A pendulum has time period `T` in air when it is made to oscillate in water it acquired a time period `T = sqrt(2)T` The density of the pendulum bob is equal to (density) of water `= 1)`

Options

A. `sqrt(2)`
B. `2`
C. `2sqrt(2)`
D. None of these

Explanation

The effective acceleration of a bob in water `=g'=g(1-(d)/(D))`, where `d` and `D` are the density of water and the bob respectively.
`(D)/(d)=`specific gravity of the bob.
Since the period of oscillation of the bob in air and water are given as
`T=2pisqrt((l)/(g))` and `T'=2pisqrt((l)/(g'))`
`:. T//T'=sqrt((g')/(g))=sqrt((g(1-d//D))/(g))=sqrt(1-(d)/(D))=sqrt(1-(1)/(s))`
Putting `T//T'=1sqrt(2)`, we obtain
`1//2=1-1//s`
`rArr (1)/(s)=(1)/(2)rArrs-2`