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JEE Main 2020 (6 Sep, Shift 2) · Heat Conduction

Heat Conduction question from JEE Main 2020 (6 Sep, Shift 2)

Q43JEE Main 2020 · 6 SEP · SHIFT 2Medium

Heat Conduction

Three rods of identical cross-section and lengths are made of three different materials of thermal conductivity K1, K2 and K3, respectively. They are joined together at their ends to make a long rod (see figure). One end of the long rod is maintained at 100oC and the other at 0oC (see figure). If the joints of the rod are at 70oC and 20oC in steady state and there is no loss of energy from the surface of the rod, the correct relationship between K1, K2 and K3 is:

Figure

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Answer: (a)

  1. Same LL, AA and heat current: K1(100−70)=K2(70−20)=K3(20−0)K_1(100-70) = K_2(70-20) = K_3(20-0)
  2. 30K1=50K2=20K330K_1 = 50K_2 = 20K_3
  3. K1K3=2030=23\frac{K_1}{K_3} = \frac{20}{30} = \frac{2}{3}, so the answer is (a) K1:K3=2:3K_1:K_3 = 2:3.
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