Loudness of sound from on isotopic point source at distance of 10 m is 20 dB. Find the loudness (in dB) at a point at a distance `10^(7//4)` m from the source.

Options

A. 3
B. 5
C. 7
D. 9

Explanation

To solve the problem, we need to find the loudness (in dB) at a distance of \(10^{7/4}\) m from an isotropic point source of sound, given that the loudness at a distance of 10 m is 20 dB.

### Step-by-Step Solution:

1. **Understanding the Relationship Between Intensity and Distance**:
The intensity of sound from a point source decreases with the square of the distance from the source. Mathematically, this can be expressed as:
\[ \frac{I_2}{I_1} = \left(\frac{R_1}{R_2}\right)^2 \]
where \(I_1\) is the intensity at distance \(R_1\) and \(I_2\) is the intensity at distance \(R_2\).

2. **Given Values**:
- \(R_1 = 10\) m
- \(R_2 = 10^{7/4}\) m
- Loudness at \(R_1\) is 20 dB.

3. **Calculating the Ratio of Intensities**:
Substitute the values into the intensity ratio equation:
\[ \frac{I_2}{I_1} = \left(\frac{10}{10^{7/4}}\right)^2 \]
Simplifying this:
\[ \frac{I_2}{I_1} = \left(10^{1 - 7/4}\right)^2 = \left(10^{-3/4}\right)^2 = 10^{-3/2} \]

4. **Finding the Sound Level at Distance \(R_2\)**:
The sound level in decibels is given by the formula:
\[ L = 10 \log_{10}\left(\frac{I}{I_0}\right) \]
where \(I_0\) is the reference intensity. We know:
\[ L_1 = 10 \log_{10}\left(\frac{I_1}{I_0}\right) = 20 \text{ dB} \]
Therefore, we can express \(L_2\) as:
\[ L_2 = 10 \log_{10}\left(\frac{I_2}{I_0}\right) = 10 \log_{10}\left(\frac{I_2}{I_1} \cdot \frac{I_1}{I_0}\right) \]
This can be rewritten as:
\[ L_2 = 10 \log_{10}\left(\frac{I_2}{I_1}\right) + L_1 \]

5. **Substituting the Known Values**:
We substitute \(L_1 = 20\) dB and \(\frac{I_2}{I_1} = 10^{-3/2}\):
\[ L_2 = 10 \log_{10}(10^{-3/2}) + 20 \]
Simplifying further:
\[ L_2 = 10 \cdot \left(-\frac{3}{2}\right) + 20 = -15 + 20 = 5 \text{ dB} \]

### Final Answer:
The loudness at a distance of \(10^{7/4}\) m from the source is **5 dB**.