The bulk modulus of water is `2.0xx10^(9) N//m^(2)`. The pressure required to increase the density of water by `0.1%` is

Options

A. `2.0xx10^(3) N//m^(2)`
B. `2.0xx10^(6) N//m^(2)`
C. `2.0xx10^(5) N//m^(2)`
D. `2.0xx10^(7)N//m^(2)`

Explanation

To find the pressure required to increase the density of water by 0.1%, we can use the relationship defined by the bulk modulus (B). The bulk modulus is defined as:

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

In terms of density, this can be rewritten as:

\[ B = \frac{P}{\frac{\Delta \rho}{\rho}} \]

Where:
- \( P \) is the pressure applied,
- \( \Delta \rho \) is the change in density,
- \( \rho \) is the original density.

Rearranging the equation gives us:

\[ P = B \cdot \frac{\Delta \rho}{\rho} \]

Now, we know:
- The bulk modulus of water, \( B = 2.0 \times 10^9 \, \text{N/m}^2 \),
- The percentage change in density, \( \frac{\Delta \rho}{\rho} = 0.1\% = \frac{0.1}{100} = 0.001 \).

Substituting these values into the equation:

\[ P = 2.0 \times 10^9 \cdot 0.001 \]

Calculating this gives:

\[ P = 2.0 \times 10^9 \cdot 0.001 = 2.0 \times 10^6 \, \text{N/m}^2 \]

Thus, the pressure required to increase the density of water by 0.1% is:

\[ P = 2.0 \times 10^6 \, \text{N/m}^2 \]

### Final Answer:
The pressure required is \( 2.0 \times 10^6 \, \text{N/m}^2 \).