The bob of a simple pendulum performs SHM with period T in air and with period `T_(1)` in water. Relation between T and `T_1` is (neglect friction due to water, density of the material of the bob is = `9/8xx 10^3 kgm^3`, density of water = `1gcc^-1`)

Options

A. `T_1 = 3T`
B. `T_1 = 2T`
C. `T_1 = T`
D. `T_1 = T/2`

Answer

A

Explanation

Time period of simple pendulum in water is given by
`T=2pisqrt(l/g_(eff))`
[Symbols have their usual meanings and `g_(eff)` = acceleration due to gravity in water]
As we khow, `g_(eff)=g((sigma-rho)/sigma)`
[Where,`sigma` = density of bob, p = density of water]
`9.8((9/8xx10^3-10^3)/(9/8xx10^3))`
`9.8((9/8-1)/(9/8))=9.8(1/8xx8/9)`
`(9.8)/9`
So, `T_1=2pisqrt((lxx9)/9.8) Rightarrow T_1=3T`

[therefore Time period of simple pendulum in air, `T_1=2pisqrt((l)/9.8)` ]