A transverse wave of frequency 250 Hz is travelling with a speed of 180 m/s. The path difference between any two point is `x xx 10^(-2)` m. When two points are `30^(@)` out of phase. Find the value of x.
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Explanation

To solve the problem, we need to find the value of \( x \) in the path difference \( x \times 10^{-2} \) m, given that the transverse wave has a frequency of 250 Hz, a speed of 180 m/s, and a phase difference of 30 degrees between two points.

### Step-by-Step Solution:

1. **Identify the given values:**
- Frequency \( f = 250 \) Hz
- Speed of wave \( v = 180 \) m/s
- Phase difference \( \Delta \phi = 30^\circ \)

2. **Calculate the wavelength \( \lambda \):**
The wavelength can be calculated using the formula:
\[ \lambda = \frac{v}{f} \]
Substituting the values:
\[ \lambda = \frac{180 \, \text{m/s}}{250 \, \text{Hz}} = 0.72 \, \text{m} \]

3. **Convert the phase difference from degrees to radians:**
Since the phase difference is given in degrees, we need to convert it to radians:
\[ \Delta \phi = 30^\circ = \frac{\pi}{6} \, \text{radians} \]

4. **Use the relationship between path difference and phase difference:**
The path difference \( \Delta x \) can be calculated using the formula:
\[ \Delta x = \frac{\lambda}{2\pi} \Delta \phi \]
Substituting the values of \( \lambda \) and \( \Delta \phi \):
\[ \Delta x = \frac{0.72 \, \text{m}}{2\pi} \times \frac{\pi}{6} \]

5. **Simplify the expression:**
\[ \Delta x = \frac{0.72}{2 \times 6} = \frac{0.72}{12} = 0.06 \, \text{m} \]

6. **Express the path difference in the required form:**
The path difference is given as \( x \times 10^{-2} \) m. Therefore, we can express \( 0.06 \) m as:
\[ 0.06 \, \text{m} = 6 \times 10^{-2} \, \text{m} \]
Thus, we find:
\[ x = 6 \]

### Final Answer:
The value of \( x \) is \( 6 \).