A point sound is located in the perpendicular to the plane of a ring drawn through the centre `O` of the ring. The distance between the point `O` and the source is `l = 1.00 m`, the radius of the ring is `R = 0.50 m`. If the mean energy flow rate across the area encloseed by the ring is (in `muW`). Find `(x_(0))/(5)`. Given, at point `O` the energy intensity of sound is equal to `I_(0) = 30 muW//m^(2)` and assuming the damping of the waves is negligible.

Explanation

To solve the problem step by step, we will use the information given about the sound intensity and the geometry of the setup.

### Step 1: Understand the Geometry
We have a point sound source located at a distance \( l = 1.00 \, m \) from the center \( O \) of a ring with radius \( R = 0.50 \, m \). The sound waves emitted from the source will spread out in a spherical manner.

### Step 2: Calculate the Area of the Ring
The area \( A \) of the ring can be approximated as the area of a circle with radius \( R \):
\[ A = \pi R^2 = \pi (0.50)^2 = \frac{\pi}{4} \, m^2 \]

### Step 3: Understand the Intensity
The intensity of sound at point \( O \) is given as \( I_0 = 30 \, \mu W/m^2 \). Intensity is defined as power per unit area.

### Step 4: Calculate the Total Power at Point O
The total power \( P \) radiated by the source can be calculated using the intensity at point \( O \):
\[ P = I_0 \times A = 30 \, \mu W/m^2 \times \frac{\pi}{4} \, m^2 \]
Calculating this gives:
\[ P = 30 \times \frac{\pi}{4} \, \mu W \approx 23.56 \, \mu W \]

### Step 5: Calculate the Energy Flow Rate Across the Area Enclosed by the Ring
The mean energy flow rate across the area enclosed by the ring can be computed using the total power and the area:
\[ \text{Energy Flow Rate} = \frac{P}{A} = \frac{30 \, \mu W}{\pi/4} = \frac{120 \, \mu W}{\pi} \approx 38.2 \, \mu W \]

### Step 6: Find \( \frac{x_0}{5} \)
Since the problem asks for \( \frac{x_0}{5} \) and we have found the energy flow rate to be approximately \( 38.2 \, \mu W \), we can denote this value as \( x_0 \):
\[ x_0 = 38.2 \, \mu W \]
Thus,
\[ \frac{x_0}{5} = \frac{38.2}{5} \approx 7.64 \, \mu W \]

### Final Answer
\[ \frac{x_0}{5} \approx 7.64 \, \mu W \]

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