Due to point isotropic sound source, theintensity at a point is observed as 40 dB. The density of air is `rho=((15)/(11))(kg)/(m^3)` and velocity of sound in air is `330(m)/(s)`. Based on this information answer the following questions.
Q. The pressure amplitude at the observation point is

Options

A. `3(N)/(m^2)`
B. `3xx10^3(N)/(m^2)`
C. `3xx10^-3(N)/(m^2)`
D. `6xx10^-2(N)/(m^2)`

Explanation

To find the pressure amplitude at the observation point given the intensity level in decibels, we can follow these steps:

### Step 1: Convert Decibels to Intensity
The intensity level \( L \) in decibels is given by the formula:
\[ L = 10 \log_{10} \left( \frac{I}{I_0} \right) \]
where \( I_0 = 10^{-12} \, \text{W/m}^2 \) is the reference intensity. Given \( L = 40 \, \text{dB} \), we can rearrange the formula to find \( I \):
\[ 40 = 10 \log_{10} \left( \frac{I}{10^{-12}} \right) \]
Dividing both sides by 10:
\[ 4 = \log_{10} \left( \frac{I}{10^{-12}} \right) \]
Now, exponentiating both sides gives:
\[ 10^4 = \frac{I}{10^{-12}} \]
Multiplying both sides by \( 10^{-12} \):
\[ I = 10^4 \times 10^{-12} = 10^{-8} \, \text{W/m}^2 \]

### Step 2: Use the Intensity to Find Pressure Amplitude
The relationship between intensity \( I \), pressure amplitude \( \Delta P_m \), density \( \rho \), and velocity \( v \) is given by:
\[ I = \frac{(\Delta P_m)^2}{2 \rho v} \]
We can rearrange this to solve for \( \Delta P_m \):
\[ \Delta P_m = \sqrt{2 I \rho v} \]

### Step 3: Substitute Known Values
We have:
- \( I = 10^{-8} \, \text{W/m}^2 \)
- \( \rho = \frac{15}{11} \, \text{kg/m}^3 \)
- \( v = 330 \, \text{m/s} \)

Substituting these values into the equation:
\[ \Delta P_m = \sqrt{2 \times 10^{-8} \times \frac{15}{11} \times 330} \]

### Step 4: Calculate the Pressure Amplitude
Calculating the expression inside the square root:
1. Calculate \( 2 \times 10^{-8} = 2 \times 10^{-8} \)
2. Calculate \( \frac{15}{11} \approx 1.3636 \)
3. Calculate \( 1.3636 \times 330 \approx 450 \)
4. Now multiply:
\[ 2 \times 10^{-8} \times 450 \approx 9 \times 10^{-6} \]
5. Finally, take the square root:
\[ \Delta P_m = \sqrt{9 \times 10^{-6}} = 3 \times 10^{-3} \, \text{N/m}^2 \]

### Final Answer
The pressure amplitude at the observation point is:
\[ \Delta P_m = 3 \times 10^{-3} \, \text{N/m}^2 \]