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A wall has two layers $A$ and $B$ , each made of a different material. Both the layers have the same thickness. The thermal conductivity of the material of $A$ is twice that of $B$ . Under thermal equilibrium, the temperature difference across the wall is $36\,^oC$. The temperature difference across the layer $A$ is ......... $^oC$
$6$
$12$
$18$
$24$
Solution
$\frac{\mathrm{K}_{\mathrm{A}} \mathrm{A} \cdot\left(\Delta \mathrm{T}_{\mathrm{A}}\right)}{\ell}=\frac{\mathrm{K}_{\mathrm{B}} \cdot \mathrm{A} \cdot\left(\Delta \mathrm{T}_{\mathrm{B}}\right)}{\ell}$
$\because \mathrm{K}_{\mathrm{A}}=2 \mathrm{K}_{\mathrm{B}}$
$\therefore \Delta \mathrm{T}_{\mathrm{B}}=2 \Delta \mathrm{T}_{\mathrm{A}}$
$\Delta \mathrm{T}_{\mathrm{A}}+\Delta \mathrm{T}_{\mathrm{B}}=36$
$3 \Delta T_{A}=36$
$\Delta \mathrm{T}_{\mathrm{A}}=12^{\circ} \mathrm{C}$