The average depth of indian Ocean is about 3000 m. The fractional compression, `(triangleV)/(V)` of water at the bottom of the ocean is (Given Bulk modulus of the water `=2.2xx10^9Nm^-2` and `g=10ms^-2`)

Options

A. `0.82%`
B. `0.91%`
C. `1.36%`
D. `1.24%`

Explanation

To solve the problem of finding the fractional compression \((\Delta V)/V\) of water at the bottom of the Indian Ocean, we can use the relationship between pressure, bulk modulus, and volume change.

### Step-by-Step Solution:

1. **Identify Given Values**:
- Depth of the Indian Ocean, \( h = 3000 \, \text{m} \)
- Bulk modulus of water, \( K = 2.2 \times 10^9 \, \text{N/m}^2 \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

2. **Calculate the Pressure at the Bottom of the Ocean**:
The pressure at a depth \( h \) in a fluid is given by the formula:
\[ P = \rho g h \]
where \( \rho \) is the density of water. The average density of water is approximately \( 1000 \, \text{kg/m}^3 \).

Substituting the values:
\[ P = 1000 \, \text{kg/m}^3 \times 10 \, \text{m/s}^2 \times 3000 \, \text{m} \]
\[ P = 1000 \times 10 \times 3000 = 30000000 \, \text{N/m}^2 = 3 \times 10^7 \, \text{N/m}^2 \]

3. **Use the Bulk Modulus to Find the Volume Change**:
The bulk modulus \( K \) is defined as:
\[ K = -\frac{P}{\Delta V / V} \]
Rearranging this gives:
\[ \frac{\Delta V}{V} = -\frac{P}{K} \]

4. **Substituting the Values**:
Now, substituting the pressure \( P \) and bulk modulus \( K \):
\[ \frac{\Delta V}{V} = -\frac{3 \times 10^7 \, \text{N/m}^2}{2.2 \times 10^9 \, \text{N/m}^2} \]
\[ \frac{\Delta V}{V} = -\frac{3}{2.2} \times 10^{-2} \]
\[ \frac{\Delta V}{V} \approx -0.01364 \]

5. **Final Result**:
The fractional compression of water at the bottom of the Indian Ocean is approximately:
\[ \frac{\Delta V}{V} \approx -0.01364 \]