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MHT CET 2025 (19 Apr, Shift 2) · Heat Conduction

Heat Conduction question from MHT CET 2025 (19 Apr, Shift 2)

Q14MHT CET 2025 · 19 APR · SHIFT 2Medium

Heat Conduction

A composite slab consists of two materials having coefficients of thermal conductivity K and 2K2K, thickness xx and 4x4x respectively. The temperatures of two outer surfaces of a composite slab are T2T_2 and T1T_1 respectively (T2>T1)(T_2 > T_1). The rate of heat transfer through the slab in a steady state is [A(T2−T1)Kx]f\left[\frac{A(T_2-T_1)K}{x}\right] f, where ff is equal to

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Answer: (d)

  1. Series resistances: R=xKA+4x2KA=3xKAR = \frac{x}{KA} + \frac{4x}{2KA} = \frac{3x}{KA}
  2. H=T2−T1R=A(T2−T1)Kx⋅13H = \frac{T_2-T_1}{R} = \frac{A(T_2-T_1)K}{x}\cdot\frac{1}{3}
  3. So f=13f = \frac{1}{3}, answer (d).
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