Question

The ends of two rods of different materials with their thermal conductivities, radii of cross sections and lengths all in the ratio 1 : 2 are maintained at the same temperature difference, if the rate of flow of heat in the larger rod is 4 cal/ sec, the rate of flow of heat in the shorter rod will be

• A. 1 cal/sec
• B. 2 cal/sec
• C. 8 cal/sec
• D. 16 cal/sec

Answer 1

The correct option is A 1 cal/sec

Rate of flow of heat $=\frac{Q}{t}=\frac{KA\Delta T}{d}$
$\therefore \frac{H_1}{H_2}=\frac{\frac{K_1{A}_1\Delta T}{d_1}}{\frac{K_2{A}_2\Delta T}{d_2}}=\frac{K_1}{K_2}\frac{A_1}{A_2}\frac{d_2}{d_1}$
$=\frac{1}{2}\times \frac{\pi r_1^2}{\pi r_2^2}\times \frac{2}{1}=\frac{1}{2}{\left(\frac{1}{2}\right)}^2.\frac{2}{1}=\frac{1}{4}$
[Since $K_1:K_2=1:2$]
$A_1:A_2=1:2$
$d_1:d_2=1:2$
$\therefore H_1=H_2\times \frac{1}{4}=4.\frac{1}{4}=1cal/sec.$
$[\because H_2=4cal/sec]$