The sound intensity is `0.008W//m^( 2)` at a distance of 10 m from an isotropic point source of sound. The power of the source is

Options

A. `2.5` watt
B. `0.8` watt
C. `8` watt
D. `10` watt

Answer

To find the power of the sound source given the intensity and distance from the source, we can use the formula for intensity of sound from an isotropic point source:

### Step-by-Step Solution:

1. **Understand the relationship between intensity and power**:
The intensity \( I \) of sound at a distance \( r \) from an isotropic point source is given by the formula:
\[ I = \frac{P}{4 \pi r^2} \]
where \( P \) is the power of the source.

2. **Identify the given values**:
From the problem, we have:
- Intensity \( I = 0.008 \, \text{W/m}^2 \)
- Distance \( r = 10 \, \text{m} \)

3. **Rearrange the formula to solve for power \( P \)**:
To find the power \( P \), we can rearrange the formula:
\[ P = I \times 4 \pi r^2 \]

4. **Substitute the known values into the equation**:
Now, substitute the values of \( I \) and \( r \):
\[ P = 0.008 \, \text{W/m}^2 \times 4 \pi (10 \, \text{m})^2 \]

5. **Calculate \( 4 \pi (10)^2 \)**:
First, calculate \( (10)^2 = 100 \):
\[ 4 \pi (10)^2 = 4 \pi \times 100 = 400 \pi \]

6. **Now calculate \( P \)**:
Substitute \( 400 \pi \) back into the equation:
\[ P = 0.008 \times 400 \pi \]
Using \( \pi \approx 3.14 \):
\[ P \approx 0.008 \times 400 \times 3.14 \]
\[ P \approx 0.008 \times 1256 \approx 10.048 \, \text{W} \]

7. **Round to the appropriate significant figures**:
Since we are looking for the power in watts, we can round \( 10.048 \) to \( 10 \, \text{W} \).

### Final Answer:
The power of the source is approximately \( 10 \, \text{W} \).

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