A stone of density `2.5 xx10^3 kg m^(-3)` completely immersed in sea water is allowed to sink from rest in 2 s. Neglect the effect of vicosity. Relative density of sea water is 1.025.

Explanation

To solve the problem step by step, we will follow these calculations:

### Step 1: Understand the given values
- Density of the stone, \( \rho_s = 2.5 \times 10^3 \, \text{kg/m}^3 \)
- Relative density of seawater = 1.025
- Therefore, the density of seawater, \( \rho_w = 1.025 \times 1000 \, \text{kg/m}^3 = 1025 \, \text{kg/m}^3 \)
- Time to sink, \( t = 2 \, \text{s} \)

### Step 2: Calculate the volume of the stone
Let the volume of the stone be \( V \). The mass of the stone can be expressed as:
\[ m = \rho_s \cdot V \]

### Step 3: Determine the buoyant force
The buoyant force \( F_b \) acting on the stone when it is submerged in seawater is given by Archimedes' principle:
\[ F_b = \rho_w \cdot V \cdot g \]
where \( g \) is the acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \)).

### Step 4: Calculate the weight of the stone
The weight \( W \) of the stone is given by:
\[ W = m \cdot g = \rho_s \cdot V \cdot g \]

### Step 5: Set up the equation of motion
Since the stone is sinking, the net force \( F \) acting on it can be expressed as:
\[ F = W - F_b \]
Substituting the expressions for weight and buoyant force:
\[ F = \rho_s \cdot V \cdot g - \rho_w \cdot V \cdot g \]
Factoring out \( V \cdot g \):
\[ F = V \cdot g \cdot (\rho_s - \rho_w) \]

### Step 6: Apply Newton's second law
According to Newton's second law, \( F = m \cdot a \):
\[ V \cdot g \cdot (\rho_s - \rho_w) = \rho_s \cdot V \cdot a \]
Cancelling \( V \) from both sides (assuming \( V \neq 0 \)):
\[ g \cdot (\rho_s - \rho_w) = \rho_s \cdot a \]

### Step 7: Solve for acceleration \( a \)
Rearranging the equation gives:
\[ a = \frac{g \cdot (\rho_s - \rho_w)}{\rho_s} \]
Substituting the values:
\[ a = \frac{9.8 \cdot (2.5 \times 10^3 - 1.025 \times 10^3)}{2.5 \times 10^3} \]
Calculating the density difference:
\[ \rho_s - \rho_w = 2.5 \times 10^3 - 1.025 \times 10^3 = 1.475 \times 10^3 \, \text{kg/m}^3 \]
Now substituting this back:
\[ a = \frac{9.8 \cdot 1.475 \times 10^3}{2.5 \times 10^3} \]
Calculating:
\[ a \approx 5.782 \, \text{m/s}^2 \]

### Step 8: Calculate the distance sunk in 2 seconds
Using the equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
Since the stone starts from rest, \( u = 0 \):
\[ s = 0 + \frac{1}{2} \cdot 5.782 \cdot (2^2) \]
Calculating:
\[ s = \frac{1}{2} \cdot 5.782 \cdot 4 = 11.564 \, \text{m} \]

### Final Answer
The stone sinks approximately \( 11.564 \, \text{m} \) in 2 seconds.

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