Two lead spheres have radii in the ratio 1 : 2 and have the same linear momentum. The ratio of their kinetic energy is

$$\begin{gathered} Given that two lead spheres have same linear momentum \\ {r}_1:r_2=1:2,p_1=p_2,K_1:K_2=? \\ We have \\ K=\frac{p^2}{2m}K\infty \frac{1}{m} \\ \frac{K_1}{K_2}=\frac{m_2}{m_1}=\frac{d{v}_2}{d{v}_1}\left[\therefore density =\frac{mass}{volume}\right] \\ \therefore \frac{K_1}{K_2}=\frac{v_2}{v_2}=\frac{{\displaystyle \frac{4}{3}}{{\mathrm{\pi r}}_2}^3}{{\displaystyle \frac{4}{3}}{{\mathrm{\pi r}}_1}^3}={\left(\frac{2}{1}\right)}^3=8 \\ \therefore K_1:K_2=8:1 \end{gathered}$$