Question

The ends of the two rods of different materials with their lengths, diameters of cross-section and thermal conductivities all in the ratio 1:2 are maintained at the same temperature difference. The rate of flow of heat in the shorter rod is 1 cal s^–1. What is the rate of flow of heat in the larger rod ?

• A. 1 cal s^–1
• B. 4 cal s^–1
• C. 8 cal s^–1
• D. 16 cal s^–1

Answer 1

Answer: (B)

4 cal s^–1

Answer 2

$\left( \frac{Q}{t} \right)_{1}$ = $\frac{KA(\theta_{1}–\theta_{2})}{d}$ = 1 cal s^–1

$\left( \frac{Q}{t} \right)_{2}$ = $\frac{2k(4A)(\theta_{1}–\theta_{2})}{2d}$

= $\frac{4kA(\theta_{1}–\theta_{2})}{d}$

= 4 cal s^–1