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A simple pendulum with a bob of mass m o...

A simple pendulum with a bob of mass m oscillates from A to C and back to A such that PB is H. IF the acceleration due to gravity is g, the velocity of bob as it passes through B is

A

zero

B

2g H

C

mgh

D

`sqrt((2gH))`

Text Solution

Verified by Experts

The correct Answer is:
D
`(1//2) mv_(B)^(2) = mg + 1, v_(B) = sqrt(2gH)`
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