The intensity of sound from a public loud speaker is `1mu "watt m"^(-2)` at a distance of 5 m. What is the intensity at a distance of 100 m ?

Answer

To solve the problem of finding the intensity of sound from a public loudspeaker at a distance of 100 m, we can use the inverse square law of intensity. Here’s a step-by-step solution:

### Step 1: Understand the relationship between intensity and distance
The intensity of sound (I) is inversely proportional to the square of the distance (R) from the source. This can be expressed mathematically as:
\[ I \propto \frac{1}{R^2} \]

### Step 2: Set up the equation
Let \( I_1 \) be the intensity at distance \( R_1 \) and \( I_2 \) be the intensity at distance \( R_2 \). According to the inverse square law:
\[ \frac{I_2}{I_1} = \frac{R_1^2}{R_2^2} \]

### Step 3: Substitute the known values
From the problem:
- \( I_1 = 1 \, \mu \text{W/m}^2 = 1 \times 10^{-6} \, \text{W/m}^2 \)
- \( R_1 = 5 \, \text{m} \)
- \( R_2 = 100 \, \text{m} \)

Now substituting these values into the equation:
\[ \frac{I_2}{1 \times 10^{-6}} = \frac{5^2}{100^2} \]

### Step 4: Calculate the ratio of distances
Calculating the squares:
\[ 5^2 = 25 \]
\[ 100^2 = 10000 \]
Thus:
\[ \frac{I_2}{1 \times 10^{-6}} = \frac{25}{10000} \]

### Step 5: Simplify the fraction
Now simplify \( \frac{25}{10000} \):
\[ \frac{25}{10000} = \frac{1}{400} \]

### Step 6: Solve for \( I_2 \)
Now we can express \( I_2 \):
\[ I_2 = 1 \times 10^{-6} \times \frac{1}{400} \]
\[ I_2 = \frac{1 \times 10^{-6}}{400} \]

### Step 7: Perform the final calculation
Calculating \( I_2 \):
\[ I_2 = 2.5 \times 10^{-9} \, \text{W/m}^2 \]

### Final Answer
The intensity of sound from the loudspeaker at a distance of 100 m is:
\[ I_2 = 2.5 \times 10^{-9} \, \text{W/m}^2 \]

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