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(C) The radius of a nucleus is given by $R = R_0 A^{1/3}$,where $A$ is the mass number.The surface area $S$ of a spherical nucleus is $S = 4 \pi R^2 =

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$125$

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(C) The radius of a nucleus is given by $R = R_0 A^{1/3}$,where $A$ is the mass number.The surface area $S$ of a spherical nucleus is $S = 4 \pi R^2 = 4 \pi R_0^2 A^{2/3}$.Thus,$S \propto A^{2/3}$.Initial mass number $A_i = 8$.Initial surface area $S_i \propto (8)^{2/3} = (2^3)^{2/3} = 2^2 = 4$.After absorbing $10$ protons and $9$ neutrons (electrons do not contribute to the mass number),the new mass number $A_f = 8 + 10 + 9 = 27$.Final surface area $S_f \propto (27)^{2/3} = (3^3)^{2/3} = 3^2 = 9$.The percentage increase in surface area is given by $\frac{S_f - S_i}{S_i} 	imes 100$.Percentage increase $= \frac{9 - 4}{4} 	imes 100 = \frac{5}{4} 	imes 100 = 125\%$.

An atom ${}_3^8 X$ is bombarded by a shower of fundamental particles. In $10 \ s$,this atom absorbs $10$ electrons,$10$ protons,and $9$ neutrons. The percentage growth in the surface area of the nucleus is recorded as: (in $\%$)

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