Rotational Motion

JEE Main level, single correct, medium

254 of 280

Question 254 of 280

A solid cylinder and a solid sphere, having same mass MM and radius RR, roll down the same inclined plane from top without slipping. They start from rest. The ratio of velocity of the solid cylinder to that of the solid sphere, with which they reach the ground, will be

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Answer: (D) 1415\sqrt{\dfrac{14}{15}}

During pure rolling, the work done by the friction is zero. Therefore, we can apply conservation of mechanical energy.
mgh=12mv2+12I(vR)2mgh=\dfrac{1}{2}m{v}^{2}+\dfrac{1}{2}I{\left(\dfrac{v}{R}\right)}^{2}
⇒mgh=12(m+IR2)v2\Rightarrow mgh=\dfrac{1}{2}\left(m+\dfrac{I}{{R}^{2}}\right){v}^{2}
⇒v=2gh1+ImR2\Rightarrow v=\sqrt{\dfrac{2gh}{1+\dfrac{I}{m{R}^{2}}}}
For a solid cylinder ImR2=12\dfrac{I}{m{R}^{2}}=\dfrac{1}{2} and for a solid sphere, ImR2=25\dfrac{I}{m{R}^{2}}=\dfrac{2}{5}
vCylvSph=1+251+12\dfrac{{v}_{Cyl}}{{v}_{Sph}}=\sqrt{\dfrac{1+\dfrac{2}{5}}{1+\dfrac{1}{2}}}=1415=\sqrt{\dfrac{14}{15}}

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