The level of sound is attenuated by 30 dB by a sound absorber. The intensity of sound will decrease by a factor of

Options

A. 100
B. 1000
C. 2000
D. 30

Explanation

To solve the problem of how the intensity of sound decreases when the level of sound is attenuated by 30 dB, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Decibel Formula**:
The loudness level \( L \) in decibels (dB) is given by the formula:
\[ L = 10 \log \left( \frac{I}{I_0} \right) \]
where \( I \) is the intensity of the sound and \( I_0 \) is the reference intensity.

2. **Setting Up the Initial and Final Intensities**:
Let \( I_1 \) be the initial intensity of the sound and \( I_2 \) be the final intensity after attenuation. The change in loudness level due to attenuation is given as:
\[ L_1 - L_2 = 30 \, \text{dB} \]

3. **Applying the Decibel Formula**:
We can express the initial and final loudness levels using the formula:
\[ L_1 = 10 \log \left( \frac{I_1}{I_0} \right) \]
\[ L_2 = 10 \log \left( \frac{I_2}{I_0} \right) \]

4. **Setting Up the Equation**:
Substituting \( L_1 \) and \( L_2 \) into the attenuation equation:
\[ 10 \log \left( \frac{I_1}{I_0} \right) - 10 \log \left( \frac{I_2}{I_0} \right) = 30 \]

5. **Simplifying the Equation**:
Using the property of logarithms:
\[ 10 \left( \log \left( \frac{I_1}{I_0} \right) - \log \left( \frac{I_2}{I_0} \right) \right) = 30 \]
This simplifies to:
\[ 10 \log \left( \frac{I_1}{I_2} \right) = 30 \]
Dividing both sides by 10 gives:
\[ \log \left( \frac{I_1}{I_2} \right) = 3 \]

6. **Exponentiating to Solve for the Intensity Ratio**:
To eliminate the logarithm, we exponentiate both sides:
\[ \frac{I_1}{I_2} = 10^3 \]
This means:
\[ \frac{I_1}{I_2} = 1000 \]

7. **Finding the Factor of Decrease**:
Therefore, the intensity of sound decreases by a factor of:
\[ I_2 = \frac{I_1}{1000} \]
This indicates that the intensity of sound decreases by a factor of 1000.

### Final Answer:
The intensity of sound will decrease by a factor of **1000**.

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