The compressibility of water is `5 xx 10^(-10) m^(2)//N` . If it is subjected to a pressure of 15 MPa, the fractional decrease in volume will be-

Options

A. 10 cc
B. 24 cc
C. 15 cc
D. 12 cc

Explanation

To find the fractional decrease in volume of water when subjected to a pressure of 15 MPa, we can use the relationship between compressibility and the change in volume. Here’s the step-by-step solution:

### Step 1: Understand the formula for fractional decrease in volume
The fractional decrease in volume (ΔV/V) can be calculated using the formula:
\[ \Delta V/V = -\beta \cdot \Delta P \]
where:
- \(\Delta V\) is the change in volume,
- \(V\) is the original volume,
- \(\beta\) is the compressibility,
- \(\Delta P\) is the change in pressure.

### Step 2: Identify the values given in the problem
From the problem, we have:
- Compressibility of water, \(\beta = 5 \times 10^{-10} \, \text{m}^2/\text{N}\)
- Pressure applied, \(\Delta P = 15 \, \text{MPa} = 15 \times 10^6 \, \text{N/m}^2\)

### Step 3: Substitute the values into the formula
Now, we can substitute the values into the formula:
\[ \Delta V/V = -\beta \cdot \Delta P = - (5 \times 10^{-10} \, \text{m}^2/\text{N}) \cdot (15 \times 10^6 \, \text{N/m}^2) \]

### Step 4: Calculate the result
Now, perform the multiplication:
\[ \Delta V/V = - (5 \times 15) \times (10^{-10} \times 10^6) = -75 \times 10^{-4} \]
This simplifies to:
\[ \Delta V/V = -7.5 \times 10^{-3} \]

### Step 5: Interpret the result
The negative sign indicates a decrease in volume. Therefore, the fractional decrease in volume is:
\[ \Delta V/V = 7.5 \times 10^{-3} \]

### Final Answer
The fractional decrease in volume of water when subjected to a pressure of 15 MPa is \(7.5 \times 10^{-3}\).

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