Question 40 of 130

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

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Answer: 2

Given here, $\dfrac{{v}_{1}}{{v}_{2}}=\dfrac{3}{2}$
Using conservation of momentum during disintegration, we have
${m}_{1}{v}_{1}={m}_{2}{v}_{2}\Rightarrow \dfrac{{m}_{1}}{{m}_{2}}=\dfrac{2}{3}$
Since, $m=\dfrac{4}{3}\pi {R}^{3}\rho$
We have $m\propto {R}^{3}\left(∵\text{nuclear mass density},\rho =\text{constant}\right)$
$\dfrac{{m}_{1}}{{m}_{2}}={\left(\dfrac{{R}_{1}}{{R}_{2}}\right)}^{3}$
$\Rightarrow \dfrac{{R}_{1}}{{R}_{2}}={\left(\dfrac{2}{3}\right)}^{\dfrac{1}{3}}$
So, $x=2$.

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