Two stretched membranes of areas `2cm^(2)` and `3cm^(2)` are placed in a liquid at the same depth. The ratio of the pressures on them is:

Options

A. `1:1`
B. `2:3`
C. `3:2`
D. `2^(2):3^(2)`

Explanation

To solve the problem, we need to determine the ratio of the pressures on two stretched membranes of different areas placed at the same depth in a liquid.

### Step-by-Step Solution:

1. **Understand the Concept of Pressure in Fluids**:
Pressure at a certain depth in a fluid is given by the formula:
\[ P = P_0 + \rho g h \]
where:
- \( P \) is the pressure at depth,
- \( P_0 \) is the atmospheric pressure,
- \( \rho \) is the density of the liquid,
- \( g \) is the acceleration due to gravity,
- \( h \) is the depth of the liquid.

2. **Identify the Given Information**:
- The areas of the membranes are \( 2 \, \text{cm}^2 \) and \( 3 \, \text{cm}^2 \).
- Both membranes are at the same depth in the liquid.

3. **Analyze the Pressure on Each Membrane**:
Since both membranes are at the same depth, the pressure acting on both membranes will be the same. This is because pressure in a fluid depends only on the depth and not on the area of the surface.

4. **Set Up the Ratio of Pressures**:
Let \( P_1 \) be the pressure on the first membrane (area = \( 2 \, \text{cm}^2 \)) and \( P_2 \) be the pressure on the second membrane (area = \( 3 \, \text{cm}^2 \)). Since both are at the same depth:
\[ P_1 = P_2 \]

5. **Calculate the Ratio**:
The ratio of the pressures \( P_1 \) to \( P_2 \) is:
\[ \frac{P_1}{P_2} = 1 \]

### Conclusion:
The ratio of the pressures on the two stretched membranes is \( 1:1 \).