13.Nuclei
hard

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

A

$8 : 27$

B

$2 : 3$

C

$3 : 2$

D

$4:9$

(JEE MAIN-2018)

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) = $ ${\left( {\frac{{27}}{8}} \right)^{\frac{1}{3}}} = {\left( {\frac{3}{2}} \right)^{3 \times \frac{1}{3}}}$

$\therefore \frac{R_{1}}{R_{2}}=\frac{3}{2}$

Standard 12
Physics

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