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.

Explanation

To solve the problem step by step, we will analyze the motion of a small body of density \( \rho' \) dropped from a height \( h \) into a lake of density \( \rho \) (where \( \rho > \rho' \)). We will determine the correct statements regarding the motion of the body in the lake, neglecting all dissipative effects.

### Step 1: Determine the velocity just before entering the lake
The body is dropped from rest, so its initial velocity \( u = 0 \). We can use the equation of motion to find the velocity \( v \) just before it enters the lake:

\[ v^2 = u^2 + 2gh \]

Substituting \( u = 0 \):

\[ v^2 = 0 + 2gh \implies v = \sqrt{2gh} \]

### Step 2: Analyze the forces acting on the body in the lake
Once the body enters the lake, two forces act on it:
1. The weight of the body \( W = \rho' V g \) (where \( V \) is the volume of the body).
2. The buoyant force \( F_b = \rho V g \).

The net force acting on the body when it is submerged is:

\[ F_{net} = W - F_b = \rho' V g - \rho V g = V g (\rho' - \rho) \]

Since \( \rho' < \rho \), the net force is downward.

### Step 3: Determine the upward acceleration of the body
Using Newton's second law, we can express the net force in terms of acceleration \( a \):

\[ F_{net} = m a \implies V g (\rho - \rho') = \rho' V a \]

Dividing both sides by \( V \):

\[ g (\rho - \rho') = \rho' a \implies a = \frac{g (\rho - \rho')}{\rho'} \]

### Step 4: Determine how deep the body will sink
To find the depth \( h' \) to which the body sinks, we can use the kinematic equation:

\[ v^2 = u^2 + 2a h' \]

At the maximum depth, the final velocity \( v = 0 \) (the body comes to rest momentarily), and the initial velocity \( u \) is the velocity just before entering the lake \( \sqrt{2gh} \):

\[ 0 = (2gh) + 2 \left(-\frac{g (\rho - \rho')}{\rho'}\right) h' \]

Solving for \( h' \):

\[ 0 = 2gh - \frac{2g (\rho - \rho')}{\rho'} h' \implies h' = \frac{h \rho'}{\rho - \rho'} \]

### Step 5: Conclusion
Based on the analysis, we can conclude the following:
1. The velocity just before entering the lake is \( v = \sqrt{2gh} \).
2. The upward acceleration of the body in the lake is \( a = \frac{g (\rho - \rho')}{\rho'} \).
3. The depth to which the body sinks is \( h' = \frac{h \rho'}{\rho - \rho'} \).
4. The body will eventually rise back to the surface because it is less dense than the water.

### Summary of Correct Statements
- Statement 1: Correct (velocity just before entering the lake).
- Statement 2: Correct (upward acceleration).
- Statement 3: Correct (depth of sinking).
- Statement 4: Incorrect (the body will rise back to the surface).