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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 ....... $K$
$10$
$20$
$30$
$40$
Solution
(b) Suppose conductivity of layer $B$ is $K$, then it is $2K$ for layer $A$. Also conductivity of
combination layers $A$ and $B$ is $KS$ $ = \frac{{2 \times 2K \times K}}{{(2K + K)}} = \frac{4}{3}K$
Hence ${\left( {\frac{Q}{t}} \right)_{Combination}} = {\left( {\frac{Q}{t}} \right)_A}$
==> $\frac{4}{3}\frac{{KA \times 60}}{{2x}} = \frac{{2K.A \times {{(\Delta \theta )}_A}}}{x}$==> ${(\Delta \theta )_A} = 20K$
