A cylindrical rod with one end in a steam chamber and the other end in ice results in melting of $0.1$ gm 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 $\frac{1}{4}$ that of first, the rate at which ice melts in $gm/\sec $will be
Medium
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(c) $\frac{Q}{t} = \frac{{KA\Delta \theta }}{l}$
$⇒$ $\frac{{mL}}{t}$ $ = \frac{{K(\pi {r^2})\Delta \theta }}{l}$
$⇒$ Rate of melting of ice $\left( {\frac{m}{t}} \right) \propto \frac{{K{r^2}}}{l}$
Since for second rod $K$ becomes $\frac{1}{4}th$ $r$ becomes double and length becomes half, so rate of melting will be twice i.e. ${\left( {\frac{m}{t}} \right)_2} = 2\,{\left( {\frac{m}{t}} \right)_1} = 2 \times 0.1 = 0.2\,gm/sec.$
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