AIPMT PRELIMS2009Physics-Thermodynamics

AIPMT PRELIMS 2009 Physics Heat Transfer MCQ Question

Type: MCQ-conceptual-Medium-Class 11

The two ends of a rod of length L and a uniform cross-sectional area A are kept at two temperatures T1T_1 and T2T_2 (T1>T2T_1 > T_2). The rate of heat transfer, dQdt\frac{dQ}{dt}, through the rod in a steady state is given by:

A

k(T¹T2)LA\frac{k(T_¹⁻ T_2)}{LA}

B

kLΔ(T¹T2)kL \Delta (T_¹⁻ T_2)

C

kA(T¹T2)L\frac{kA(T_¹⁻ T_2)}{L}

D

kL(T¹T2)A\frac{kL(T_¹⁻ T_2)}{A}

Correct Answer

Option C

Detailed Explanation

The rate of heat transfer through a rod in a steady state is given by Fourier's law of heat conduction: dQdt=kA(T¹T2)L\frac{dQ}{dt} = \frac{kA(T_¹⁻ T_2)}{L}, where k is the thermal conductivity.

Found an issue with this question?

Related Questions