JEE Main20199 Apr 2019Evening ShiftPhysicsThermal Properties of MatterActual
Two materials having coefficients of thermal conductivity 3 K and K and thickness d and 3 d respectively, are joined to form a slab as shown in the figure. The temperatures of the outer surfaces are θ 2 and θ 1 respectively, θ 2 > θ 1 . The temperature at the interface is
Options
- Aθ 2 + θ 1 2
- Bθ 1 6 + 5 θ 2 6
- Cθ 1 3 + 2 θ 2 3
- Dθ 1 10 + 9 θ 2 10
Correct answer
D. θ 1 10 + 9 θ 2 10
Step-by-step solution
Let the temperature of the junction T ° C . Rate of heat flow in Rod 1 = rate of heat flow in Rod 2 3 kA d θ 2 - T = kA 3 d T - θ 1 ⇒ 9 θ 2 - T = T - θ 1 ⇒ 10 T = 9 θ 2 + θ 1 ⇒ T = 9 θ 2 + θ 1 10 = θ 1 10 + 9 θ 2 10