Rotational Motion

JEE Main level, single correct, medium

274 of 280

Question 274 of 280

A solid spherical ball is rolling on a frictionless horizontal plane surface about its axis of symmetry. The ratio of rotational kinetic energy of the ball to its total kinetic energy is

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Answer: (C) 27\dfrac{2}{7}

Under pure rolling condition: v=ωRv=\omega R
Translational K.E=12mv2K.E=\dfrac{1}{2}m{v}^{2}
Rotational K.E=12×25mR2(vR)2=15mv2K.E=\dfrac{1}{2}\times \dfrac{2}{5}m{R}^{2}{\left(\dfrac{v}{R}\right)}^{2}=\dfrac{1}{5}m{v}^{2}
Total K.E=12mv2+15mv2=710mv2K.E=\dfrac{1}{2}m{v}^{2}+\dfrac{1}{5}m{v}^{2}=\dfrac{7}{10}m{v}^{2}
So, RotationalK⋅ETotalK⋅E=15mv2710mv2=27\dfrac{RotationalK\cdot E}{TotalK\cdot E}=\dfrac{\dfrac{1}{5}m{v}^{2}}{\dfrac{7}{10}m{v}^{2}}=\dfrac{2}{7}

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