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. `2 pi sqrt((I rho)/((rho - d)g))`
B. `2 pi sqrt((Id)/(rhog))`
C. `2pi sqrt((Irho)/(dg))`
D. `2 pi sqrt((Id)/((rho-d)g))`

Answer

B

Explanation

Let at any instant, cube is at a depth x from the equilibrium position, then net force acting on the cube = upthrust on the portion of length x,
`F = -rhol^(2)xg =-rhol^(2)gx" "...(i)`

Negative sign shows that, force is opposite to x. Hence equation of SHM `F = - kx" "...(ii)`
Comparing Eqs. (i) and (ii), we get, `k = rhol^(2)g`
`T = 2pi sqrt((m)/(k))=2pi sqrt((l^(3)d)/(rhol^(2)g))=2pi sqrt((ld)/(rhog))`