Question

The ends of two rods of different materials with their thermal conductivities, radii of cross-sections and lengths all are in the ratio $1:2$ are maintained at the same temperature difference. If the rate of flow of heat in the longer rod is $4\text{cal/sec}$, then that in the shorter rod in $\text{cal/sec}$ will be

• A. $1$
• B. $2$
• C. $8$
• D. $16$

Answer 1

The correct option is A $1$

Given that,
$\frac{K_s}{K_l}=\frac{1}{2};\frac{l_s}{l_l}=\frac{1}{2};\frac{r_s}{r_l}=\frac{1}{2}$
${\left(\frac{dQ}{dt}\right)}_l=4\text{cal/sec}$
We know that,
$\frac{dQ}{dt}=\frac{K\left(\pi r^2\right)dT}{dl}$
$\frac{{\left(\frac{dQ}{dt}\right)}_s}{{\left(\frac{dQ}{dt}\right)}_l}=\frac{K_s}{K_l}\times \frac{r_s^2}{r_l^2}\times \frac{l_l}{l_s}$
$=\frac{1}{2}\times \frac{1}{4}\times 2$
$\Rightarrow {\left(\frac{dQ}{dt}\right)}_s=\frac{{\left(\frac{dQ}{dt}\right)}_l}{4}=\frac{4}{4}=1\text{cal/sec}$