A nucleus at rest splits into two nuclear par | vedclass

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(D) Given the ratio of radii $\frac{r_1}{r_2} = \frac{1}{2}$.Since the density $\rho$ is the same,the mass $m$ is proportional to the volume $V = \fra

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(D) Given the ratio of radii $\frac{r_1}{r_2} = \frac{1}{2}$.Since the density $\rho$ is the same,the mass $m$ is proportional to the volume $V = \frac{4}{3} \pi r^3$.Therefore,the ratio of masses is $\frac{m_1}{m_2} = \frac{r_1^3}{r_2^3} = (\frac{1}{2})^3 = \frac{1}{8}$.Since the nucleus is initially at rest,the total momentum is zero. According to the law of conservation of linear momentum:$m_1 v_1 + m_2 v_2 = 0$$m_1 v_1 = -m_2 v_2$Taking the magnitude,$|\frac{v_1}{v_2}| = \frac{m_2}{m_1} = \frac{8}{1}$.Thus,the ratio of their velocities is $8 : 1$.

$A$ nucleus at rest splits into two nuclear parts having the same density and radii in the ratio $1 : 2$. Their velocities are in the ratio: (in $: 1$)

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