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JEE Main 20206 SeptemberEvening Shift
Exam
JEE Main
Year
2020
Date
6 September
Shift
Evening Shift
Subject
Physics
Chapter
Thermal Properties of Matter

JEE Main 2020 · 6 September · Evening Shift · Physics · Thermal Properties of Matter

Three rods of identical cross-section and length are made of three...

Three rods of identical cross-section and length are made of three different materials of thermal conductivity K1,K2{K}_{1}, {K}_{2} and K3{K}_{3}, respectively. They are joined together at their ends to make a long rod (see figure). One end of the long rod is maintained at 100°C100 ° C and the other at 0°C0 ° C (see figure). If the joints of the rod are at 70°C70 ° C and 20°C20 ° C in steady and there is no loss of energy from the surface of the rod, the correct relationship between K1,K2{K}_{1}, {K}_{2} and K3{K}_{3} is :

Three rods of identical cross-section and length are made of three different materials of thermal conductivity K _ 1 ,

Options

  1. AK1:K3=2:3{K}_{1}: {K}_{3} = 2: 3,
    K2:K3=2:5{K}_{2}: {K}_{3} = 2: 5
  2. BK1<K2<K3{K}_{1} < {K}_{2} < {K}_{3}
  3. CK1:K2=5:2{K}_{1}: {K}_{2} = 5: 2,
    K1:K3=3:5{K}_{1}: {K}_{3} = 3: 5
  4. DK1>K2>K3{K}_{1} > {K}_{2} > {K}_{3}

Correct answer

A. K1:K3=2:3{K}_{1}: {K}_{3} = 2: 3,
K2:K3=2:5{K}_{2}: {K}_{3} = 2: 5

Step-by-step solution

K1(100−70)=K2(70−20)=K3(20−0){K}_{1} ( 100 - 70 ) = {K}_{2} ( 70 - 20 ) = {K}_{3} ( 20 - 0 )

K130=K2(50)=K3(20){K}_{1}30 = {K}_{2} ( 50 ) = {K}_{3} ( 20 )

⇒K1:K2:K3=10:6:15⇒ {K}_{1}: {K}_{2}: {K}_{3} = 10: 6: 15

K1:K3=2:3{K}_{1}: {K}_{3} = 2: 3

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