Q 12-13-131JEE MainJEE Main 2018 (15 Apr, Shift 2)Easy
An unstable heavy nucleus at rest breaks into two nuclei which move away with velocities in the ratio of $8 : 27$. The ratio of the radii of the nuclei (assumed to be spherical) is:
Answer: (C) $3 : 2$
Momentum is conserved and was zero, so $m_1v_1 = m_2v_2$:
$$\frac{m_1}{m_2} = \frac{v_2}{v_1} = \frac{27}{8}$$
Nuclear radius $R = R_0A^{1/3}$, so
$$\frac{R_1}{R_2} = \left(\frac{27}{8}\right)^{1/3} = \frac32$$
Solution by Sreeraj P, M.Sc Physics