Question

A partition wall has two layers A and B in contact, each made of a different material. They have the same thickness but the thermal conductivity of layer A is twice that of layer B. If the steady state temperature difference across the wall is 60k, then the corresponding difference across the layer A is

• A. 10 K
• B. 20 K
• C. 30 K
• D. 40 K

Answer 1

The correct option is B 20 K

Suppose conductivity of layer B is K, then it is 2K for layer A. Also conductivity of combination of layers A and B is



$K_s=\frac{2\times 2K\times K}{(2K+K)}=\frac{4}{3}K$
Hence ${\left(\frac{Q}{t}\right)}_{combination}={\left(\frac{Q}{t}\right)}_A$
$\Rightarrow \frac{4}{3}\frac{KA\times 60}{2x}=\frac{2K.A\times (\Delta \theta {)}_A}{x}\Rightarrow (\Delta \theta {)}_A=20K$