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JEE Main 2004 (AIEEE) · Heat Conduction

Heat Conduction question from JEE Main 2004 (AIEEE)

Q54JEE Main 2004 · AIEEEMedium

Heat Conduction

The temperature of the two outer surfaces of a composite slab, consisting of two materials having coefficients of thermal conductivity KK and 2K2K and thickness xx and 4x,4x, respectively, are T2T_2 and T1 (T2>T1).T_1\ (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, with ff equal to

Figure

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Answer: (d)

  1. Slabs in series: thermal resistances add, R=xKA+4x2KA=3xKAR = \dfrac{x}{KA} + \dfrac{4x}{2KA} = \dfrac{3x}{KA}
  2. H=T2−T1R=A(T2−T1)Kx⋅13H = \dfrac{T_2-T_1}{R} = \dfrac{A(T_2-T_1)K}{x}\cdot\dfrac{1}{3}
  3. So the answer is (d) f=13f = \dfrac{1}{3}.
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