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(D) The rate of heat flow through a rod is given by the formula $\frac{Q}{t} = \frac{KA \Delta T}{L}$,where $K$ is the thermal conductivity,$A$ is the

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$1 : 8$

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(D) The rate of heat flow through a rod is given by the formula $\frac{Q}{t} = \frac{KA \Delta T}{L}$,where $K$ is the thermal conductivity,$A$ is the cross-sectional area,$\Delta T$ is the temperature difference,and $L$ is the length.Since the material is the same,$K_1 = K_2$. Given $\Delta T_1 = \Delta T_2$,the ratio of the rates of heat flow is $\frac{(Q/t)_1}{(Q/t)_2} = \frac{A_1}{A_2} 	imes \frac{L_2}{L_1}$.The area $A = \pi r^2 = \pi (d/2)^2$,so the ratio of areas is $\frac{A_1}{A_2} = \left(\frac{d_1}{d_2}ight)^2 = \left(\frac{1}{2}ight)^2 = \frac{1}{4}$.Given the ratio of lengths $\frac{L_1}{L_2} = \frac{2}{1}$,we have $\frac{L_2}{L_1} = \frac{1}{2}$.Substituting these values: $\frac{(Q/t)_1}{(Q/t)_2} = \frac{1}{4} 	imes \frac{1}{2} = \frac{1}{8}$.

The heat is flowing through two cylindrical rods of the same material. The diameters of the rods are in the ratio $1 : 2$ and their lengths are in the ratio $2 : 1$. If the temperature difference between their ends is the same,the ratio of the rates of flow of heat through them will be

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