A wooden cube (density of wood d) of side l floats in a liquid of density with its upper and lower surfaces horizontal. If the cube is pushed slightly down and released, it performs simple harmonic motion period, T. Then, T is equal to :

Options

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

Answer

B

Explanation

By using `T = 2 pisqrt((m)/(Arhog))`
Where `m = l^(3)d` and `A = l^(2)`
`T = 2pisqrt((l^(3)d)/(l^(2)rhog)) rArr T = 2pi sqrt((ld)/(rhog))`