The pressure required to reduce the given volume of water by `1%` is,
(Bulk modulus of water `=2xx10^(9)N//m^(2)`)

Options

A. `2xx10^(7)N//m^(2)`
B. `2xx10^(8)N//m^(2)`
C. `2xx10^(10)N//m^(2)`
D. `2xx10^(10)N//m^(2)`

Explanation

To find the pressure required to reduce the volume of water by 1%, we can use the concept of bulk modulus. The bulk modulus (B) is defined as the ratio of the change in pressure (ΔP) to the relative change in volume (ΔV/V). The formula for bulk modulus is given by:

\[ B = -\frac{\Delta P}{\frac{\Delta V}{V}} \]

Where:
- \( B \) is the bulk modulus,
- \( \Delta P \) is the change in pressure,
- \( \Delta V \) is the change in volume,
- \( V \) is the original volume.

### Step-by-Step Solution:

1. **Identify the given values:**
- Bulk modulus of water, \( B = 2 \times 10^9 \, \text{N/m}^2 \)
- Change in volume, \( \Delta V = -0.01V \) (since we are reducing the volume by 1%)

2. **Express the relative change in volume:**
- The relative change in volume is given by:
\[ \frac{\Delta V}{V} = \frac{-0.01V}{V} = -0.01 \]

3. **Substitute the values into the bulk modulus formula:**
- Rearranging the bulk modulus formula gives:
\[ \Delta P = -B \cdot \frac{\Delta V}{V} \]
- Substituting the known values:
\[ \Delta P = - (2 \times 10^9) \cdot (-0.01) \]

4. **Calculate the change in pressure:**
- Performing the multiplication:
\[ \Delta P = 2 \times 10^9 \times 0.01 = 2 \times 10^7 \, \text{N/m}^2 \]

5. **Final answer:**
- The pressure required to reduce the volume of water by 1% is:
\[ \Delta P = 2 \times 10^7 \, \text{N/m}^2 \]