A body of density `rho` is dropped from rest from height h (from the surface of water) into a lake of density of water `sigma(sigmagt rho)`. Neglecting all disspative effects, the acceleration of body while it is in the take is

Options

A. `g((sigma)/(rho)-1)` upwards
B. `g((sigma)/(rho)-1)` downwards
C. `g((sigma)/(rho))` upwards
D. `g((sigma)/(rho))` downwards

Explanation

To solve the problem of determining the acceleration of a body of density \( \rho \) when it is dropped into a lake of density \( \sigma \) (where \( \sigma > \rho \)), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Body**:
- When the body is submerged in the water, two main forces act on it:
- The weight of the body acting downwards: \( F_{\text{weight}} = V \cdot \rho \cdot g \)
- The buoyant force acting upwards: \( F_{\text{buoyant}} = V \cdot \sigma \cdot g \)
- Here, \( V \) is the volume of the body, \( g \) is the acceleration due to gravity, \( \rho \) is the density of the body, and \( \sigma \) is the density of the water.

2. **Set Up the Equation of Motion**:
- According to Newton's second law, the net force acting on the body is equal to the mass of the body times its acceleration:
\[ F_{\text{net}} = F_{\text{buoyant}} - F_{\text{weight}} = m \cdot a \]
- The mass \( m \) of the body can be expressed as \( m = V \cdot \rho \).

3. **Substitute the Forces into the Equation**:
- Substitute the expressions for the forces into the equation:
\[ V \cdot \sigma \cdot g - V \cdot \rho \cdot g = V \cdot \rho \cdot a \]

4. **Simplify the Equation**:
- Factor out \( V \cdot g \) from the left side:
\[ V \cdot g (\sigma - \rho) = V \cdot \rho \cdot a \]
- Cancel \( V \) from both sides (assuming \( V \neq 0 \)):
\[ g (\sigma - \rho) = \rho \cdot a \]

5. **Solve for Acceleration \( a \)**:
- Rearranging the equation gives:
\[ a = \frac{g (\sigma - \rho)}{\rho} \]

6. **Final Expression for Acceleration**:
- Thus, the acceleration of the body while it is in the lake is:
\[ a = g \left( \frac{\sigma}{\rho} - 1 \right) \]

### Conclusion:
The acceleration of the body while it is in the lake is directed upwards and is given by the formula:
\[ a = g \left( \frac{\sigma}{\rho} - 1 \right) \]