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. `sqrt2`
B. 2
C. `2sqrt2`
D. none of these

Answer

B

Explanation

The effective acceleration of a bob in water `=g. = g (1-(d)/(D))` where d & D are the density of water & the bob respectively.
`(D)/(d)=` specific gravity of the bob.
Since the period of oscillation of the bob in air & water are given as
`T = 2pi sqrt((l)/(g)) & T. = 2pi sqrt((l)/(g.))`
`:. T//T. = sqrt((g.)/(g)) = sqrt((g (1-d//D))/(g)) = sqrt(1-(d)/(D)) = sqrt(1 - (1)/(s))`
Putting `T//T. = 1//sqrt2`, we obtain
`1//2 = 1-1//s`
`rArr (1)/(s) = (1)/(2) rArr s = 2`
Hence B is correct