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Q.
A body rolls down an inclined plane. If its kinetic energy of rotation is $40\%$ of its kinetic energy of translation motion, then the body is
KEAMKEAM 2015System of Particles and Rotational Motion
Solution:
The ratio of kinetic energies
$\frac{ KE _{ rot }}{ KE _{\text {trans }}}=\frac{\frac{1}{2} m v^{2}\left(\frac{K^{2}}{R^{2}}\right)}{\frac{1}{2} m v^{2}} =\frac{K^{2}}{R^{2}}=0.4=\frac{2}{5}$
$\Rightarrow K^{2} =\frac{2}{5} R^{2}$
$\Rightarrow m K^{2}=\frac{2}{5} m R^{2}$
$\Rightarrow $ Body is a solid sphere