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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 :

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$\frac{θ_{1}}{10} + \frac{9 θ_{2}}{10}$

$\frac{θ_{2} + θ_{1}}{2}$

$\frac{θ_{1}}{3} + \frac{2 θ_{2}}{3}$

$\frac{θ_{1}}{6} + \frac{5 θ_{1}}{6}$

answer is A.

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Detailed Solution

$H_{1} = H_{2}$
or $(3 k) A (\frac{θ_{2} - θ}{d}) = k A (\frac{θ - θ_{1}}{3 d})$
or $θ = (\frac{θ_{1} + 9 θ_{2}}{10})$

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