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MHT CET 2026 (17 Apr, Shift 2) · Heat Conduction

Heat Conduction question from MHT CET 2026 (17 Apr, Shift 2)

Q8MHT CET 2026 · 17 APR · SHIFT 2Medium

Heat Conduction

In a composite slab there are two materials having coefficients of thermal conductivity K and 2K, thickness x and 4x repectively. The temperature of the two outer surfaces of a composite slab are T2T_2 and T1 (T2>T1). T2T_1\ (T_2 > T_1).\ T_2 is on side K and T1T_1 is on side 2K. 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] \cdot f, where f 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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