Two sound waves of frequencies `100Hz` and `102Hz` and having same amplitude `'A'` are interfering. At a stationary detector, which can detect resultant amplitude greater than or equal to `A`. So, in a given time interval of 12 seconds, finds the total duration in which detector is active.

Explanation

To solve the problem, we need to determine the total duration in which the detector is active when two sound waves of frequencies 100 Hz and 102 Hz interfere with each other. The detector is active when the resultant amplitude is greater than or equal to the amplitude \( A \).

### Step-by-Step Solution:

1. **Determine the Beat Frequency**:
The beat frequency \( f_b \) is given by the difference in frequencies of the two sound waves:
\[ f_b = |f_2 - f_1| = |102 \, \text{Hz} - 100 \, \text{Hz}| = 2 \, \text{Hz} \]

2. **Calculate the Time Period of Beats**:
The time period \( T_b \) of the beats is the inverse of the beat frequency:
\[ T_b = \frac{1}{f_b} = \frac{1}{2 \, \text{Hz}} = 0.5 \, \text{seconds} \]

3. **Determine the Duration of Active Time in One Beat Cycle**:
In one cycle of beats (0.5 seconds), the amplitude will be equal to or greater than \( A \) for a certain duration. The amplitude reaches its maximum when the cosine term in the interference equation is equal to 1, and it is zero when the cosine term is equal to -1.

The condition for the resultant amplitude \( R \) to be at least \( A \) can be expressed as:
\[ R = 2A \cos\left(\frac{2\pi}{T_b} t\right) \geq A \]
This simplifies to:
\[ \cos\left(\frac{2\pi}{T_b} t\right) \geq \frac{1}{2} \]

4. **Find the Angles for the Condition**:
The cosine function is equal to \( \frac{1}{2} \) at angles \( \frac{\pi}{3} \) and \( \frac{5\pi}{3} \). Therefore, we can find the time intervals:
\[ \frac{2\pi}{T_b} t = \frac{\pi}{3} \quad \Rightarrow \quad t = \frac{T_b}{6} = \frac{0.5}{6} = \frac{1}{12} \, \text{seconds} \]
\[ \frac{2\pi}{T_b} t = \frac{5\pi}{3} \quad \Rightarrow \quad t = \frac{5T_b}{6} = \frac{5 \times 0.5}{6} = \frac{5}{12} \, \text{seconds} \]

5. **Calculate the Duration of Active Time**:
The duration during which the amplitude is at least \( A \) in one cycle is:
\[ \text{Active Time} = t_{\text{max}} - t_{\text{min}} = \left(\frac{5}{12} - \frac{1}{12}\right) = \frac{4}{12} = \frac{1}{3} \, \text{seconds} \]

6. **Determine the Number of Cycles in 12 Seconds**:
The number of cycles in 12 seconds is:
\[ \text{Number of cycles} = \frac{12}{T_b} = \frac{12}{0.5} = 24 \]

7. **Calculate Total Active Time in 12 Seconds**:
The total active time in 12 seconds is:
\[ \text{Total Active Time} = \text{Number of cycles} \times \text{Active Time per cycle} = 24 \times \frac{1}{3} = 8 \, \text{seconds} \]

### Final Answer:
The total duration in which the detector is active in 12 seconds is **8 seconds**.