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

Answer

B

Explanation

When wooden cube is floating in liquid, then
`Aldg=Alrhog ` …(i)
If wooden cube is pushed downward through a small distance `x`, then the restoring force the cube is
`-F=` upward force `-` weight of cubr
or `-F=A(l+x)rhog-Aldg=Arhogx`
`[:' Alrhog =Aldg]`
or `F=-Arhogx`....(ii)
i.e., `F prop x` and it is directed towards equilibrium position of floating cube.Therefore, it pushing force is removed, the cub will execute linear SHM.
Comparing (ii), with `F=-Kx,` we have,
spring factor `=Arhog.`
Here, Inertia factor `=Ald`
`:. ` time period, `T=2pisqrt((i n ertia fact o r )/(spri ng fact o r))`
or `T=2pisqrt((Ald)/(Arhog))=2pisqrt((ld)/(rhog))`