Q 12-13-118JEE MainJEE Main 2020 (3 Sep, Shift 2)Medium
The radius $R$ of a nucleus of mass number $A$ can be estimated by the formula $R = (1.3\times10^{-15})A^{1/3}\ \text{m}$. It follows that the mass density of a nucleus is of the order of: $(M_{prot} \cong M_{neut} \simeq 1.67\times10^{-27}\ \text{kg})$
Answer: (D) $10^{17}\ \text{kg m}^{-3}$
$$\rho = \frac{A\times1.67\times10^{-27}}{\tfrac43\pi(1.3\times10^{-15})^{3}A} = \frac{1.67\times10^{-27}}{9.2\times10^{-45}} \approx 1.8\times10^{17}\ \text{kg m}^{-3}$$
The density is independent of $A$ and of the order of $10^{17}\ \text{kg m}^{-3}$.
Solution by Sreeraj P, M.Sc Physics