A small body of density `rho'` is dropped from rest at a height `h` into a lake of density `rho`, where `rhogtrho’`. Which of the following statements is/are correct if all dissipative effects are neglected (neglect viscosity)?

Options

A. The speed of the body just entering the lake is `sqrt(2gh)`
B. The body in the lake experiences upward acceleration equal to `{(rho//rho')-1}g`
C. The maximum depth to which the body sinks in the lake is `hrho'(rho-rho')`
D. The body does not come back to the surface of the lake.

Answer

A, B, C

Explanation

By the principle of conservation of energy, we have
`mgh=(1/2)mv^(2)`
or `v=sqrt(2gh)`
Since `rhogtp'`, the buoyant force of the lake is greater than the weight of the body. If a is the upward acceleation, then
`ma=rhogV-rho'gV` or `(rho'V)a=rhogV-rho'gV`
Cancelling `V` on the both the sides, we get
`a=g((rho-rho')/rho)=g(rho/rho'-1)`
Now `v^(2)=v_(0)^(2)+2ay`
where `v_(0)` is the velocity at `y=0, v=0` is the velocity at the maximm depth and `y` is the negative displacement from the surface on the maximum depth. With these get `v_(0)^(2)=-2ay`.
`-y(v_(0)^(2))/(2a)=(2gh)/(2g((rho-rho'))/(rho')=(hrho')/(rho-rho')`