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Let heavy nucleus breaks into two nuclei of mass \(m_{1}\) and \(m_{2}\) and move away with velocities \({V}_{1}^1\) and \(V_{2}^2\) respectively.

According to question, \(\frac{V_{1}}{V_{2}}=\frac{8}{27}\) \(m_{1} V_{1}=m_{2} V_{2}\) (Law of momentum conservation)

\(\Rightarrow\) \(\frac{{{m_1}}}{{{m_2}}} = \frac{{{V_2}}}{{{V_1}}} = \frac{{27}}{8}\)

\(\rho  \times \frac{4}{3}\pi R_1^3\)     \(\left( {\because {\text{ density }}\rho  = \frac{{{\text{ mass }}}}{{{\text{ volume }}}}} \right)\)

\(\rho  \times \frac{4}{3}\pi R_2^3\)  

\( \Rightarrow \left( {\frac{{{R_1}}}{{{R_2}}}} \right) = \) \({\left( {\frac{{27}}{8}} \right)^{\frac{1}{3}}} = {\left( {\frac{3}{2}} \right)^{3 \times \frac{1}{3}}}\)

\(\therefore \frac{R_{1}}{R_{2}}=\frac{3}{2}\)