A rectangular block of mass m and area of cross-section A floats in a liquid of density `rho`. If it is given a small vertical displacement from equilibrium, it undergoes oscillation with a time period T. Then

Options

A. `Tprop(1)/(sqrt(m))`
B. `Tpropsqrt(rho)`
C. `Tprop(1)/(sqrt(A))`
D. `Tprop(1)/(rho)`

Answer

C

Explanation

Refer figure, Let h to be the height of block immersed in liquid, when the block si floating.
`thereforemg=Ahrhog` . . . (i)
If the block is given a vertical displacement y, then the effective restoring force is
`F=-[A(h+y)rhog-mg]=-[A(h+y)rhog-Ahrhog]`
`=-Arhogy` . . . (ii)
i.e., `F propy and -ve ` sign shown that F is directed towards its equilibrium position of block. so, if the block is left free, it will execute SHM
for SHM, `F=-m omega^(2)y` . . . (iii)
Comparing (ii) and (iii), we get
`A rhog=momega^(2),omega=sqrt((Arhog)/(m))`
`therefore`Time period, `T=(2pi)/(omega)=2pisqrt((m)/(Arhog))impliesTprop(1)/(sqrt(A))`.