A column of water `40 cm` height supports a `30 cm` column of an unknown liquid . What is the density of the liquid ?

Answer

To find the density of the unknown liquid, we can use the principle of hydrostatic pressure. The pressure exerted by the column of water must equal the pressure exerted by the column of the unknown liquid at the same level.

### Step-by-Step Solution:

1. **Understand the Pressure Balance**:
The pressure exerted by a liquid column is given by the formula:
\[ P = \rho g h \]
where \( P \) is the pressure, \( \rho \) is the density of the liquid, \( g \) is the acceleration due to gravity, and \( h \) is the height of the liquid column.

2. **Set Up the Equation**:
For the water column and the unknown liquid column, we have:
\[ P_{\text{water}} = P_{\text{unknown liquid}} \]
Therefore, we can write:
\[ \rho_{\text{water}} g h_{\text{water}} = \rho_{\text{unknown}} g h_{\text{unknown}} \]

3. **Substitute Known Values**:
- The height of the water column \( h_{\text{water}} = 40 \, \text{cm} = 0.4 \, \text{m} \)
- The height of the unknown liquid column \( h_{\text{unknown}} = 30 \, \text{cm} = 0.3 \, \text{m} \)
- The density of water \( \rho_{\text{water}} = 1000 \, \text{kg/m}^3 \)

Substituting these values into the equation gives:
\[ 1000 \, \text{kg/m}^3 \cdot g \cdot 0.4 \, \text{m} = \rho_{\text{unknown}} \cdot g \cdot 0.3 \, \text{m} \]

4. **Cancel Out \( g \)**:
Since \( g \) appears on both sides of the equation, we can cancel it out:
\[ 1000 \, \text{kg/m}^3 \cdot 0.4 = \rho_{\text{unknown}} \cdot 0.3 \]

5. **Solve for \( \rho_{\text{unknown}} \)**:
Rearranging the equation to solve for \( \rho_{\text{unknown}} \):
\[ \rho_{\text{unknown}} = \frac{1000 \cdot 0.4}{0.3} \]
\[ \rho_{\text{unknown}} = \frac{400}{0.3} \approx 1333.33 \, \text{kg/m}^3 \]

### Final Answer:
The density of the unknown liquid is approximately \( 1333.33 \, \text{kg/m}^3 \).

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