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A point mass M is released from rest and slides down a spherical bowl of radius R from a height H as shown in the figure below. The surface of the bowl is smooth (no friction). The velocity of the mass at the bottom of the bowl is
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- √gH
- √2gR
- √2gH
- 0
Correct Option: C
Loss in potential energy = Gain in kinetic energy
| Mgh = | MV² | |
| 2 |
V = √2gH
