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The time period of a simple pendulum of ...

The time period of a simple pendulum of length L as measured in an elevator descending with acceleration g / 3 is

A

`2pisqrt((3L)/(g))`

B

`pisqrt(((3L)/(g)))`

C

`2pisqrt(((3L)/(2g)))`

D

`2pisqrt(((3L)/(3g)))`

Text Solution

Verified by Experts

The correct Answer is:
C
`t=2pisqrt((L)/((T//m)))`
`T=Mg-(Mg)/(3)=(2Mg)/(3)Rightarrow (T)/(M)=((2g)/(3))`
`t=2pisqrt((3L)/(2g))`
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