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A wall is made up of two layers $A$ and $B.$ The thickness of the two layers is the same, but materials are different. The thermal conductivity of $A$ is double than that of $B.$ In thermal equilibrium the temperature difference between the two ends is $36\,^oC.$ Then the difference of temperature at the two surfaces of $A$ will be......... $^oC$
$6$
$12$
$18$
$24$
Solution
Suppose thickness of each wall is $x$ then
$\left(\frac{\mathrm{Q}}{\mathrm{t}}\right)_{\text {combination }}=\left(\frac{\mathrm{Q}}{\mathrm{t}}\right)_{\mathrm{A}} \Rightarrow \frac{\mathrm{K}_{\mathrm{s}} \mathrm{A}\left(\theta_{1}-\theta_{2}\right)}{2 \mathrm{x}}=\frac{2 \mathrm{KA}\left(\theta_{1}-\theta\right)}{\mathrm{x}}$
$\because K_{S}=\frac{2 \times 2 \mathrm{K} \times \mathrm{K}}{(2 \mathrm{K}+\mathrm{K})}=\frac{4}{3} \mathrm{K}$ and $\left(\theta_{1}-\theta_{2}\right)=36^{\circ}$
$\Rightarrow \frac{\frac{4}{3} \mathrm{KA} \times 36}{2 \mathrm{x}}=\frac{2 \mathrm{KA}\left(\theta_{1}-\theta\right)}{\mathrm{x}}$
Hence temperature difference across wall $A$ is
$\left(\theta_{1}-\theta\right)=12^{\circ} \mathrm{C}$
