MHT CET · Physics · Rotational Motion
A body slides down a smooth inclined plane of inclination \(\theta\) and reaches the bottom with velocity V. If the same body is a ring which rolls down the same inclined plane then linear velocity at the bottom of plane is
- A \(\frac{\mathrm{V}}{\sqrt{2}}\)
- B V
- C 2 V
- D \(\frac{V}{2}\)
Answer & Solution
Correct Answer
(A) \(\frac{\mathrm{V}}{\sqrt{2}}\)
Step-by-step Solution
Detailed explanation
\(mgh = \frac{1}{2}mV^2\) (Sliding body) \(gh = \frac{1}{2}V^2\)
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