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(C) The rate of heat flow through a cylindrical rod is given by $\frac{Q}{t} = \frac{KA\Delta 	heta}{l}$.Since the heat is used to melt ice,$\frac{Q}

Vedclass · Accepted Answer

$0.2$

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(C) The rate of heat flow through a cylindrical rod is given by $\frac{Q}{t} = \frac{KA\Delta 	heta}{l}$.Since the heat is used to melt ice,$\frac{Q}{t} = \frac{mL}{t}$,where $L$ is the latent heat of fusion.Thus,$\frac{mL}{t} = \frac{K(\pi r^2)\Delta 	heta}{l}$.This implies that the rate of melting of ice,$\frac{m}{t} \propto \frac{Kr^2}{l}$.Let the initial rate be $R_1 = \left(\frac{m}{t}ight)_1 = 0.1 \ gm/sec$.For the second rod,$K_2 = \frac{1}{4}K_1$,$r_2 = 2r_1$,and $l_2 = \frac{1}{2}l_1$.The new rate $R_2$ is given by $R_2 = R_1 	imes \left(\frac{K_2}{K_1}ight) 	imes \left(\frac{r_2}{r_1}ight)^2 	imes \left(\frac{l_1}{l_2}ight)$.Substituting the values: $R_2 = 0.1 	imes \left(\frac{1}{4}ight) 	imes (2)^2 	imes (2) = 0.1 	imes \frac{1}{4} 	imes 4 	imes 2 = 0.2 \ gm/sec$.

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