If the intensity of sound is increased by a factor of 30, by how many decibels in the sound level increased :-

Options

A. 12 dB
B. 14.77 dB
C. 10dB
D. 13dB

Answer

To solve the problem of how many decibels the sound level increases when the intensity of sound is increased by a factor of 30, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Formula for Sound Level**:
The sound level in decibels (β) is given by the formula:
\[ \beta = 10 \log \left( \frac{I}{I_0} \right) \]
where \( I \) is the intensity of the sound and \( I_0 \) is the reference intensity, typically \( 10^{-12} \, \text{W/m}^2 \).

2. **Define Initial and Final Intensities**:
Let:
- \( I_1 \) be the initial intensity.
- \( I_2 \) be the final intensity after the increase.
According to the problem, the intensity increases by a factor of 30:
\[ I_2 = 30 I_1 \]

3. **Calculate the Initial and Final Sound Levels**:
- The initial sound level \( \beta_1 \) is:
\[ \beta_1 = 10 \log \left( \frac{I_1}{I_0} \right) \]
- The final sound level \( \beta_2 \) is:
\[ \beta_2 = 10 \log \left( \frac{I_2}{I_0} \right) = 10 \log \left( \frac{30 I_1}{I_0} \right) \]

4. **Find the Increase in Sound Level**:
To find the increase in sound level, we calculate \( \beta_2 - \beta_1 \):
\[ \beta_2 - \beta_1 = 10 \log \left( \frac{I_2}{I_1} \right) \]
Substituting \( I_2 = 30 I_1 \):
\[ \beta_2 - \beta_1 = 10 \log \left( \frac{30 I_1}{I_1} \right) = 10 \log(30) \]

5. **Calculate \( \log(30) \)**:
We can express \( \log(30) \) as:
\[ \log(30) = \log(10 \times 3) = \log(10) + \log(3) \]
Since \( \log(10) = 1 \) and \( \log(3) \approx 0.477 \):
\[ \log(30) \approx 1 + 0.477 = 1.477 \]

6. **Final Calculation**:
Now, substituting back into the equation for the increase in sound level:
\[ \beta_2 - \beta_1 = 10 \times 1.477 \approx 14.77 \, \text{dB} \]

### Conclusion:
The increase in sound level when the intensity of sound is increased by a factor of 30 is approximately **14.77 dB**.