heat transfer, dQ/dt, through the rod in a steady state is given by:
A) dQ/dt = k(T₁ − T₂)/(LA)
B) dQ/dt = kLA(T₁ − T₂)
C) dQ/dt = kA(T₁ − T₂)/L
D) dQ/dt = kL(T₁ − T₂)/A
Explanation
[Solution from source] Fourier's law for steady conduction gives dQ/dt = kA(T₁ − T₂)/L: the heat current grows with area and temperature difference and falls with length. Forms with L in the numerator or A in the denominator have the wrong dependence. Answer (3).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.