JEE Main · 2019 · 09 Apr 2019 · Shift 2 · Heat & Thermodynamics

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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_{2}

and

θ_{1}

respectively

(θ_{2} > θ_{1})

. The temperature at the interface is:

(A) $\frac{θ _{2} + θ _{1}}{2}$
(B) $\frac{θ _{1}}{10} + \frac{9 θ _{2}}{10}$ ✓ Correct
(C) $\frac{θ _{1}}{3} + \frac{2 θ _{2}}{3}$
(D) $\frac{θ _{1}}{6} + \frac{5 θ _{2}}{6}$

Answer: (B)

Solution

In the steady state the same heat current flows through both layers, so

\frac{3 K A ( θ _{2} - θ )}{d} = \frac{K A ( θ - θ _{1} )}{3 d}

, where

θ

is the interface temperature. Cross-multiplying,

9 (θ_{2} - θ) = θ - θ_{1}

, so

10 θ = 9 θ_{2} + θ_{1}

and

θ = \frac{θ _{1}}{10} + \frac{9 θ _{2}}{10}

JEE Main · 2019 · 09 Apr 2019 · Shift 2 · Single correct option

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JEE Main · 2019 · 09 Apr 2019 · Shift 2 · Heat & Thermodynamics

Solution

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