Two identical conducting rods are first connected independently to two vessels, one containing water at 100$℃$ and the other containing ice at 0$℃$. In the second case, the rods are joined end to end and connected to the same vessels. Let $q_{1}$ and $q_{2}$ ${gs}^{- 1}$ be the rate of melting of ice in the two cases respectively. The ratio $\frac{q_{1}}{q_{2}}$ is

Question

Dr. Abubacker · Accepted Answer

(c) Heat current, $\frac{d\mathcal{Q}}{dt} = L.\left( \frac{dm}{dt} ight)$
Or $\frac{Temperature\ difference}{Thermal\ resistance} = \ L.\left( \frac{dm}{dt} ight)$
Or $\left( \frac{dm}{dt} ight) \propto \ \frac{1}{Thermal\ resistance}\ $
Or $q \propto \ \frac{1}{R}$
In the first case rods are in parallel and thermal resistance is $\frac{\ \ R}{2}$ while in second case rods are in series and thermal resistance is $2R$.
$	herefore$ $\frac{q_{1}}{q_{2}} = \ \frac{2R}{\frac{R}{2}} = \ \frac{4}{1}\ $

Explanation

Reviewed by Dr. Abubacker

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