JEE Main · 2022 · 25 Jul 2022 · Shift 1 · Rotational Mechanics

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A solid cylinder and a solid sphere, having same mass M and radius R, 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 :

(A) $\frac{5}{3}$
(B) $\frac{4}{5}$
(C) $\frac{3}{5}$
(D) $\frac{14}{15}$ ✓ Correct

Answer: (D)

Solution

For a rolling object down an incline, the velocity at the bottom is:

v = \frac{2 g H}{1 + k ^{2} / R ^{2}}

where

k

is the radius of gyration.

For a solid cylinder:

k^{2} / R^{2} = 1/2

For a solid sphere:

k^{2} / R^{2} = 2/5

The ratio:

\frac{v _{cylinder}}{v _{sphere}} = \frac{1 + 2/5}{1 + 1/2} = \frac{7/5}{3/2} = \frac{7}{5} \times \frac{2}{3} = \frac{14}{15}

JEE Main · 2022 · 25 Jul 2022 · Shift 1 · Single correct option

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