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(C) The rate of heat flow $H$ is given by $H = \frac{KA \Delta T}{l}$.Since the rate of melting of ice is proportional to the heat flow,we have $H \pr

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$0.2$

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(C) The rate of heat flow $H$ is given by $H = \frac{KA \Delta T}{l}$.Since the rate of melting of ice is proportional to the heat flow,we have $H \propto \frac{KA}{l}$.Let the initial parameters be $K_1, A_1, l_1$. Then $H_1 \propto \frac{K_1 A_1}{l_1} = 0.1 \, g/s$.For the second rod,$l_2 = \frac{l_1}{2}$,$r_2 = 2r_1 \implies A_2 = \pi (2r_1)^2 = 4A_1$,and $K_2 = \frac{K_1}{4}$.The new rate of heat flow $H_2$ is proportional to $\frac{K_2 A_2}{l_2} = \frac{(K_1/4) \cdot (4A_1)}{l_1/2} = 2 \cdot \frac{K_1 A_1}{l_1}$.Thus,$H_2 = 2 H_1 = 2 	imes 0.1 \, g/s = 0.2 \, g/s$.

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