A cylindrical rod with one end in a steam cha | vedclass

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Question

(c) $\frac{Q}{t} = \frac{{KA\Delta 	heta }}{l}$

$&rArr;$ $\frac{{mL}}{t}$ $ = \frac{{K(\pi {r^2})\Delta 	heta }}{l}$

$&rArr;$ Rate of melting

Vedclass · Accepted Answer

$0.2$

Vedclass · Answer

(c) $\frac{Q}{t} = \frac{{KA\Delta 	heta }}{l}$

$&rArr;$ $\frac{{mL}}{t}$ $ = \frac{{K(\pi {r^2})\Delta 	heta }}{l}$

$&rArr;$ Rate of melting of ice $\left( {\frac{m}{t}} ight) \propto \frac{{K{r^2}}}{l}$
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. ${\left( {\frac{m}{t}} ight)_2} = 2\,{\left( {\frac{m}{t}} ight)_1} = 2 	imes 0.1 = 0.2\,gm/sec.$

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