The intensity of sound from a radio at a distance of 2 metres from its speaker is `1xx10^(-2) mu W//m^(2)`. The intensity at a distance of 10 meters would be

Options

A. `10^-8 W/m^2`
B. `4×10^-10 W/m^2`
C. `2×10^-8 W/m^2`
D. `5×10^-9 W/m^2`

Answer

To find the intensity of sound at a distance of 10 meters from the speaker, we can use the inverse square law of intensity. The intensity of sound is inversely proportional to the square of the distance from the source.

### Step-by-step Solution:

1. **Given Data**:
- Intensity at distance \( r_1 = 2 \) m:
\[ I_1 = 1 \times 10^{-2} \, \mu W/m^2 = 1 \times 10^{-2} \times 10^{-6} \, W/m^2 = 1 \times 10^{-8} \, W/m^2 \]
- Distance \( r_2 = 10 \) m.

2. **Using the Inverse Square Law**:
The relationship between intensity and distance is given by:
\[ \frac{I_2}{I_1} = \left(\frac{r_1}{r_2}\right)^2 \]
Rearranging this gives:
\[ I_2 = I_1 \left(\frac{r_1}{r_2}\right)^2 \]

3. **Substituting the Values**:
Substitute \( I_1 \), \( r_1 \), and \( r_2 \):
\[ I_2 = 1 \times 10^{-8} \, W/m^2 \left(\frac{2}{10}\right)^2 \]
\[ I_2 = 1 \times 10^{-8} \, W/m^2 \left(\frac{2}{10}\right)^2 = 1 \times 10^{-8} \, W/m^2 \left(\frac{4}{100}\right) \]

4. **Calculating \( I_2 \)**:
\[ I_2 = 1 \times 10^{-8} \, W/m^2 \times \frac{4}{100} = 1 \times 10^{-8} \, W/m^2 \times 0.04 \]
\[ I_2 = 4 \times 10^{-10} \, W/m^2 \]

5. **Final Answer**:
The intensity at a distance of 10 meters is:
\[ I_2 = 4 \times 10^{-10} \, W/m^2 \]