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

`T=2pi sqrt((l)/(g)) and T_(1)=2pi sqrt((l)/(g'))`
where `g'=g(1-(sigma)/(rho)) or (g')/(g)=1-(sigma)/(rho)`
`therefore (T)/(T_(1))=sqrt((g')/(g))=sqrt(1-(sigma)/(rho))`
where `sigma`= density of liquid (water)`=1g//c c =10^(3) kg//m^(3) and rho`=density of the material of the bob
`=(9)/(8)xx10^(3)kg//m^(3)`
`=(T)/(T_(1))=sqrt(1-(10^(3))/((9)/(8)xx10^(3)))=sqrt(1-(8)/(9))=sqrt((1)/(9))=(1)/(3)`
`therefore T_(1)=3T`