A sound has an intensity of `2 xx 10^(-8) W//m^(2)`. Its intensity level in decibels is `(log_(10) 2 = 0.3)`

Options

A. `23`
B. `3`
C. `43`
D. `4.3`

Explanation

To find the intensity level of a sound in decibels given its intensity, we can use the formula for sound intensity level:

\[ L = 10 \log_{10} \left( \frac{I}{I_0} \right) \]

where:
- \( L \) is the sound intensity level in decibels (dB),
- \( I \) is the intensity of the sound in watts per square meter (W/m²),
- \( I_0 \) is the reference intensity, which is typically \( 1 \times 10^{-12} \, \text{W/m}^2 \) (the threshold of hearing).

### Step-by-step Solution:

1. **Identify the given values**:
- Intensity \( I = 2 \times 10^{-8} \, \text{W/m}^2 \)
- Reference intensity \( I_0 = 1 \times 10^{-12} \, \text{W/m}^2 \)

2. **Substitute the values into the formula**:
\[ L = 10 \log_{10} \left( \frac{2 \times 10^{-8}}{1 \times 10^{-12}} \right) \]

3. **Calculate the fraction inside the logarithm**:
\[ \frac{2 \times 10^{-8}}{1 \times 10^{-12}} = 2 \times 10^{4} \]

4. **Now substitute this back into the formula**:
\[ L = 10 \log_{10} (2 \times 10^{4}) \]

5. **Use the logarithmic property**:
\[ \log_{10} (a \times b) = \log_{10} a + \log_{10} b \]
So,
\[ L = 10 \left( \log_{10} 2 + \log_{10} (10^{4}) \right) \]

6. **Calculate \( \log_{10} (10^{4}) \)**:
\[ \log_{10} (10^{4}) = 4 \]

7. **Now substitute \( \log_{10} 2 \)**:
Given \( \log_{10} 2 = 0.3 \),
\[ L = 10 \left( 0.3 + 4 \right) \]

8. **Calculate the final value**:
\[ L = 10 \times 4.3 = 43 \, \text{dB} \]

### Final Answer:
The intensity level of the sound is **43 decibels**.