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A simple pendulum has time period T = 2s...

A simple pendulum has time period T = 2s in air. If the whole arrangement is placed in a non viscous liquid whose density is 1/2 times the density of bob. The time period in the liquid will be

A

`(2)/(sqrt2)s`

B

4s

C

`2sqrt2s`

D

`4sqrt4s`

Text Solution

Verified by Experts

The correct Answer is:
C
`T=2pisqrt((I)/(g_(eff)))`, `g_(eff)=(mg-f_(B))/(m)=(pvg-(p)/(2)vg)/(pvg)=(1-(1)/(2))g=(g)/(2)` [If is buoyant force]
Also `2pisqrt((I)/(g))=2` (given) `Rightarrow` `T'=2sqrt2s`
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