The sound level at a point 5.0 m away from a point source is 40 dB. What will be the level at a point 50 m away from the source ?

Explanation

To solve the problem of finding the sound level at a point 50 m away from a point source, given that the sound level at a point 5 m away is 40 dB, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship between sound level (in dB) and intensity**:
The sound level in decibels (β) is given by the formula:
\[ \beta = 10 \log_{10} \left( \frac{I}{I_0} \right) \]
where \(I\) is the intensity of the sound and \(I_0\) is the reference intensity.

2. **Set up the equations for the two points**:
For point A (5 m away), we have:
\[ \beta_A = 40 \text{ dB} \]
For point B (50 m away), we need to find:
\[ \beta_B \]

3. **Express intensities in terms of sound levels**:
Rearranging the formula gives us:
\[ \frac{I}{I_0} = 10^{\frac{\beta}{10}} \]
For point A:
\[ \frac{I_A}{I_0} = 10^{\frac{40}{10}} = 10^4 \]
For point B:
\[ \frac{I_B}{I_0} = 10^{\frac{\beta_B}{10}} \]

4. **Divide the intensity equations**:
Dividing the intensity at point A by the intensity at point B:
\[ \frac{I_A}{I_B} = \frac{10^4}{10^{\frac{\beta_B}{10}}} \]

5. **Use the relationship between intensity and distance**:
The intensity is inversely proportional to the square of the distance from the source:
\[ I \propto \frac{1}{r^2} \]
Therefore:
\[ \frac{I_A}{I_B} = \frac{r_B^2}{r_A^2} \]
Substituting \(r_A = 5 \text{ m}\) and \(r_B = 50 \text{ m}\):
\[ \frac{I_A}{I_B} = \frac{50^2}{5^2} = \frac{2500}{25} = 100 \]

6. **Equate the two expressions for intensity ratio**:
From the previous steps, we have:
\[ 100 = \frac{10^4}{10^{\frac{\beta_B}{10}}} \]
Rearranging gives:
\[ 10^{\frac{\beta_B}{10}} = \frac{10^4}{100} = 10^2 \]

7. **Solve for \(\beta_B\)**:
Taking the logarithm:
\[ \frac{\beta_B}{10} = 2 \implies \beta_B = 20 \text{ dB} \]

### Final Answer:
The sound level at a point 50 m away from the source is **20 dB**.