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)/(rhog))`
B. `2pisqrt((lrho)/(dg))`
C. `2pisqrt((ld)/((rho-d)g)`
D. `2pisqrt((lrho)/((rho-d)g)`

Answer

A

Explanation

At equilibrium
`F_(b) = mg`
`rhoAl_(0)g = dAlg….(i)`

Restoring force, `F = mg - F_(b)'`
`F = mg - rhoA(l_(0) + x)g`
`dAla = dAlg - rhoAl_(0)g - rhogAx`
`a = -(rhog)/(dl) xx`
`omega = sqrt((rhog)/(dl))`
`T = 2pisqrt((ld)/(rhog)) ......(i)`