A nucleus disintegrates into two smaller parts, which have their velocities in the ratio 3 : 2. The ratio of their nuclear sizes will be $$(\frac{x}{3})^{\frac{1}{3}}$$. The value of '$$x$$' is:

A nucleus disintegrates into two smaller parts with velocities in the ratio $$3:2$$.

Since the nucleus is initially at rest, the total momentum after disintegration must be zero. This implies $$m_1 v_1 = m_2 v_2$$ and hence $$\frac{m_1}{m_2} = \frac{v_2}{v_1} = \frac{2}{3}$$.

Now, because nuclear mass is proportional to volume at constant density, $$m \propto r^3$$, we have $$\frac{m_1}{m_2} = \frac{r_1^3}{r_2^3} = \frac{2}{3}$$.

Substituting and taking cube roots gives $$\frac{r_1}{r_2} = \left(\frac{2}{3}\right)^{1/3} = \left(\frac{x}{3}\right)^{1/3}$$. Comparing the expressions yields $$\frac{x}{3} = \frac{2}{3}$$ and therefore $$x = \mathbf{2}$$.