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 (kg)/m^3`, density of water = `10^3(kg)/m^3`)

Options

A. `T_(1) = 3T`
B. `T_(1) = 2T`
C. `T_(1) = T`
D. `T_(1) = (T)/(2)`

Answer

A

Explanation

According to question, time period of simple pendulum in water is given by
`T = 2pi sqrt((l)/(g_(eff))) " "["Symbols have their usual meaning and g"_(eff)="acceleration due to gravity in water"]`
As we know, `g_(eff)=g((sigma-rho)/(sigma))" "["Where,"sigma="density of bob,"rho="density of water"]`
`=9.8(((9)/(8)xx10^(3)-10^(3))/((9)/(8)xx 10^(3)))`
`= 9.8 (((9)/(8)-1)/((9)/(8)))=9.8((1)/(8)xx(8)/(9))`
`= (9.8)/(9)`
So, `T_(1)=2pi sqrt((l xx 9)/(9.8)) rArr T_(1) = 3T" "[because"Time period of simple pendulum in air,"T = 2pi sqrt((l)/(9.8))]`