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

A cord is wound round the circumference of wheel of radius \(r\). The axis of the wheel is horizontal and moment of inertia about it is \(I\). A weight \(mg\) is attached to the end of the cord and falls from rest. After falling through a distance \(h\), the angular velocity of the wheel will be

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Answer & Solution

Correct Answer

(B) \(\sqrt {\frac{{2mgh}}{{I + m{r^2}}}} \)

Step-by-step Solution

Detailed explanation

According to law of conservation of energy

\(mgh = \frac{1}{2}(I + m{r^2}){\omega ^2}\)

\( \Rightarrow \) \(\omega  = \sqrt {\frac{{2mgh}}{{I + m{r^2}}}} \).

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