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JEE PHYSICS MCQ: THERMAL PROPERTIES OF MATTER

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The two ends of a rod of length LL and a uniform cross-sectional area AA are kept at two temperatures T1T_1 and T2T_2 (T1>T2)\left(T_1 > T_2\right). The rate of heat transfer, dQdt\frac{d Q}{d t} through the rod in a steady state is given by:
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Correct option: C
kA(T1−T2)L\frac{{kA}\left({T}_1-{T}_2\right)}{{L}}
Option C (kA(T1−T2)L\frac{{kA}\left({T}_1-{T}_2\right)}{{L}}) is correct because it represents the rate of heat transfer through a rod in steady state according to Fourier's law of heat conduction.

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