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

Text Solution

Verified by Experts

The correct Answer is:
C
`g_("eff")=g-(g)/(3)=(2g)/(3)`
`T=2pisqrt((3L)/(2g))`.
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