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JEE Main 20245 AprilEvening Shift
Exam
JEE Main
Year
2024
Date
5 April
Shift
Evening Shift
Subject
Physics
Chapter
Rotational Motion

JEE Main 2024 · 5 April · Evening Shift · Physics · Rotational Motion

A hollow sphere is rolling on a plane surface about its axis of...

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 x5\frac{x}{5}. The value of xx is _____ .

Correct answer

0

Step-by-step solution

12Iω212Iω2+12mv2=(12)(23mR2)ω2(12)(23mR2)ω2+12 m(Rω)2=2323+1=25x=2\begin{aligned} & \frac{\frac{1}{2} \mathrm{I} \omega^2}{\frac{1}{2} \mathrm{I} \omega^2+\frac{1}{2} \mathrm{mv}^2}=\frac{\left(\frac{1}{2}\right)\left(\frac{2}{3} \mathrm{mR}^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\end{aligned}
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