A plane longitudinal wave of angular frequency `omega`=1000 rad Is is travelling along positive x direction in a homogeneous gaseous medium of density `d = 1kgm^(-3)` . Intensity of wave is `I = 10^(-10) W.m^(-2)` and maximum pressure change is `(Delta P)_m = 2 xx10^(-4) Nm^(-2)` . Assuming at x = 0, initial phase of medium particles to be zero
Velocity of the wave is

Options

A. `500 ms^(-1)`
B. `100 ms^(-1)`
C. `300 ms^(-1)`
D. `200 ms^(-1)`

Answer

To find the velocity of the wave, we can use the information provided in the question and apply the relevant formulas for wave motion.

### Step-by-Step Solution:

1. **Understanding the Given Information**:
- Angular frequency, \(\omega = 1000 \, \text{rad/s}\)
- Density of the medium, \(d = 1 \, \text{kg/m}^3\)
- Intensity of the wave, \(I = 10^{-10} \, \text{W/m}^2\)
- Maximum pressure change, \(\Delta P_m = 2 \times 10^{-4} \, \text{N/m}^2\)

2. **Relating Intensity to Wave Properties**:
The intensity \(I\) of a wave can be expressed in terms of the displacement amplitude \(A\), angular frequency \(\omega\), and density \(d\) as:
\[ I = \frac{1}{2} d v A^2 \]
where \(v\) is the velocity of the wave.

3. **Using the Maximum Pressure Change**:
The maximum pressure change in a longitudinal wave is given by:
\[ \Delta P_m = d v A \omega \]
Rearranging this gives:
\[ A = \frac{\Delta P_m}{d v \omega} \]

4. **Substituting \(A\) into the Intensity Equation**:
We can substitute the expression for \(A\) into the intensity equation:
\[ I = \frac{1}{2} d v \left(\frac{\Delta P_m}{d v \omega}\right)^2 \]
Simplifying this gives:
\[ I = \frac{1}{2} \frac{\Delta P_m^2}{d v \omega^2} \]

5. **Solving for Velocity \(v\)**:
Rearranging the above equation to solve for \(v\):
\[ v = \frac{\Delta P_m^2}{2 I d \omega^2} \]
Plugging in the known values:
- \(\Delta P_m = 2 \times 10^{-4} \, \text{N/m}^2\)
- \(I = 10^{-10} \, \text{W/m}^2\)
- \(d = 1 \, \text{kg/m}^3\)
- \(\omega = 1000 \, \text{rad/s}\)

\[ v = \frac{(2 \times 10^{-4})^2}{2 \times 10^{-10} \times 1 \times (1000)^2} \]

6. **Calculating \(v\)**:
\[ v = \frac{4 \times 10^{-8}}{2 \times 10^{-10} \times 1 \times 10^6} = \frac{4 \times 10^{-8}}{2 \times 10^{-4}} = 200 \, \text{m/s} \]

### Final Answer:
The velocity of the wave is \(v = 200 \, \text{m/s}\).