Consider two nuclei, A of mass number 27 and B of mass number 64. Considering them as liquid-drops, the ratio of their densities $$(\frac{d_A}{d_B})$$​ will be:

Correct option is C

​$$\begin{aligned}\bullet\ \text{Mass number of nucleus A} &= 27 \\\bullet\ \text{Mass number of nucleus B} &= 64\end{aligned}$$

$$\begin{aligned}&\text{According to the liquid drop model, the radius } R \text{ of a nucleus is proportional to the cube root of its mass number:} \\&\qquad R = R_0 A^{1/3} \\&\text{So,} \\&\bullet\ \text{For nucleus A:} \quad R_A = R_0 (27)^{1/3} = 3R_0 \\&\bullet\ \text{For nucleus B:} \quad R_B = R_0 (64)^{1/3} = 4R_0\end{aligned}$$​$$\begin{aligned}&\text{Volume of a nucleus is proportional to } R^3, \text{ so:} \\&\bullet\ \text{For nucleus A:} \\&\quad V_A = \frac{4}{3} \pi (3R_0)^3 = \frac{4}{3} \pi 27R_0^3 \\&\quad \rho_A = \frac{\text{mass}}{\text{volume}} = \frac{27x}{\frac{4}{3} \pi 27R_0^3} = \frac{3x}{4 \pi R_0^3} \\&\bullet\ \text{For nucleus B:} \\&\quad V_B = \frac{4}{3} \pi (4R_0)^3 = \frac{4}{3} \pi 64R_0^3 \\&\quad \rho_B = \frac{64x}{\frac{4}{3} \pi 64R_0^3} = \frac{3x}{4 \pi R_0^3} \\&\quad \frac{\rho_A}{\rho_B} = \frac{3x / 4 \pi R_0^3}{3x / 4 \pi R_0^3} = 1\end{aligned}$$​​​