The bulk modulus of water if its volume changes from `100litres` to `99.5` litre under a pressure of `100atm` is (Take `1atm=10^5Nm^-2`)

Options

A. `2xx10^(7)Nm&^(-2)`
B. `2xx10^(8)Nm&^(-2)`
C. `2xx10^(9)Nm&^(-2)`
D. `2xx10^(10)Nm&^(-2)`

Explanation

To find the bulk modulus of water when its volume changes from 100 liters to 99.5 liters under a pressure of 100 atm, we can follow these steps:

### Step 1: Convert the given pressure from atm to N/m²
Given:
- Pressure \( P = 100 \, \text{atm} \)
- Conversion factor: \( 1 \, \text{atm} = 10^5 \, \text{N/m}^2 \)

Calculating the pressure in N/m²:
\[ P = 100 \, \text{atm} \times 10^5 \, \text{N/m}^2/\text{atm} = 10^7 \, \text{N/m}^2 \]

### Step 2: Calculate the change in volume
Given:
- Initial volume \( V = 100 \, \text{liters} \)
- Final volume \( V' = 99.5 \, \text{liters} \)

Calculating the change in volume \( \Delta V \):
\[ \Delta V = V' - V = 99.5 \, \text{liters} - 100 \, \text{liters} = -0.5 \, \text{liters} \]

### Step 3: Convert the change in volume to cubic meters
Since \( 1 \, \text{liter} = 10^{-3} \, \text{m}^3 \):
\[ \Delta V = -0.5 \, \text{liters} = -0.5 \times 10^{-3} \, \text{m}^3 = -5 \times 10^{-4} \, \text{m}^3 \]

### Step 4: Calculate the bulk modulus
The formula for bulk modulus \( K \) is given by:
\[ K = -\frac{P}{\frac{\Delta V}{V}} \]

Where:
- \( V = 100 \, \text{liters} = 100 \times 10^{-3} \, \text{m}^3 = 0.1 \, \text{m}^3 \)

Substituting the values into the formula:
\[ K = -\frac{10^7 \, \text{N/m}^2}{\frac{-5 \times 10^{-4} \, \text{m}^3}{0.1 \, \text{m}^3}} \]

Calculating \( \frac{\Delta V}{V} \):
\[ \frac{\Delta V}{V} = \frac{-5 \times 10^{-4}}{0.1} = -5 \times 10^{-3} \]

Now substituting this back into the bulk modulus formula:
\[ K = -\frac{10^7}{-5 \times 10^{-3}} = \frac{10^7}{5 \times 10^{-3}} = 2 \times 10^9 \, \text{N/m}^2 \]

### Final Answer
The bulk modulus of water is \( K = 2 \times 10^9 \, \text{N/m}^2 \).

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