A small speaker delivers `2 W` of audio will one detect `120 dB` intensity sound ? [Given reference intensity of sound as `10^(-12) W//m^(2)`]

Options

A. 40 cm
B. 20 cm
C. 10 cm
D. 30 cm

Answer

To determine whether a small speaker delivering 2 W of audio can produce a sound intensity of 120 dB, we can follow these steps:

### Step 1: Understand the relationship between intensity and decibels
The sound intensity level (in decibels) is given by the formula:
\[ L = 10 \log \left( \frac{I}{I_0} \right) \]
where:
- \( L \) is the sound level in decibels (dB),
- \( I \) is the intensity of the sound in watts per square meter (W/m²),
- \( I_0 \) is the reference intensity, which is \( 10^{-12} \, \text{W/m}^2 \).

### Step 2: Set up the equation for 120 dB
Given that \( L = 120 \, \text{dB} \), we can set up the equation:
\[ 120 = 10 \log \left( \frac{I}{10^{-12}} \right) \]

### Step 3: Solve for intensity \( I \)
Dividing both sides by 10:
\[ 12 = \log \left( \frac{I}{10^{-12}} \right) \]
To eliminate the logarithm, we exponentiate both sides:
\[ 10^{12} = \frac{I}{10^{-12}} \]
Multiplying both sides by \( 10^{-12} \):
\[ I = 10^{12} \times 10^{-12} = 1 \, \text{W/m}^2 \]

### Step 4: Relate intensity to power and area
The intensity \( I \) can also be expressed in terms of power \( P \) and area \( A \):
\[ I = \frac{P}{A} \]
For a point source, the area \( A \) is given by the surface area of a sphere:
\[ A = 4 \pi r^2 \]
Thus, we can write:
\[ 1 = \frac{2}{4 \pi r^2} \]

### Step 5: Solve for the radius \( r \)
Rearranging the equation gives:
\[ 4 \pi r^2 = 2 \]
\[ r^2 = \frac{2}{4 \pi} = \frac{1}{2 \pi} \]
Taking the square root:
\[ r = \sqrt{\frac{1}{2 \pi}} \approx 0.399 \, \text{m} \]

### Step 6: Conclusion
Thus, the radius at which the sound intensity is 120 dB is approximately \( 0.399 \, \text{m} \) or \( 39.9 \, \text{cm} \). Therefore, a small speaker delivering 2 W of audio can indeed produce a sound intensity of 120 dB.

---