An increases in pressure required to decreases the 200 litres volume of a liquid by `0.004%` in container is: (bul modulus of the liquid `=2100MPa`)

Options

A. 188 kPa
B. 8.4 kPa
C. 18.8 kPa
D. 84 kPa

Answer

To solve the problem, we need to calculate the increase in pressure required to decrease the volume of a liquid by 0.004% using the bulk modulus of the liquid.

### Step-by-Step Solution:

1. **Understand the Given Data**:
- Initial volume of the liquid, \( V = 200 \) liters
- Percentage decrease in volume, \( \Delta V\% = 0.004\% \)
- Bulk modulus of the liquid, \( B = 2100 \) MPa

2. **Convert the Percentage Decrease into Volume**:
- The decrease in volume, \( \Delta V \), can be calculated as:
\[ \Delta V = \frac{0.004}{100} \times 200 \text{ liters} \]
- Calculating this gives:
\[ \Delta V = 0.008 \text{ liters} \]

3. **Convert the Volume from Liters to Cubic Meters**:
- Since \( 1 \text{ liter} = 0.001 \text{ m}^3 \):
\[ \Delta V = 0.008 \text{ liters} = 0.008 \times 0.001 \text{ m}^3 = 8 \times 10^{-6} \text{ m}^3 \]
- The initial volume in cubic meters is:
\[ V = 200 \text{ liters} = 200 \times 0.001 \text{ m}^3 = 0.2 \text{ m}^3 \]

4. **Use the Bulk Modulus Formula**:
- The formula relating pressure change (\( \Delta P \)), bulk modulus (\( B \)), and volume change (\( \Delta V \)) is:
\[ B = -\frac{\Delta P}{\frac{\Delta V}{V}} \]
- Rearranging gives:
\[ \Delta P = -B \cdot \frac{\Delta V}{V} \]

5. **Substituting the Values**:
- Substitute the known values into the equation:
\[ \Delta P = -2100 \times 10^6 \text{ Pa} \cdot \frac{8 \times 10^{-6} \text{ m}^3}{0.2 \text{ m}^3} \]

6. **Calculate the Pressure Change**:
- First, calculate \( \frac{\Delta V}{V} \):
\[ \frac{\Delta V}{V} = \frac{8 \times 10^{-6}}{0.2} = 4 \times 10^{-5} \]
- Now calculate \( \Delta P \):
\[ \Delta P = 2100 \times 10^6 \cdot 4 \times 10^{-5} = 84,000 \text{ Pa} = 84 \text{ kPa} \]

### Final Answer:
The increase in pressure required to decrease the volume of the liquid by 0.004% is **84 kPa**.