A streamlined body falls through air from a height h on the surface of a liquid. If d and (D > d) represents the densities of the material of the body and liquid respectively, then the time after which the body will be instantaneously at rest, is

The correct answer is Option 4: $\frac{d}{D-d}\sqrt{\frac{2h}{g}}$.

Let us break down the motion of the body into two parts: the free fall in the air, and the motion inside the liquid until the body comes to rest.

Part 1: Motion in the air (Free Fall)
The body falls from a height _h under gravity. Let _t1 be the time of fall in air and _v be the velocity of the body just before it hits the liquid surface.
Using the equations of motion under constant acceleration _g:
$h=\frac{1}{2}g{t}_1^2$
This gives the time of fall in air as:
$t_1=\sqrt{\frac{2h}{g}}$
The velocity of the body just before striking the liquid surface is:
$v=g{t}_1=\sqrt{2gh}$

Part 2: Motion inside the liquid
Let _V be the volume of the body. Since the density of the material of the body is _d, its mass is _{m = V · d}.
When the body is inside the liquid, two main vertical forces act on it (neglecting viscous drag as the body is streamlined):
1. Downward gravitational force: $W=mg=Vdg$
2. Upward buoyant force (upthrust): $F=VDg$, where _D is the density of the liquid.
Since _{D > d}, the upward buoyant force is greater than the downward gravitational force, resulting in a net upward retarding force.
The net upward force on the body is:
$F_{\text{net}}=F-W=VDg-Vdg=V(D-d)g$
Therefore, the upward deceleration _a of the body inside the liquid is:
$a=\frac{F_{\text{net}}}{m}=\frac{V(D-d)g}{Vd}=\frac{(D-d)}{d}g$

Let _t2 be the time after entering the liquid during which the body comes to instantaneous rest.
Using the equation of motion:
$v_{\mathrm{final}}=u-a{t}_2$
Since the final velocity _vfinal = 0 and the initial velocity upon entering the liquid is _{u = v}:
$0=v-a{t}_2$
$t_2=\frac{v}{a}$
Substitute the values of _v and _a:
$t_2=\frac{g{t}_1}{\frac{(D-d)}{d}g}=\frac{d}{D-d}{t}_1$
Substituting the value of _t1:
$t_2=\left(\frac{d}{D-d}\right)\sqrt{\frac{2h}{g}}$

The question asks for the time after which the body will be instantaneously at rest. This refers to the duration of its motion inside the liquid, which is _t2.
Therefore, the time after entering the liquid until it comes to rest is:
$t=\left(\frac{d}{D-d}\right)\sqrt{\frac{2h}{g}}$