The bob of simple pendulum executes SHM in water with a period T, while the period of oscillation of the bob is `T_(0)` in air. Neglecting frictional force of water and given that the density of the bob is `(4000)/(3)kgm^(-3)` , find the ration between T and `T_(0)`.

Options

A. `T=T_(0)`
B. `T=4T_(0)`
C. `T=2T_(0)`
D. `T=(T_(0))/(2)`

Answer

A

Explanation

Here, density of bob, `rho=(4)/(3)xx1000kgm^(-3)`and density of water, `sigma=1000kgm^(-3)`
In air, `T_(0)=2pisqrt((l)/(g))" "...(i)`
In water, `T=2pisqrt((l)/(g(1-(sigma)/(rho))))=2pisqrt((l)/(g(1-(3)/(4))))`
`=2(sqrt(2pi)sqrt((l)/(g)))=2T_(0)` (Using (i))