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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
$8 : 27$
$2 : 3$
$3 : 2$
$4:9$
Solution
Let heavy nucleus breaks into two nuclei of mass $m_{1}$ and $m_{2}$ and move away with velocities ${V}_{1}^1$ and $V_{2}^2$ respectively.
According to question, $\frac{V_{1}}{V_{2}}=\frac{8}{27}$ $m_{1} V_{1}=m_{2} V_{2}$ (Law of momentum conservation)
$\Rightarrow$ $\frac{{{m_1}}}{{{m_2}}} = \frac{{{V_2}}}{{{V_1}}} = \frac{{27}}{8}$
$\rho \times \frac{4}{3}\pi R_1^3$ $\left( {\because {\text{ density }}\rho = \frac{{{\text{ mass }}}}{{{\text{ volume }}}}} \right)$
$\rho \times \frac{4}{3}\pi R_2^3$
$ \Rightarrow \left( {\frac{{{R_1}}}{{{R_2}}}} \right) = $ ${\left( {\frac{{27}}{8}} \right)^{\frac{1}{3}}} = {\left( {\frac{3}{2}} \right)^{3 \times \frac{1}{3}}}$
$\therefore \frac{R_{1}}{R_{2}}=\frac{3}{2}$
