A fiber of length `10`cm is illuminated with light from an light emitting diode (LED) which is turned on and off repeatedely for equal amount of time. The speed of the pulse of light are `2.00xx10^(8)m//s` and `2.1xx10^(8)m//s` in fiber. Maximum frequency of LED os that pulse arrive without overlapping is `60X(KHz)`. Calculate `X`.

Explanation

To solve the problem, we need to calculate the time taken by the light pulses to travel through the fiber and find the maximum frequency of the LED such that the pulses do not overlap.

### Step-by-Step Solution:

1. **Convert the length of the fiber to meters**:
\[ \text{Length of fiber} = 10 \text{ cm} = 0.1 \text{ m} \]

2. **Calculate the time taken for the pulse to travel through the fiber for both speeds**:
- For the first speed \( v_1 = 2.00 \times 10^8 \text{ m/s} \):
\[ T_1 = \frac{\text{Length}}{\text{Speed}} = \frac{0.1 \text{ m}}{2.00 \times 10^8 \text{ m/s}} = 5.0 \times 10^{-10} \text{ s} \]

- For the second speed \( v_2 = 2.1 \times 10^8 \text{ m/s} \):
\[ T_2 = \frac{0.1 \text{ m}}{2.1 \times 10^8 \text{ m/s}} \approx 4.76 \times 10^{-10} \text{ s} \]

3. **Calculate the time difference between the two pulses**:
\[ \Delta T = T_1 - T_2 = (5.0 \times 10^{-10} - 4.76 \times 10^{-10}) \text{ s} = 0.24 \times 10^{-10} \text{ s} = 2.4 \times 10^{-11} \text{ s} \]

4. **Determine the maximum frequency of the LED**:
The maximum frequency \( f \) is given by the inverse of the time difference:
\[ f = \frac{1}{\Delta T} = \frac{1}{2.4 \times 10^{-11}} \approx 4.17 \times 10^{10} \text{ Hz} \]

5. **Convert frequency to kilohertz**:
\[ f \approx 4.17 \times 10^{10} \text{ Hz} = 4.17 \times 10^{7} \text{ kHz} \]

6. **Set the maximum frequency equal to \( 60x \) kHz**:
\[ 60x = 4.17 \times 10^{7} \text{ kHz} \]

7. **Solve for \( x \)**:
\[ x = \frac{4.17 \times 10^{7}}{60} \approx 6.95 \times 10^{5} \]

Rounding to the nearest whole number, we find:
\[ x \approx 7 \]

### Final Answer:
The value of \( X \) is \( 7 \).