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 analyze the forces acting on the stone and apply the equations of motion.

### Step 1: Identify the forces acting on the stone
The stone is subjected to two main forces:
- The weight of the stone (downward): \( F_g = mg \)
- The buoyant force (upward): \( F_b = \rho_{water} \cdot V \cdot g \)

Where:
- \( m \) is the mass of the stone
- \( g \) is the acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \))
- \( \rho_{water} \) is the density of water (approximately \( 1000 \, \text{kg/m}^3 \))
- \( V \) is the volume of the stone

### Step 2: Write the equation of motion
According to Newton's second law, the net force acting on the stone is equal to the mass of the stone times its acceleration:
\[ F_{net} = mg - F_b = ma \]
Substituting the expressions for weight and buoyant force:
\[ mg - \rho_{water} \cdot V \cdot g = ma \]

### Step 3: Express mass in terms of density and volume
The mass of the stone can be expressed as:
\[ m = \rho_{stone} \cdot V \]
Where \( \rho_{stone} = 2000 \, \text{kg/m}^3 \).

### Step 4: Substitute and simplify the equation
Substituting \( m \) into the equation gives:
\[ \rho_{stone} \cdot V \cdot g - \rho_{water} \cdot V \cdot g = \rho_{stone} \cdot V \cdot a \]
Factoring out \( V \) (assuming \( V \neq 0 \)):
\[ (\rho_{stone} - \rho_{water}) \cdot g = \rho_{stone} \cdot a \]

### Step 5: Solve for acceleration \( a \)
Rearranging the equation:
\[ a = \frac{(\rho_{stone} - \rho_{water}) \cdot g}{\rho_{stone}} \]
Substituting the known values:
\[ a = \frac{(2000 - 1000) \cdot 9.8}{2000} = \frac{1000 \cdot 9.8}{2000} = \frac{9.8}{2} = 4.9 \, \text{m/s}^2 \]

### Step 6: Use the equations of motion to find the depth after 4 seconds
Using the equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
Where:
- \( u = 0 \) (initial velocity)
- \( a = 4.9 \, \text{m/s}^2 \)
- \( t = 4 \, \text{s} \)

Substituting the values:
\[ s = 0 + \frac{1}{2} \cdot 4.9 \cdot (4^2) = \frac{1}{2} \cdot 4.9 \cdot 16 = 2.45 \cdot 16 = 39.2 \, \text{m} \]

### Final Answer
The stone will reach a depth of **39.2 meters** after 4 seconds.

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