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 follow these steps:

### Step 1: Calculate the Pressure at the Depth of the Ocean
The pressure \(P\) at a depth \(h\) in a fluid is given by the formula:
\[ P = \rho g h \]
where:
- \(\rho\) is the density of water (approximately \(1000 \, \text{kg/m}^3\)),
- \(g\) is the acceleration due to gravity (given as \(10 \, \text{m/s}^2\)),
- \(h\) is the depth (given as \(3000 \, \text{m}\)).

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

### Step 2: Use the Bulk Modulus to Find the Fractional Compression
The bulk modulus \(B\) is defined as:
\[ B = -\frac{P}{\Delta V/V} \]
Rearranging this gives:
\[ \frac{\Delta V}{V} = -\frac{P}{B} \]
Since we are interested in the magnitude, we can ignore the negative sign:
\[ \frac{\Delta V}{V} = \frac{P}{B} \]

### Step 3: Substitute the Values into the Equation
We know:
- \(P = 3 \times 10^7 \, \text{N/m}^2\),
- \(B = 2.2 \times 10^9 \, \text{N/m}^2\).

Substituting these values:
\[ \frac{\Delta V}{V} = \frac{3 \times 10^7}{2.2 \times 10^9} \]

### Step 4: Calculate the Fractional Compression
Now, performing the division:
\[ \frac{\Delta V}{V} = \frac{3}{2.2} \times 10^{-2} = 1.36 \times 10^{-2} \]

### Step 5: Convert to Percentage
To express this as a percentage, we multiply by 100:
\[ \frac{\Delta V}{V} \times 100 = 1.36 \times 10^{-2} \times 100 = 1.36\% \]

### Final Answer
The fractional compression \((\Delta V)/V\) of water at the bottom of the Indian Ocean is approximately **1.36%**.

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