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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, then the velocity of the bob as it passes through B is

A

mgH

B

`sqrt(2gH)`

C

`2gH`

D

Zero

Text Solution

Verified by Experts

The correct Answer is:
B
Law of conservation of energy
`(1)/(2)mv^(2)=mgHimplies v=sqrt(2gH)`.
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