If the bulk modulus of water is 4000 MPa, what is the speed of sound in water ?

Explanation

To find the speed of sound in water given the bulk modulus, we can use the formula:

\[ v = \sqrt{\frac{B}{\rho}} \]

where:
- \( v \) is the speed of sound,
- \( B \) is the bulk modulus,
- \( \rho \) is the density of the medium.

### Step 1: Identify the given values
- Bulk modulus of water, \( B = 4000 \, \text{MPa} = 4000 \times 10^6 \, \text{Pa} \)
- Density of water, \( \rho = 1000 \, \text{kg/m}^3 \)

### Step 2: Substitute the values into the formula
We substitute the values of \( B \) and \( \rho \) into the speed of sound formula:

\[ v = \sqrt{\frac{4000 \times 10^6 \, \text{Pa}}{1000 \, \text{kg/m}^3}} \]

### Step 3: Simplify the expression
First, simplify the fraction inside the square root:

\[ \frac{4000 \times 10^6}{1000} = 4000 \times 10^3 = 4 \times 10^6 \]

### Step 4: Calculate the square root
Now, take the square root of \( 4 \times 10^6 \):

\[ v = \sqrt{4 \times 10^6} = \sqrt{4} \times \sqrt{10^6} = 2 \times 10^3 \]

### Step 5: Convert to meters per second
Thus, we have:

\[ v = 2 \times 10^3 \, \text{m/s} = 2000 \, \text{m/s} \]

### Final Answer
The speed of sound in water is \( 2000 \, \text{m/s} \).

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