A body of density `rho` is dropped from reat from a height h into a lake of density `sigma(sigmagtrho)`. The maximum depth the body sinks inside the liquid is (neglect viscous effect of liquid)

Options

A. `(hrho)/(sigma-rho)`
B. `(h sigma)/(sigma-rho)`
C. `(h rho)/(sigma)`
D. `(h sigma)/(rho)`

Explanation

To solve the problem of finding the maximum depth a body of density \( \rho \) sinks into a lake of density \( \sigma \) after being dropped from a height \( h \), we can follow these steps:

### Step 1: Calculate the velocity of the body just before it hits the surface of the lake.
When the body is dropped from height \( h \), it accelerates due to gravity. The velocity \( v \) just before it hits the lake can be calculated using the equation of motion:
\[ v = \sqrt{2gh} \]

### Step 2: Analyze the forces acting on the body when it is submerged.
Once the body enters the water, two main forces act on it:
- The buoyant force (upward) given by \( F_b = \sigma V g \), where \( V \) is the volume of the body.
- The weight of the body (downward) given by \( F_w = \rho V g \).

### Step 3: Determine the net force and acceleration.
The net force \( F \) acting on the body when it is submerged is:
\[ F = F_b - F_w = \sigma V g - \rho V g = V g (\sigma - \rho) \]
Using Newton's second law, the acceleration \( a \) of the body can be expressed as:
\[ ma = V g (\sigma - \rho) \]
Since \( m = \rho V \), we can substitute this into the equation:
\[ \rho V a = V g (\sigma - \rho) \]
Dividing both sides by \( V \) (assuming \( V \neq 0 \)):
\[ a = \frac{g (\sigma - \rho)}{\rho} \]

### Step 4: Use kinematic equations to find the maximum depth.
The body will continue to sink until its velocity becomes zero. We can use the kinematic equation:
\[ v^2 = u^2 + 2as \]
Where:
- \( v = 0 \) (final velocity at maximum depth),
- \( u = \sqrt{2gh} \) (initial velocity just before entering the water),
- \( a = -\frac{g (\sigma - \rho)}{\rho} \) (negative because it is deceleration),
- \( s = h_{max} \) (the maximum depth we want to find).

Substituting the values:
\[ 0 = (2gh) - 2\left(-\frac{g (\sigma - \rho)}{\rho}\right)h_{max} \]
This simplifies to:
\[ 2gh = \frac{2g (\sigma - \rho)}{\rho} h_{max} \]

### Step 5: Solve for \( h_{max} \).
Cancelling \( 2g \) from both sides:
\[ h = \frac{(\sigma - \rho)}{\rho} h_{max} \]
Rearranging gives:
\[ h_{max} = \frac{\rho h}{\sigma - \rho} \]

### Final Answer:
The maximum depth \( h_{max} \) that the body sinks into the lake is:
\[ h_{max} = \frac{\rho h}{\sigma - \rho} \]

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