A sound absorber attenuates the sound level by `20 dB`. The intensity decreases by a factor of

Options

A. (a)10000
B. (b)10
C. (c)100
D. (d)1000

Explanation

To solve the problem of how much the intensity of sound decreases when the sound level is attenuated by 20 dB, we can follow these steps:

### Step 1: Understand the relationship between sound level (L) and intensity (I)
The sound level in decibels (dB) is given by the formula:
\[ L = 10 \log_{10} \left( \frac{I}{I_0} \right) \]
where:
- \( L \) is the sound level in decibels,
- \( I \) is the intensity of the sound,
- \( I_0 \) is the reference intensity (usually \( 10^{-12} \, \text{W/m}^2 \)).

### Step 2: Set up the equation for the initial and final sound levels
Let the initial sound level be \( L \) and the final sound level after attenuation be \( L' \). Given that the sound level is attenuated by 20 dB, we can express this as:
\[ L' = L - 20 \]

### Step 3: Write the equations for the initial and final intensities
Using the formula for sound level, we can write:
1. For the initial sound level:
\[ L = 10 \log_{10} \left( \frac{I}{I_0} \right) \]
2. For the final sound level:
\[ L' = 10 \log_{10} \left( \frac{I'}{I_0} \right) \]

### Step 4: Substitute \( L' \) into the equation
Substituting \( L' = L - 20 \) into the second equation gives:
\[ L - 20 = 10 \log_{10} \left( \frac{I'}{I_0} \right) \]

### Step 5: Rearrange the equations
From the first equation, we can express \( L \) in terms of \( I \):
\[ L = 10 \log_{10} \left( \frac{I}{I_0} \right) \]
Now we can substitute this into our equation for \( L' \):
\[ 10 \log_{10} \left( \frac{I}{I_0} \right) - 20 = 10 \log_{10} \left( \frac{I'}{I_0} \right) \]

### Step 6: Divide through by 10
Dividing the entire equation by 10 gives:
\[ \log_{10} \left( \frac{I}{I_0} \right) - 2 = \log_{10} \left( \frac{I'}{I_0} \right) \]

### Step 7: Rewrite the equation
Rearranging gives:
\[ \log_{10} \left( \frac{I}{I_0} \right) = \log_{10} \left( \frac{I'}{I_0} \right) + 2 \]

### Step 8: Exponentiate to eliminate the logarithm
Exponentiating both sides results in:
\[ \frac{I}{I_0} = 100 \cdot \frac{I'}{I_0} \]

### Step 9: Solve for the ratio of intensities
This simplifies to:
\[ I = 100 \cdot I' \]
Thus, we find that:
\[ \frac{I'}{I} = \frac{1}{100} \]

### Conclusion
The intensity decreases by a factor of 100 when the sound level is attenuated by 20 dB.

### Final Answer
The intensity decreases by a factor of **100**.

---