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

Options

A. `(2)/(sqrt2)s`
B. 4s
C. `2sqrt2s`
D. `4sqrt4s`

Answer

C

Explanation

`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`