Question

The end of two rods of different material with their thermal conductivity, area of cross-section and lengths all in the ratio $1:2$ are maintained at the same temperature difference. If the rate of flow of heat in the first rod is $4$ cal/s. Then in the second-rod rate of heat flow in cal/s will be

• A. 1
• B. 2
• C. 8
• D. 16

Answer 1

The correct option is C 8

Here , $Q_1=\frac{K_1{A}_1({\theta }_1-{\theta }_2)}{l_1}$ and $Q_2=\frac{K_2{A}_2({\theta }_1-{\theta }_2)}{l_2}$

$\therefore \frac{Q_2}{Q_1}=\frac{K_2}{K_1}\times \frac{A_2}{A_1}\times \frac{l_1}{l_2}=2\times 2\times \frac{1}{2}=2$

so, $Q_2=2Q_1=2\times 4=8$ cal/s