If the bulk modulus of water is 2100 M Pa, what is the speed of sound in water ?

Options

A. `1450ms^(-1)`
B. `2100ms^(-1)`
C. `1400ms^(-1)`
D. `1200ms^(-1)`

Explanation

To find the speed of sound in water given its 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-by-Step Solution:

**Step 1: Identify the given values.**
- Bulk modulus of water, \( B = 2100 \, \text{MPa} = 2100 \times 10^6 \, \text{Pa} \)
- Density of water, \( \rho = 1000 \, \text{kg/m}^3 = 10^3 \, \text{kg/m}^3 \)

**Step 2: Substitute the values into the formula.**
\[ v = \sqrt{\frac{2100 \times 10^6 \, \text{Pa}}{10^3 \, \text{kg/m}^3}} \]

**Step 3: Simplify the expression.**
\[ v = \sqrt{2100 \times 10^6 \div 10^3} \]
\[ v = \sqrt{2100 \times 10^{6-3}} \]
\[ v = \sqrt{2100 \times 10^3} \]

**Step 4: Calculate the square root.**
\[ v = \sqrt{2100} \times \sqrt{10^3} \]
\[ v = \sqrt{2100} \times 10^{1.5} \]
\[ v = \sqrt{2100} \times 31.62 \]

**Step 5: Estimate \( \sqrt{2100} \).**
- We know \( 45^2 = 2025 \) and \( 46^2 = 2116 \), so \( \sqrt{2100} \) is approximately between 45 and 46. A closer approximation gives us around 45.83.

**Step 6: Calculate the final speed.**
\[ v \approx 45.83 \times 31.62 \]
\[ v \approx 1450 \, \text{m/s} \]

### Final Answer:
The speed of sound in water is approximately \( 1450 \, \text{m/s} \).

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