1.

A wooden cube (density of wood 'd') of side 'l' floats in a liquid of density 'rho' with its upper and lower surfaces horizontal. If the cube is pushed slightly down and released, it performs simple harmonic motion of period 'T'. Then, 'T' is equal to :

Answer»

`2pi SQRT((ld)/(rho g))`
`2pi sqrt((L rho)/(dg))`
`2pi sqrt((ld)/((rho-d)g))`
`2pi sqrt((ld)/((rho-d)g))`

Solution :
Restoring force,
`F=mg-F_(B).` br> `F="mg"-rhoA(l_(0)+x)g`
`DA l a=dAlg-rho A l_(0)g-rho g A x`
`a=-(rho g)/(d l)x`
`omega=sqrt((rho g)/(d l))`
`T=2pi sqrt((l d)/(rho g))`
So, correct CHOICE is (a).


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