Two stretched membranes of area `3 cm^2` and. `4 cm^2` are placed in a liquid at the same depth. The ratio of the pressure on them is :

Options

A. `1:1`
B. `3:4 `
C. `4:3`
D. none of these

Explanation

To solve the problem, we need to understand how pressure works in a liquid. The pressure at a certain depth in a fluid is given by the formula:

\[ P = \rho \cdot g \cdot h \]

where:
- \( P \) is the pressure,
- \( \rho \) is the density of the liquid,
- \( g \) is the acceleration due to gravity,
- \( h \) is the depth of the liquid.

### Step-by-step Solution:

1. **Identify the Given Information:**
- Area of the first membrane, \( A_1 = 3 \, \text{cm}^2 \)
- Area of the second membrane, \( A_2 = 4 \, \text{cm}^2 \)
- Both membranes are at the same depth in the liquid.

2. **Understand the Concept of Pressure:**
- Pressure in a fluid at a given depth is independent of the area of the surface. It only depends on the height of the liquid column above the surface, the density of the liquid, and the acceleration due to gravity.

3. **Apply the Pressure Formula:**
- Since both membranes are at the same depth, the pressure on both membranes can be expressed as:
\[ P_1 = \rho \cdot g \cdot h \]
\[ P_2 = \rho \cdot g \cdot h \]
- Here, \( P_1 \) is the pressure on the first membrane and \( P_2 \) is the pressure on the second membrane.

4. **Calculate the Ratio of Pressures:**
- Since both pressures are equal, we have:
\[ P_1 = P_2 \]
- Therefore, the ratio of the pressures is:
\[ \frac{P_1}{P_2} = \frac{\rho \cdot g \cdot h}{\rho \cdot g \cdot h} = 1 \]

5. **Conclusion:**
- The ratio of the pressure on the two membranes is \( 1:1 \).

### Final Answer:
The ratio of the pressure on the two membranes is \( 1:1 \).

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