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(B) The radius $R$ of a nucleus is related to its mass number $A$ by the formula $R = R_0 A^{1/3}$, where $R_0$ is a constant.The surface area $S$ of

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$9 : 25$

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(B) The radius $R$ of a nucleus is related to its mass number $A$ by the formula $R = R_0 A^{1/3}$, where $R_0$ is a constant.The surface area $S$ of a spherical nucleus is given by $S = 4 \pi R^2$.Substituting the expression for $R$, we get $S = 4 \pi (R_0 A^{1/3})^2 = 4 \pi R_0^2 A^{2/3}$.Thus, the surface area is proportional to $A^{2/3}$, i.e., $S \propto A^{2/3}$.Given the ratio of mass numbers $A_1 : A_2 = 27 : 125$, the ratio of their surface areas is $S_1 : S_2 = (A_1 / A_2)^{2/3}$.Substituting the values: $S_1 / S_2 = (27 / 125)^{2/3} = ((3^3) / (5^3))^{2/3} = (3/5)^2 = 9/25$.Therefore, the ratio of their surface areas is $9 : 25$.

If the ratio of the mass numbers of two nuclei is $27 : 125$, then the ratio of their surface areas is

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