A sound of intensity `I` is greater by `3.0103 dB` from anoterh sound of intersity `10 nW cm^(-2)`. The absoulte value of intensity of sound level `I` is `Wm^(-2)` is :

Options

A. `2.5 xx 10^(-4)`
B. `2 xx 10^(-4)`
C. `2.0 xx 10^(-2)`
D. `2.5 xx 10^(-2)`

Answer

To solve the problem step by step, we will follow the reasoning outlined in the video transcript.

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

### Step 2: Identify the given values
We are given:
- The intensity of another sound \(I' = 10 \, \text{nW/cm}^2\)
- The difference in sound levels is \(3.0103 \, \text{dB}\)

### Step 3: Convert \(I'\) to W/m²
First, we convert \(I'\) from nanowatts per centimeter squared to watts per meter squared:
\[ I' = 10 \, \text{nW/cm}^2 = 10 \times 10^{-9} \, \text{W/cm}^2 \]
Since \(1 \, \text{cm}^2 = 10^{-4} \, \text{m}^2\), we have:
\[ I' = 10 \times 10^{-9} \, \text{W/cm}^2 \times \frac{1}{10^{-4} \, \text{m}^2/\text{cm}^2} = 10 \times 10^{-9} \times 10^{4} \, \text{W/m}^2 = 10^{-5} \, \text{W/m}^2 \]

### Step 4: Set up the equation for the difference in decibels
The difference in sound levels can be expressed as:
\[ \beta = \beta_1 - \beta_2 = 10 \log_{10} \left( \frac{I}{I_0} \right) - 10 \log_{10} \left( \frac{I'}{I_0} \right) \]
This simplifies to:
\[ \beta = 10 \log_{10} \left( \frac{I}{I'} \right) \]

### Step 5: Substitute the known values
We know that \(\beta = 3.0103 \, \text{dB}\) and \(I' = 10^{-5} \, \text{W/m}^2\). Therefore:
\[ 3.0103 = 10 \log_{10} \left( \frac{I}{10^{-5}} \right) \]

### Step 6: Solve for \(I\)
Dividing both sides by 10:
\[ 0.30103 = \log_{10} \left( \frac{I}{10^{-5}} \right) \]
Now, converting from logarithmic form to exponential form:
\[ 10^{0.30103} = \frac{I}{10^{-5}} \]
Calculating \(10^{0.30103}\) gives approximately \(2\):
\[ 2 = \frac{I}{10^{-5}} \]
Multiplying both sides by \(10^{-5}\):
\[ I = 2 \times 10^{-5} \, \text{W/m}^2 \]

### Final Answer
The absolute value of intensity \(I\) is:
\[ I = 2 \times 10^{-5} \, \text{W/m}^2 \]

---