Pressure `3 m` below the free surface of a liquid is `15KN//m^(2)` in excess of atmosphere pressure. Datermine its density and specific gravity. `[g = 10m//sec^(2)]`

Explanation

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

### Given:
- Pressure at a depth of 3 m below the free surface of a liquid, \( P = 15 \, \text{kN/m}^2 \) (which is equal to \( 15 \times 10^3 \, \text{N/m}^2 \))
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)
- Depth, \( h = 3 \, \text{m} \)

### Step 1: Understand the relationship between pressure, density, gravity, and height.
The pressure at a certain depth in a fluid can be expressed by the formula:
\[ P = P_0 + \rho g h \]
Where:
- \( P \) is the pressure at depth,
- \( P_0 \) is the atmospheric pressure (which we can consider as 0 for this problem since we are given excess pressure),
- \( \rho \) is the density of the liquid,
- \( g \) is the acceleration due to gravity,
- \( h \) is the depth below the surface.

### Step 2: Substitute the known values into the equation.
Since we are given that \( P_0 = 0 \) (considering excess pressure), we can simplify the equation:
\[ P = \rho g h \]
Substituting the known values:
\[ 15 \times 10^3 = \rho \times 10 \times 3 \]

### Step 3: Solve for the density \( \rho \).
Rearranging the equation to solve for \( \rho \):
\[ \rho = \frac{15 \times 10^3}{10 \times 3} \]
Calculating the right-hand side:
\[ \rho = \frac{15 \times 10^3}{30} = \frac{15,000}{30} = 500 \, \text{kg/m}^3 \]

### Step 4: Calculate the specific gravity.
Specific gravity (SG) is defined as the ratio of the density of the liquid to the density of water. The density of water is approximately \( 1000 \, \text{kg/m}^3 \).
\[ SG = \frac{\rho}{\text{Density of water}} = \frac{500}{1000} = 0.5 \]

### Final Answers:
- Density of the liquid, \( \rho = 500 \, \text{kg/m}^3 \)
- Specific gravity, \( SG = 0.5 \)