An unknown nucleus has a nuclear density of \(2.29 \times 10^{17} kg / m ^3\) and mass of \(19.926 \times 10^{-27} kg\). Its mass number A is approximately :
(Take \(R_0=1.2 \times 10^{-15} m ; 4 \pi=12.56\) )

(A) 12
The nuclear density \(\rho\) is given by the ratio of the mass of the nucleus to its volume:
\(\rho=\frac{m}{\frac{4}{3} \pi R^3}\)
The radius of the nucleus is \(R=R_0 A^{1 / 3}\), which gives \(R^3=R_0^3 A\).
Substituting this into the density formula:
\(\rho=\frac{3 m}{4 \pi R_0^3 A}\)
Rearranging for the mass number \(A\) :
\(A=\frac{3 m}{4 \pi R_0^3 \rho}\)
Substituting the given values:
\(A=\frac{3 \times 19.926 \times 10^{-27}}{12.56 \times\left(1.2 \times 10^{-15}\right)^3 \times 2.29 \times10^{17}}\)
\(A=\frac{59.778 \times 10^{-27}}{12.56 \times 1.728 \times 10^{-45} \times 2.29 \times 10^{17}}\)
\(A=\frac{59.778 \times 10^{-27}}{49.7 \times 10^{-28}}\)
\(A=\frac{597.78}{49.7} \approx 12\)