The thickness of a metallic plate is $0.4 cm$ . The temperature between its two surfaces is ${20^o}C$. The quantity of heat flowing per second is $50$ calories from $5c{m^2}$ area. In $CGS$ system, the coefficient of thermal conductivity will be
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(c) $\frac{Q}{t} = \frac{{KA(\Delta \theta )}}{l}$
==> $50 = \frac{{5 \times 20\;K}}{{0.4}} \Rightarrow K = \frac{1}{5} = 0.2$
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