A nucleus at rest splits into two nuclear par | vedclass

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(D) The correct answer is $D$. $8:1$.According to the law of conservation of linear momentum,the initial momentum of the nucleus at rest is zero.There

Vedclass · Accepted Answer

$8$

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(D) The correct answer is $D$. $8:1$.According to the law of conservation of linear momentum,the initial momentum of the nucleus at rest is zero.Therefore,the magnitude of the momentum of the two parts must be equal: $M_1 v_1 = M_2 v_2$.Since the density $\rho$ is the same for both parts,the mass $M$ is given by $M = \rho \cdot V = \rho \cdot (\frac{4}{3} \pi r^3)$.Substituting this into the momentum equation:$(\rho \cdot \frac{4}{3} \pi r_1^3) v_1 = (\rho \cdot \frac{4}{3} \pi r_2^3) v_2$.Canceling the common terms $(\rho \cdot \frac{4}{3} \pi)$,we get $r_1^3 v_1 = r_2^3 v_2$.Rearranging for the ratio of velocities: $\frac{v_1}{v_2} = (\frac{r_2}{r_1})^3$.Given the ratio of radii $\frac{r_1}{r_2} = \frac{1}{2}$,we have $\frac{r_2}{r_1} = \frac{2}{1}$.Thus,$\frac{v_1}{v_2} = (\frac{2}{1})^3 = \frac{8}{1}$.

$A$ 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 $\qquad$ . (in $:1$)

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