MHT CET · Physics · Rotational Motion
A disc of mass ' \(m\) ' and radius ' \(r\) ' rolls down an inclined plane of height ' h '. When it reaches the bottom of the plane, its rotational kinetic energy is ( \(\mathrm{g}=\) acceleration due to gravity)
- A \(\frac{\mathrm{mgh}}{3}\)
- B \(\frac{\mathrm{mgh}}{6}\)
- C \(\frac{\mathrm{mgh}}{2}\)
- D \(\frac{\mathrm{mgh}}{4}\)
Answer & Solution
Correct Answer
(A) \(\frac{\mathrm{mgh}}{3}\)
Step-by-step Solution
Detailed explanation
\(mgh = K_{trans} + K_{rot}\) \(I_{disc} = \frac{1}{2}mr^2\)
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