Final answer:
The body's momentary rest occurs when the upward buoyant force balances the downward gravitational force, influenced by the densities of the body (d) and liquid (D). The correct formula for the time of rest involves the ratio of these densities as √(2h/g · d/D).
Step-by-step explanation:
The question at hand relates to the dynamics of a body falling through a fluid, with an application of physics principles concerning fluid resistance and buoyancy. Upon falling into a liquid, a body experiences a buoyant force in addition to the force of gravity. The time at which the body will be momentarily at rest can be found by considering the net force on the body as it enters the liquid. The initial velocity of the body as it enters the liquid is determined by its fall through air, which can be calculated using the equation v² = 2gh.
However, the scenario also involves the deceleration due to buoyant force upon entering the liquid. This buoyant force is dependent on the densities of the liquid (D) and the body (d), and the acceleration due to gravity (g). The body will be instantaneously at rest when the upward buoyant force equals the downward gravitational force on the body.
Therefore, considering buoyancy and gravity, the correct time after which the body will be instantaneously at rest is not directly given by the simple free-fall equation √(2h/g); instead, it involves the ratio of the densities. The correct option would be Option C: √(2h/g · d/D), which accounts for the buoyancy effects as the body decelerates in the liquid.