A vessel of `4m^(3)` volume contains an oil which weight 30.2kN. Determine the specific gravity of the oil.

Answer

To determine the specific gravity of the oil in the vessel, we will follow these steps:

### Step 1: Understand the Given Information
We are given:
- Volume of the vessel (V) = 4 m³
- Weight of the oil (W) = 30.2 kN

### Step 2: Convert Weight to Newtons
Since weight is given in kilonewtons, we need to convert it to newtons for our calculations.
\[ W = 30.2 \, \text{kN} = 30.2 \times 10^3 \, \text{N} = 30200 \, \text{N} \]

### Step 3: Calculate the Mass of the Oil
Using the relation between weight and mass:
\[ W = m \cdot g \]
where \( g \) (acceleration due to gravity) is approximately \( 9.8 \, \text{m/s}^2 \). Rearranging gives us:
\[ m = \frac{W}{g} = \frac{30200 \, \text{N}}{9.8 \, \text{m/s}^2} \]
Calculating this:
\[ m = \frac{30200}{9.8} \approx 3081.63 \, \text{kg} \]

### Step 4: Calculate the Density of the Oil
Density (\( \rho \)) is defined as mass per unit volume:
\[ \rho_{\text{oil}} = \frac{m}{V} = \frac{3081.63 \, \text{kg}}{4 \, \text{m}^3} \]
Calculating this:
\[ \rho_{\text{oil}} = \frac{3081.63}{4} \approx 770.41 \, \text{kg/m}^3 \]

### Step 5: Calculate the Specific Gravity of the Oil
Specific gravity (SG) is the ratio of the density of the substance to the density of water (which is \( 1000 \, \text{kg/m}^3 \)):
\[ SG = \frac{\rho_{\text{oil}}}{\rho_{\text{water}}} = \frac{770.41 \, \text{kg/m}^3}{1000 \, \text{kg/m}^3} \]
Calculating this:
\[ SG \approx 0.770 \]

### Final Answer
The specific gravity of the oil is approximately **0.770**.

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