The approximate depth of an ocean is `2700m`. The compressibility of water is `45.4xx10^(-11)Pa^-1` and density of water is `10^3(kg)/(m^3)`. What fractional compression of water will be obtained at the bottom of the ocean?

Options

A. `0.8 xx 10^(-2)`
B. `1.0 xx 10^(-2)`
C. `1.2 xx 10^(-2)`
D. `1.4 xx 10^(-2)`

Explanation

To find the fractional compression of water at the bottom of the ocean, we can use the relationship between compressibility, pressure change, and volume change. Here’s a step-by-step solution:

### Step 1: Understand the formula for compressibility
The compressibility (κ) of a substance is defined as:
\[ \kappa = -\frac{1}{V} \frac{\Delta V}{\Delta P} \]
Where:
- \( \Delta V \) is the change in volume,
- \( V \) is the original volume,
- \( \Delta P \) is the change in pressure.

### Step 2: Rearranging the formula
We can rearrange the formula to find the fractional change in volume:
\[ \frac{\Delta V}{V} = -\kappa \Delta P \]

### Step 3: Calculate the change in pressure (\( \Delta P \))
The change in pressure at a depth \( h \) in a fluid is given by:
\[ \Delta P = \rho g h \]
Where:
- \( \rho \) is the density of water (given as \( 10^3 \, \text{kg/m}^3 \)),
- \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)),
- \( h \) is the depth of the ocean (given as \( 2700 \, \text{m} \)).

Substituting the values:
\[ \Delta P = (10^3 \, \text{kg/m}^3)(9.81 \, \text{m/s}^2)(2700 \, \text{m}) \]
Calculating this gives:
\[ \Delta P = 10^3 \times 9.81 \times 2700 = 26,487,000 \, \text{Pa} \approx 2.6487 \times 10^7 \, \text{Pa} \]

### Step 4: Substitute values into the fractional change in volume formula
Now we can substitute \( \Delta P \) and the given compressibility (\( \kappa = 45.4 \times 10^{-11} \, \text{Pa}^{-1} \)) into the rearranged formula:
\[ \frac{\Delta V}{V} = -\kappa \Delta P \]
\[ \frac{\Delta V}{V} = - (45.4 \times 10^{-11}) (2.6487 \times 10^7) \]

### Step 5: Calculate the fractional change in volume
Calculating this gives:
\[ \frac{\Delta V}{V} = - (45.4 \times 10^{-11}) \times (2.6487 \times 10^7) \approx -1.202 \times 10^{-3} \]
This can be expressed as:
\[ \frac{\Delta V}{V} \approx -1.2 \times 10^{-2} \]

### Conclusion
The fractional compression of water at the bottom of the ocean is approximately:
\[ \frac{\Delta V}{V} \approx 1.2 \times 10^{-2} \]

### Final Answer
The correct option is \( 1.2 \times 10^{-2} \).

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