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JEE Mains · Physics · STD 11 - System of particles and rotational motion
  • JEE Mains 2021
  • Medium
  • Numerical

Two bodies, a ring and a solid cylinder of same material are rolling down without slipping an inclined plane. The radii of the bodies are same. The ratio of velocity of the centre of mass at the bottom of the inclined plane of the ring to that of the cylinder is \(\frac{\sqrt{x}}{2}\). Then, the value of \(x\) is .... .

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Answer & Solution

Correct Answer

(B) \(3\)

Step-by-step Solution

Detailed explanation

\({I}\) in both cases is about point of contact \(IAOR,\)

Ring,

\(m g h=\frac{1}{2} I \omega^{2}\)

\(m g h=\frac{1}{2}\left(2 m R^{2}\right) \frac{v_{R}^{2}}{R^{2}}\)

\(V_{R}=\sqrt{g h}\)

Solid, cylinder

\(m g h=\frac{1}{2} I \omega^{2}\)

\(m g h=\frac{1}{2}\left(\frac{3}{2} m R^{2}\right) \frac{v_{c}^{2}}{R^{2}}\)

\(V_{R}=\sqrt{\frac{4 g h}{3}}\)

\(\frac{v_{R}}{v_{c}}=\frac{\sqrt{3}}{2}\)

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