A cylindrical rod with one end in a steam chamber and the other end in ice results in melting of $0.1\mathrm{g}$ of ice per second. If the rod is replaced by another with half the length and double the radius of the first and if the thermal conductivity of material of second rod is $1/4$ that of first, the rate at which ice melts in $\mathrm{g / sec}$ will be

The heat conducted per second through a rod is proportional to $\frac{kA\Delta T}{L}$. Since the end temperatures remain the same, $\Delta T$ is unchanged, so the melting rate of ice is proportional to $\frac{kA}{L}$. For a cylindrical rod, $A=\pi r^2$. Hence the factor change is determined by $\frac{k r^2}{L}$. For the second rod, $k_2=\frac{k_1}{4}$, $r_2=2r_1$, and $L_2=\frac{L_1}{2}$. Therefore, the ratio of melting rates is $$\frac{R_2}{R_1}=\frac{k_2}{k_1}\cdot\frac{r_2^2}{r_1^2}\cdot\frac{L_1}{L_2}.$$ Substituting the given changes, $$\frac{R_2}{R_1}=\frac{1}{4}\cdot 4\cdot 2=2.$$ So the new melting rate is $$R_2=2\times 0.1=0.2\,\mathrm{g/sec}.$$ Comparing with the options, this matches option $\mathrm{(C)}$.