The two ends of a rod of length ' x ' and uniform cross-sectional area ' A ' are kept at temperatures ' T₁ ' and ' T₂ ' respectively ( T₁ > T₂). If the rate of heat transfer is ' Q/t ', through the rod in steady state, then the coefficient of thermal conductivity ' K ' is (a) (AQ)/(tx(T₁-T₂)) (b) (xQ)/(tA(T₁-T₂)) (c) (xAQ)/(t(T₁-T₂)) (d) (Q)/(txA(T₁-T₂))

Answer: (b). Fourier's law: Q/t = (KA(T₁-T₂))/(x) Rearranging: K = (xQ)/(tA(T₁-T₂)) So the answer is (b).

The two ends of a rod of length ' x ' and uniform cross-sectional… | MHT CET 2025 (22 Apr, Shift 2)

Q12MHT CET 2025 · 22 APR · SHIFT 2Easy

Heat Conduction

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

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Answer: (b)

Fourier's law: $\frac{Q}{t} = \frac{KA(T_1-T_2)}{x}$
Rearranging: $K = \frac{xQ}{tA(T_1-T_2)}$
So the answer is (b).

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Heat Conduction question from MHT CET 2025 (22 Apr, Shift 2)

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