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(D) The rate of heat flow is given by $Q = \frac{kA(T_1 - T_2)t}{l}$,where $Q$ is the heat required to melt the ice,$k$ is thermal conductivity,$A$ is

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$1 : 25$

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(D) The rate of heat flow is given by $Q = \frac{kA(T_1 - T_2)t}{l}$,where $Q$ is the heat required to melt the ice,$k$ is thermal conductivity,$A$ is surface area,$l$ is thickness,and $t$ is time.Since $Q = mL = ho V L$,and $V = 4\pi r^2 l$,we have $Q \propto r^2 l$.Equating the heat flow: $k \frac{r^2}{l} \Delta T t \propto r^2 l$.Thus,$k \propto \frac{l^2}{t}$.Given $r_l = 2r_s$,$l_l = \frac{l_s}{4}$,$t_l = 25 \ min$,and $t_s = 16 \ min$.Ratio $\frac{k_l}{k_s} = \left( \frac{l_l}{l_s} ight)^2 	imes \frac{t_s}{t_l} = \left( \frac{1}{4} ight)^2 	imes \frac{16}{25} = \frac{1}{16} 	imes \frac{16}{25} = \frac{1}{25}$.

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