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 find the increase in pressure required to decrease the volume of a liquid by 0.004% given that the bulk modulus of the liquid is 2100 MPa.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Volume of the liquid, \( V = 200 \) liters
- Change in volume percentage, \( \Delta V\% = 0.004\% \)
- Bulk modulus, \( K = 2100 \) MPa

2. **Convert the Bulk Modulus to Pascals:**
- Since \( 1 \text{ MPa} = 10^6 \text{ Pa} \), we convert:
\[ K = 2100 \text{ MPa} = 2100 \times 10^6 \text{ Pa} = 2.1 \times 10^9 \text{ Pa} \]

3. **Calculate the Change in Volume:**
- The change in volume, \( \Delta V \), can be calculated using the percentage change:
\[ \Delta V = V \times \frac{\Delta V\%}{100} = 200 \times \frac{0.004}{100} = 200 \times 0.00004 = 0.008 \text{ liters} \]

4. **Use the Bulk Modulus Formula:**
- The bulk modulus is defined as:
\[ K = -\frac{P}{\frac{\Delta V}{V}} \]
Rearranging gives:
\[ P = -K \times \frac{\Delta V}{V} \]

5. **Substitute the Values:**
- Substitute \( K \), \( \Delta V \), and \( V \) into the equation:
\[ P = -2.1 \times 10^9 \times \frac{0.008}{200} \]

6. **Calculate the Pressure:**
- First, calculate \( \frac{0.008}{200} \):
\[ \frac{0.008}{200} = 0.00004 \]
- Now substitute this back into the pressure equation:
\[ P = -2.1 \times 10^9 \times 0.00004 = -84,000 \text{ Pa} = -84 \text{ kPa} \]
(The negative sign indicates that pressure is applied to decrease the volume.)

7. **Final Answer:**
- The increase in pressure required is:
\[ P = 84 \text{ kPa} \]

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