A wooden cube (density of wood `'d'`) of side `'l'` flotes in a liquid of density `'rho'` with its upper and lower surfaces horizonta. If the cube is pushed slightly down and released, it performs simple harmonic motion of period `'T'`. Then, `'T'` is equal to :-

Options

A. `2pisqrt(ld)/(pg)`
B. `2pisqrt(lp)/(dg)`
C. `2pisqrt(ld)/((p-d)g)`
D. `2pisqrt(ld)/((p-d)g)`

Answer

C

Explanation

At equilibrium position
`F_(B) = Mg`
`Axrhog = (Algd) " "…(i)`
After pushing slightly down
Restoring force ` = F_(B) -Mg`
` Ma = A(x + y) rhog - Mg`
` = A(x)rhog + Ay rhog - Algd`
`Ma = A(rhog)y`
`a = ((Arhog)/(M))y " "because a = omega^(2)y`
`omega = sqrt((Arhog)/(Ald))`
`T = 2pi sqrt((ld)/(rhog))`