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The time period of a simple pendulum ins...

The time period of a simple pendulum inside a stationery lift is `T`. If the lift accelerates upwards uniformly with `(g)/(4)`, then its time period would be

A

`2sqrt(5)T`

B

`(2T)/(sqrt(5))`

C

`2T`

D

`(T)/(2)`

Text Solution

Verified by Experts

The correct Answer is:
B
`(T_(1))/(T_(2))=sqrt((g+a)/(g))`
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