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6.System of Particles and Rotational Motion
hard
A hollow sphere is rolling on a plane surface about its axis of symmetry. The ratio of rotational kinetic energy to its total kinetic energy is $\frac{x}{5}$. The value of $x$ is________.
A
$2$
B
$5$
C
$4$
D
$3$
(JEE MAIN-2024)
Solution
$\frac{\frac{1}{2} \mathrm{I} \omega^2}{\frac{1}{2} \mathrm{I} \omega^2+\frac{1}{2} m v^2}=\frac{\left(\frac{1}{2}\right)\left(\frac{2}{3} m R^2\right) \omega^2}{\left(\frac{1}{2}\right)\left(\frac{2}{3} \mathrm{mR}^2\right) \omega^2+\frac{1}{2} \mathrm{~m}(\mathrm{R} \omega)^2}$
$=\frac{\frac{2}{3}}{\frac{2}{3}+1}=\frac{2}{5}$
$\mathrm{x}=2$
Standard 11
Physics
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