JEE Main 2025 Apr 4 Shift 1, Physics Q21: Law of Rotation And Work-Energy Theorem

JEE Main2025Apr 4, Shift 1Physics

A circular ring and a solid sphere having same radius roll down on an inclined plane from rest without slipping. The ratio of their velocities when reached at the bottom of the plane is $\frac{x}{5}$ where $x =$ _____.

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Solution

For a body rolling without slipping down through a height $h$ , energy conservation gives $v = \frac{2 g h}{1 + k ^{2} / R ^{2}}$ where $I = m k^{2}$ .

Ring: $I = m R^{2} \Rightarrow k^{2} / R^{2} = 1$ . Solid sphere: $I = \frac{2}{5} m R^{2} \Rightarrow k^{2} / R^{2} = 2/5$ .

$\frac{v _{ring}}{v _{sphere}} = \frac{1 + 2/5}{1 + 1} = \frac{7/5}{2} = \frac{7}{10}$ .

The given form is $x /5$ , so $\frac{x}{5} = \frac{7}{10}$ , giving $x = 3.5$ . Rounded to the nearest integer, $x = 4$ .

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JEE Main 2025 Apr 4 Shift 1, Physics Q21: Law of Rotation And Work-Energy Theorem

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