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)

Given that
$\frac{l_{1}}{l_{2}}=\frac{1}{2} ; \frac{r_{1}}{r_{2}}=\frac{1}{2} ; \frac{k_{1}}{k_{2}}=\frac{1}{2}$
We use $-R=\frac{Q}{t}=\frac{k A \Delta T}{L}$
$\Rightarrow \frac{R_{1}}{R_{2}}=\frac{k_{1} A_{1}}{L_{1}} \times \frac{L_{2}}{k_{2} A_{2}}$
$\Rightarrow \frac{R_{1}}{R_{2}}=\frac{1}{2} \times \frac{1}{4} \times \frac{2}{1}=\frac{1}{4}$
$\Rightarrow R_{2}=4\, R_{1}=4\, cal / s$