Question 101 of 126

Two materials having coefficients of thermal conductivity $3K$ and $K$ and thickness $d$ and $3\text{d}$ respectively, are joined to form a slab as shown in the figure. The temperatures of the outer surfaces are ${\theta }_{2}$ and ${\theta }_{1}$ respectively, $\left({\theta }_{2}\gt {\theta }_{1}\right).$ The temperature at the interface is

A$\dfrac{{\theta }_{2}+{\theta }_{1}}{2}$B$\dfrac{{\theta }_{1}}{6}+\dfrac{5{\theta }_{2}}{6}$C$\dfrac{{\theta }_{1}}{3}+\dfrac{2{\theta }_{2}}{3}$D$\dfrac{{\theta }_{1}}{10}+\dfrac{9{\theta }_{2}}{10}$

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Answer: (D) $\dfrac{{\theta }_{1}}{10}+\dfrac{9{\theta }_{2}}{10}$

Let the temperature of the junction $T^{\circ}C$ .
Rate of heat flow in Rod $1$ = rate of heat flow in Rod $2$
$\dfrac{3kA}{d}\left({\theta }_{2}-T\right)=\dfrac{kA}{3d}\left(T-{\theta }_{1}\right)$
$\Rightarrow 9\left({\theta }_{2}-T\right)=\left(T-{\theta }_{1}\right)$
$\Rightarrow 10T=9{\theta }_{2}+{\theta }_{1}$
$\Rightarrow T=\dfrac{9{\theta }_{2}+{\theta }_{1}}{10}=\dfrac{{\theta }_{1}}{10}+\dfrac{9{\theta }_{2}}{10}$