A stone of density 2000 kg `m^(-3)` completely immersed in a lake is allowed to sink from rest . If the effect of friction is neglected , than after 4 seconds , the stone will reach a depth of

Options

A. 78.4 m
B. 39.2 m
C. 19.6 m
D. 9.8 m

Answer

To solve the problem step by step, we will follow the principles of physics related to buoyancy and motion.

### Step 1: Understand the forces acting on the stone
When the stone is submerged in water, two main forces act on it:
1. The gravitational force (weight) acting downward, which is given by \( F_g = mg \).
2. The buoyant force acting upward, which is given by \( F_b = \rho_{water} \cdot V \cdot g \).

### Step 2: Set up the equation of motion
According to Newton's second law, the net force acting on the stone can be expressed as:
\[ F_{net} = F_g - F_b = ma \]
Where:
- \( m \) is the mass of the stone,
- \( a \) is the acceleration of the stone.

### Step 3: Express the forces in terms of density and volume
The weight of the stone can be expressed as:
\[ F_g = \rho_{stone} \cdot V \cdot g \]
The buoyant force can be expressed as:
\[ F_b = \rho_{water} \cdot V \cdot g \]

### Step 4: Substitute the expressions into the equation of motion
Substituting the expressions for \( F_g \) and \( F_b \) into the equation of motion gives:
\[ \rho_{stone} \cdot V \cdot g - \rho_{water} \cdot V \cdot g = \rho_{stone} \cdot V \cdot a \]

### Step 5: Simplify the equation
Since the volume \( V \) is common in all terms, we can cancel it out:
\[ \rho_{stone} \cdot g - \rho_{water} \cdot g = \rho_{stone} \cdot a \]
This simplifies to:
\[ (\rho_{stone} - \rho_{water}) \cdot g = \rho_{stone} \cdot a \]

### Step 6: Solve for acceleration \( a \)
Rearranging the equation gives:
\[ a = \frac{(\rho_{stone} - \rho_{water}) \cdot g}{\rho_{stone}} \]
Substituting the given values:
- \( \rho_{stone} = 2000 \, \text{kg/m}^3 \)
- \( \rho_{water} = 1000 \, \text{kg/m}^3 \)
- \( g = 9.8 \, \text{m/s}^2 \)

We find:
\[ a = \frac{(2000 - 1000) \cdot 9.8}{2000} = \frac{1000 \cdot 9.8}{2000} = \frac{9.8}{2} = 4.9 \, \text{m/s}^2 \]

### Step 7: Calculate the distance fallen in 4 seconds
Using the kinematic equation for distance:
\[ s = ut + \frac{1}{2} a t^2 \]
Since the stone starts from rest, \( u = 0 \):
\[ s = 0 + \frac{1}{2} \cdot 4.9 \cdot (4^2) \]
\[ s = \frac{1}{2} \cdot 4.9 \cdot 16 \]
\[ s = 2.45 \cdot 16 = 39.2 \, \text{m} \]

### Conclusion
After 4 seconds, the stone will have reached a depth of **39.2 meters**.

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