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A wooden cube (density of wood 'd') of s...

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 :-

A

`2pisqrt((lrho)/(dg))`

B

`2pisqrt((ld)/((rho-d)g))`

C

`2pisqrt((lrho)/((rho-d)g))`

D

`2pisqrt((ld)/(rhog))`

Text Solution

Verified by Experts

The correct Answer is:
D
`T=2pisqrt((hrho)/(rho,g))`
Replacing h by l, `rho` by d and `rhol ` by `rho`, we get
`T=2pisqrt((ld)/(rhog))`.
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