A water column of height 30 cm is supporting a 20 cm column of a liquid of density p. Calculate the value of p.

Answer

To solve the problem of finding the density \( p \) of the liquid supported by a 30 cm water column, we can use the principle of hydrostatic pressure balance. Here’s a step-by-step solution:

### Step 1: Understand the Problem
We have a water column of height 30 cm (0.3 m) and a liquid column of height 20 cm (0.2 m) above it. We need to find the density \( p \) of the liquid.

### Step 2: Write the Pressure Balance Equation
The pressure at the bottom of the water column must equal the pressure at the bottom of the liquid column:
\[ P_{\text{water}} = P_{\text{liquid}} \]

### Step 3: Express Pressures in Terms of Density
The pressure due to a liquid column is given by the formula:
\[ P = \rho g h \]
Where:
- \( \rho \) is the density of the liquid,
- \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)),
- \( h \) is the height of the liquid column.

For the water column:
\[ P_{\text{water}} = \rho_w g h_w = \rho_w g (0.3) \]
Where \( \rho_w \) (density of water) is approximately \( 1000 \, \text{kg/m}^3 \).

For the liquid column:
\[ P_{\text{liquid}} = p g h_l = p g (0.2) \]

### Step 4: Set the Pressures Equal
Setting the two pressures equal gives us:
\[ \rho_w g (0.3) = p g (0.2) \]

### Step 5: Cancel Out \( g \)
Since \( g \) appears on both sides of the equation, we can cancel it out:
\[ \rho_w (0.3) = p (0.2) \]

### Step 6: Solve for \( p \)
Rearranging the equation to solve for \( p \):
\[ p = \frac{\rho_w (0.3)}{0.2} \]

### Step 7: Substitute the Value of \( \rho_w \)
Substituting \( \rho_w = 1000 \, \text{kg/m}^3 \):
\[ p = \frac{1000 \times 0.3}{0.2} \]
\[ p = \frac{300}{0.2} = 1500 \, \text{kg/m}^3 \]

### Final Answer
The density \( p \) of the liquid is \( 1500 \, \text{kg/m}^3 \).

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