A solid cube of side a and density `rho_(0)` floats on the surface of a liquid of density `rho`. If the cube is slightly pushed downward, then it oscillates simple harmonically with a period of

Options

A. `2pi sqrt((sigma a)/(rhog))`
B. `2pi sqrt((rhoa)/(sigma g))`
C. `2pi sqrt((rhog)/(sigma a))`
D. `2pi sqrt((sigma a)/(rhog))`

Answer

A

Explanation

As a is the side of cube `sigma` is its denesity.
Mass of cube is `a^(2) sigma` weight `= a^(2) sigma g`
let it be the height of cube immersed in liqied of density `p` in equilibrium then `F = a^(2) pg = Mg = a^(2) sigma g`
If it is pushed by `y` then the buoyant force
`F' = a^(2) ( h+y)pg`
Restoring force is `Delta F = F' - F = a^(2) (h+y) sigma g - a^(2) h sigma g`
`= a^(2) y rho g`
Restoring acceleration `= (Delta F)/(M) = - (a^(2)yrhog)/(M) = (a^(2)rhog)/(a^(2) sigma) y`
Motion is `S.H.M.`
`rArr T = 2pi sqrt((a^(2) sigma)/(a^(2) rhog)) = 2pi sqrt((a sigma)/(rhog))`