Question

A body of density $\rho$ is dropped from (at rest) height $h$ into a lake of density $\delta$ ($\delta > \rho$). The maximum depth to which the body sinks before returning to float on the surface is [Neglect all dissipative forces]

• A. $\frac{(\delta - \rho)^2 h}{\rho}$
• B. $\frac{2h\rho}{(\delta - \rho)}$
• C. $\frac{h\rho}{2(\delta - \rho)}$
• D. $\frac{h\rho}{(\delta - \rho)}$

Hint:

The Work-Energy Theorem approach is much faster and less error-prone here than calculating velocity at the surface and then setting up kinematic equations for retardation inside the fluid!

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
A body falls from a height $h$ and gains kinetic energy before hitting the water. Once in the water, it experiences a net upward buoyant force (since $\delta > \rho$) which eventually brings it to a stop at a maximum depth $d$.

Step 2: Key Formula or Approach:
We can use the Work-Energy Theorem. The total work done by gravity from height $h$ down to depth $d$ must be entirely dissipated by the negative work done by the buoyant force acting over depth $d$.

Step 3: Detailed Explanation:
Let $V$ be the volume of the body and $m = V\rho$ be its mass.
Total downward work done by gravity over the entire distance $(h + d)$:
$$W_g = mg(h + d) = V\rho g(h + d)$$
Upward buoyant force acts only while the body is in the water (over distance $d$). The work done by buoyancy is negative (opposite to motion):
$$W_b = -(\text{Buoyant Force}) \times d = -(V\delta g) \times d$$
Since the body starts at rest and comes to a momentary stop at maximum depth, the change in kinetic energy is zero ($\Delta K = 0$).
$$W_{\text{net}} = W_g + W_b = 0$$
$$V\rho g(h + d) - V\delta gd = 0$$
Divide the entire equation by $Vg$ to simplify:
$$\rho(h + d) - \delta d = 0$$
$$\rho h + \rho d - \delta d = 0$$
Move the $d$ terms to one side:
$$\rho h = \delta d - \rho d$$
$$\rho h = d(\delta - \rho)$$
Solve for $d$:
$$d = \frac{h\rho}{\delta - \rho}$$

Step 4: Final Answer:
The maximum depth is $\frac{h\rho}{(\delta - \rho)}$, matching option (D).