A nucleus disintegrates into two smaller part | vedclass

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(B) Given the ratio of velocities is $\frac{v_1}{v_2} = \frac{3}{2}$.According to the law of conservation of linear momentum,$m_1 v_1 = m_2 v_2$. Ther

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$2$

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(B) Given the ratio of velocities is $\frac{v_1}{v_2} = \frac{3}{2}$.According to the law of conservation of linear momentum,$m_1 v_1 = m_2 v_2$. Therefore,the ratio of their masses is $\frac{m_1}{m_2} = \frac{v_2}{v_1} = \frac{2}{3}$.Since the nuclear mass density is constant,the mass of a nucleus is proportional to its volume,i.e.,$m \propto r^3$,where $r$ is the nuclear radius.Thus,$\frac{m_1}{m_2} = \left(\frac{r_1}{r_2}\right)^3$.Substituting the mass ratio: $\left(\frac{r_1}{r_2}\right)^3 = \frac{2}{3}$.Taking the cube root on both sides: $\frac{r_1}{r_2} = \left(\frac{2}{3}\right)^{\frac{1}{3}}$.Comparing this with the given expression $\left(\frac{x}{3}\right)^{\frac{1}{3}}$,we get $x = 2$.

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

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