A small spherical ball of radius `r` is released from its completely submerged Position (as shown in the figure) in a liquid whose density varies with height `h` (measured from the bottom) as `rho_(L)=rho_(0)[4-(3h//h_(0))]`.The density of the ball is (`5//2)rho_(0)`. The height of the vessel is `h_(0)=12//pi^(2)` Consider `r lt lt h_(0)` and `g = 10 m//s^(2)`.

Where will the ball come in the equilibrium?

Options

A. At a depth `h_(0)//2` from top
B. At the bottom of the vessel
C. At a depth `3h_(0)//4` from top
D. The ball will never be at equilirbium

Answer

A

Explanation

Let the net force on this particle be zero at height `h`. Then weight
`=` upthrust (at `h`)
`implies4/3pir^(3)g5/2rho_(0)-4/3pir^(2)grho_(0)[4-(3h)/(h_(0))]`
`implies (3h)/(h_(0))=4-5/2=3/2impliesh=(h_(0))/2`………..i
Therefore, at `h=(h_(0))/2`, the equilibrium position, the particle has a speed in downward direction due to which it will move downward. Let at a further depth `x` from `(h_(0))/2`.

`F_("net")=` upthrust `-` weight
`=4/3pir^(3)grho_(0)[4-(3((h_(0))/2-x))/(h_(0))]-4/3pir^(3)g5/2rho_(0)`
`=4/3pir^(3)rho_(0)g[4-3/2+(3x)/(h_(0))-5/2]=(4pir^(3)rho_(0)g)/h_(0)x`
`:. F=(4pir^(3)rho_(0)g)/h_(0)x`
Since the net force on the and displacement of ball from its mean position are oppositely directed the particle executes `SHM` about its mean position i.e. `h=(h_(0))/2`
`:. F=ma=-(4pir^(3)rho_(0)g)/(h_(0))x`
`=4/3pir^(3)5/2rho_(0)a=-(4pir^(3)rho_(0)g)/(h_(0))x`
`impliesa=-(6g)/(5h_(0))x=-(12/(h_(0)))x=-omega^(2)x`
where `omega=sqrt(12/(h_(0)))=ssqrt(12/(12/pir^(2)))=pi`
Hence, time taken by the particle to reach the bottom is
`t=T/2 =1/2(2pi)/(omega)=(pi)/(omega)=(pi)/(pi)=1s`