A plane wave propagates along positive x-direction in a homogeneous medium of density `p=200 kg//m^(3)`. Due to propagation of the wave medium particle oscillate. Space density of their oscillation energy is `E=0.16pi^(2) J//m^(3)` and maximum shear strain produced in the mendium is `phi_(0)=8pixx10^(-5)`. if at an instant, phase difference between two particles located at points `(1m,1m,1m)` and `(2m, 2m, 2m,)` is `Deltatheta=144^(@)`, assuming at `t=0` phase of particle at `x=0` to be zero,
wave length is

Options

A. `2.5 m`
B. `5 m`
C. `10 m`
D. `6 m`

Explanation

To find the wavelength of the plane wave propagating in a homogeneous medium, we can follow these steps:

### Step 1: Understand the given information
- Density of the medium, \( p = 200 \, \text{kg/m}^3 \)
- Energy density of oscillation, \( E = 0.16 \pi^2 \, \text{J/m}^3 \)
- Maximum shear strain, \( \phi_0 = 8 \pi \times 10^{-5} \)
- Phase difference between two particles located at \( (1m, 1m, 1m) \) and \( (2m, 2m, 2m) \) is \( \Delta \theta = 144^\circ \)

### Step 2: Convert phase difference to radians
Convert the phase difference from degrees to radians:
\[ \Delta \theta = 144^\circ \times \frac{\pi}{180} = \frac{144 \pi}{180} = \frac{8\pi}{10} = 0.8\pi \, \text{radians} \]

### Step 3: Calculate the distance between the two points
The distance \( \Delta x \) between the two points is:
\[ \Delta x = 2m - 1m = 1m \]

### Step 4: Relate phase difference to wavelength
The phase difference \( \Delta \theta \) is related to the wavelength \( \lambda \) by the equation:
\[ \Delta \theta = \frac{2\pi \Delta x}{\lambda} \]
Rearranging this gives:
\[ \lambda = \frac{2\pi \Delta x}{\Delta \theta} \]

### Step 5: Substitute the values
Substituting \( \Delta x = 1m \) and \( \Delta \theta = 0.8\pi \):
\[ \lambda = \frac{2\pi \times 1}{0.8\pi} = \frac{2}{0.8} = 2.5 \, \text{m} \]

### Conclusion
The wavelength \( \lambda \) of the wave is:
\[ \lambda = 2.5 \, \text{m} \]