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 will follow these steps:

### Step 1: Understand the given information
We know that the loudness of sound from an isotropic point source at a distance of 10 m is 20 dB. We need to find the loudness at a distance of \(10^{7/4}\) m from the source.

### Step 2: Use the formula for loudness in decibels
The formula for loudness \(L\) in decibels is given by:
\[ L = 10 \log_{10} \left(\frac{I}{I_0}\right) \]
where \(I\) is the intensity of sound at the point and \(I_0\) is the reference intensity.

### Step 3: Relate intensity to distance
The intensity \(I\) of sound from a point source is inversely proportional to the square of the distance from the source:
\[ I \propto \frac{P}{4\pi r^2} \]
where \(P\) is the power of the source and \(r\) is the distance from the source.

### Step 4: Set up equations for the two distances
1. At distance \(r_1 = 10\) m:
\[ L_1 = 20 \text{ dB} = 10 \log_{10} \left(\frac{I_1}{I_0}\right) \]
This implies:
\[ I_1 = \frac{P}{4\pi (10)^2} \]

2. At distance \(r_2 = 10^{7/4}\) m:
\[ L_2 = 10 \log_{10} \left(\frac{I_2}{I_0}\right) \]
This implies:
\[ I_2 = \frac{P}{4\pi (10^{7/4})^2} \]

### Step 5: Relate the two intensities
Now we can express \(L_2\) in terms of \(L_1\):
\[ L_2 - L_1 = 10 \log_{10} \left(\frac{I_2}{I_1}\right) \]
Substituting the expressions for \(I_1\) and \(I_2\):
\[ L_2 - 20 = 10 \log_{10} \left(\frac{\frac{P}{4\pi (10^{7/4})^2}}{\frac{P}{4\pi (10)^2}}\right) \]
The \(P\) and \(4\pi\) cancel out:
\[ L_2 - 20 = 10 \log_{10} \left(\frac{(10)^2}{(10^{7/4})^2}\right) \]

### Step 6: Simplify the logarithm
\[ L_2 - 20 = 10 \log_{10} \left(\frac{100}{10^{7/2}}\right) \]
\[ = 10 \log_{10} \left(10^{2 - 3.5}\right) = 10 \log_{10} \left(10^{-1.5}\right) \]
\[ = 10 \times (-1.5) = -15 \]

### Step 7: Solve for \(L_2\)
\[ L_2 - 20 = -15 \implies L_2 = 20 - 15 = 5 \text{ dB} \]

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

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