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: Understand the situation.
A body is dropped from a height $h$ above a lake. It falls, hits the water, and sinks down to some deepest point. Then it floats back up. We must find the deepest depth it reaches. The lake is denser than the body, so the water pushes the body up.

Step 2: Pick the right idea.
We use the work and energy idea. The body starts at rest and, at the deepest point, it is again at rest for an instant. So the total change in kinetic energy is zero. This means the work done by all forces from start to deepest point adds up to zero.

Step 3: Name the quantities.
Let the body have volume $V$ and density $\rho$. Its mass is $m = V\rho$. The lake has density $\delta$. Let the deepest depth below the water surface be $d$.

Step 4: Work done by gravity.
Gravity pulls the body down for the whole trip, which is the fall $h$ plus the sinking $d$. So gravity does positive work:
\[ W_{g} = V\rho\, g\,(h + d) \]

Step 5: Work done by buoyancy.
The upward push of water (buoyancy) acts only while the body is inside the water, that is over the depth $d$. This push is $V\delta g$. It acts opposite to the motion while sinking, so its work is negative:
\[ W_{b} = -\,V\delta g\, d \]

Step 6: Add the works and set to zero.
Since the body starts and ends at rest:
\[ V\rho g (h + d) - V\delta g\, d = 0 \]
Divide every term by $Vg$:
\[ \rho h + \rho d - \delta d = 0 \]

Step 7: Solve for $d$.
Group the $d$ terms: $\rho h = d(\delta - \rho)$. So:
\[ d = \frac{h\rho}{\delta - \rho} \]
This matches option (4).
\[ \boxed{d = \frac{h\rho}{(\delta - \rho)}} \]