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A nucleus disintegrates into two nuclear parts which have their velocities in the ratio $2: 1$. The ratio of their nuclear size will be
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Verified Answer
The correct answer is:
$1: 2^{1 / 3}$
According to the law of conservation of momentum, we get
$m_1 V_1=m_2 V_2$
or $\left(\frac{m_1}{m_2}\right)=\left(\frac{v_2}{v_1}\right)$
But $m=\frac{4}{3} \pi r^3 \rho \text { or } m \propto r^3$
$\begin{array}{ll}\Rightarrow & \frac{m_1}{m_2}=\frac{r_1^3}{r_2^3} \\ \therefore & \frac{r_1^3}{r_2^3}=\frac{V_2}{V_1}=\frac{1}{2} \\ \Rightarrow & \frac{r_1}{r_2}=\left(\frac{1}{2}\right)^{1 / 3} \\ \therefore & r_1: r_2=1:(2)^{1 / 3}\end{array}$
$m_1 V_1=m_2 V_2$
or $\left(\frac{m_1}{m_2}\right)=\left(\frac{v_2}{v_1}\right)$
But $m=\frac{4}{3} \pi r^3 \rho \text { or } m \propto r^3$
$\begin{array}{ll}\Rightarrow & \frac{m_1}{m_2}=\frac{r_1^3}{r_2^3} \\ \therefore & \frac{r_1^3}{r_2^3}=\frac{V_2}{V_1}=\frac{1}{2} \\ \Rightarrow & \frac{r_1}{r_2}=\left(\frac{1}{2}\right)^{1 / 3} \\ \therefore & r_1: r_2=1:(2)^{1 / 3}\end{array}$
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