The bob of a simple pendulum executes SHM in water with a period t. The period of oscillation of the bob in air is `t_(0)` . What is the relation between t ant `t_(0)` ?
Given, density of the bob is `(4000)/(3) kg m^(-3)` . Neglect frictional force of water.

Options

A. `t=t_(0)`
B. `t=4t_(0)`
C. `t=2t_(0)`
D. `t=t_(0)//2`

Answer

C

Explanation

In air, the time period of the bob is `t_(0)=2pisqrt((l)/(g))`
In water apparent weight
`mg '=(4)/(3)pir^(3) p_(s)`,`g-(4)/(3)pir^(3)p_(w)g`
mg =`4/3pir^(3) Ps g(1-(pv)/Ps)=mg(1-(3)/(4))=mg/4 implies g'=g//4`
In water, the time period of the bob `t=2pisqrt(1/g)`,=`2pisqrt(41/g)=2t_(0)`