Physics Circle home Student Login

Home›Chapters›Heat Transfer and Radiation PYQs›Heat Conduction›Question 45

JEE Main 2019 (9 Apr, Shift 2) · Heat Conduction

Heat Conduction question from JEE Main 2019 (9 Apr, Shift 2)

Q45JEE Main 2019 · 9 APR · SHIFT 2Medium

Heat Conduction

Two materials having coefficients of thermal conductivity '3K' and 'K' and thickness 'd' and '3d', respectively, are joined to form a slab as shown in the figure. The temperatures of the outer surfaces are 'θ\theta2' and 'θ\theta1' respectively, (θ\theta2 > θ\theta1). The temperature at the interface is :-

Figure

Show answer and solution

Answer: (a)

  1. R=LKAR = \frac{L}{KA}: R1=d3KAR_1 = \frac{d}{3KA}, R2=3dKAR_2 = \frac{3d}{KA}, ratio 1:91:9.
  2. The drop across the 3K3K layer is 110(θ2−θ1)\frac{1}{10}(\theta_2 - \theta_1).
  3. θ=θ2−θ2−θ110=θ110+9θ210\theta = \theta_2 - \frac{\theta_2-\theta_1}{10} = \frac{\theta_1}{10} + \frac{9\theta_2}{10}
  4. So the answer is (a).
Report a mistake

More questions on Heat Conduction