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Q.
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 :
The two nuclei have velocity in ratio $8: 27$. By conservation of momentum, we have
$m_{1} v_{1}=m_{2} v_{2}$
$ \Rightarrow \frac{v_{1}}{v_{2}}=\frac{m_{2}}{m_{1}} $
$\Rightarrow \frac{m_{2}}{m_{1}}=\frac{8}{27} $
Now, since $m=\rho \frac{4}{3} \pi r^{3}$
Therefore $\frac{m_{2}}{m_{1}}=\frac{\rho \frac{4}{3} \pi r_{2}^{3}}{\rho \frac{4}{3} \pi r_{1}^{3}} $
$\Rightarrow \frac{m_{2}}{m_{1}}=\left(\frac{r_{2}}{r_{1}}\right)^{3} $
$\Rightarrow \left(\frac{r_{2}}{r_{1}}\right)^{3}=\frac{8}{27} $
$\Rightarrow \frac{r_{2}}{r_{1}}=\frac{2}{3}$
Thus, ratio of radii of nuclei $r_{1}: r_{2}=3: 2$.