The density `rho` of a liquid varies with depth h from the free surface as `rho=kh`. A small body of density `rho_(1)` is released from the surface of liquid. The body will

Options

A. come to a momentary rest at a depth `2rho_(1)//k` from the free surface
B. execute S.H.M. about a point at a depth `rho_(1)//k` from the surface
C. execute S.H.M. of amplitude `rho_(1)//k`
D. all of these

Answer

D

Explanation

Let V be the volume of body which is released on the surface of liquic. When body reaches at depth x from the surface of liquid, then upward thrust on body due to liquid `=V(kx)g`
Weight of body `=Vrho_(1)g`
At equilibrium , `V(kx)g=Vrho_(1)g or x=rho_(1)//k`
If the body comes to rest at depth h, then work done on body by the upward thrust dur to liquid
`=` change in potential energy of body
`:. -int_(0)^(h)V(kx)g dx=-Vrho_(1) gh or h =(2rho_(1))/(k)`
Now, the body will move upwards towards equilibrium position due to displacemtn x from equilibrium position . Hence the body will execute S.H.M. Choice (d) is right.