If the compressibility of water is `sigma` per unit atmospheric pressure, then the decrease in volume V due to atmospheric pressure P will be

Options

A. `sigmaP//V`
B. `sigmaPV`
C. `sigma//PV`
D. `sigmaV//P`

Answer

To solve the problem, we need to find the decrease in volume \( \Delta V \) of water due to atmospheric pressure \( P \), given that the compressibility of water is \( \sigma \) per unit atmospheric pressure.

### Step-by-Step Solution:

1. **Understand Compressibility**:
Compressibility \( \sigma \) is defined as the fractional change in volume per unit pressure. Mathematically, it can be expressed as:
\[ \sigma = -\frac{\Delta V}{V \cdot P} \]
where:
- \( \Delta V \) is the change in volume,
- \( V \) is the original volume,
- \( P \) is the pressure applied.

2. **Rearranging the Equation**:
From the definition of compressibility, we can rearrange the equation to solve for \( \Delta V \):
\[ \Delta V = -\sigma \cdot V \cdot P \]
Here, the negative sign indicates a decrease in volume.

3. **Substituting Values**:
Since the problem states that the compressibility of water is \( \sigma \) per unit atmospheric pressure, we can substitute \( P \) with the atmospheric pressure value (which we can denote as \( P \) in general).

4. **Final Expression**:
Thus, the decrease in volume \( \Delta V \) due to the atmospheric pressure \( P \) can be expressed as:
\[ \Delta V = -\sigma \cdot V \cdot P \]
However, since we are interested in the magnitude of the decrease, we can drop the negative sign:
\[ \Delta V = \sigma \cdot V \cdot P \]

### Conclusion:
The decrease in volume \( V \) due to atmospheric pressure \( P \) is given by:
\[ \Delta V = \sigma \cdot V \cdot P \]