JEE Main 2019 — Physics solution

JEE Main2019Physics

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
$\frac{θ _{1}}{6} + \frac{5 θ _{2}}{6}$
B
$\frac{θ _{1}}{10} + \frac{9 θ _{2}}{10}$
C
$\frac{θ _{1}}{3} + \frac{2 θ _{2}}{3}$
D
$\frac{θ _{2} + θ _{1}}{2}$

Solution

$H = \frac{3 K A}{d} (θ_{2} - θ) = \frac{K A}{3 d} (θ - θ_{1}) \Rightarrow θ = \frac{9 θ _{2}}{10} + \frac{θ _{1}}{10}$

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