The frequency of fork A is 3% more than the frequency of a standard fork where as the frequency of fork B is 3% less. The fork A and B produce 6 beats per second. The frequency of standard fork will be -

Options

A. 100 Hz
B. 103 Hz
C. 97 Hz
D. 106 Hz

Explanation

To solve the problem step by step, let's denote the frequency of the standard fork as \( x \).

### Step 1: Determine the frequencies of forks A and B
- The frequency of fork A is 3% more than the frequency of the standard fork:
\[ f_A = x + 0.03x = 1.03x \]

- The frequency of fork B is 3% less than the frequency of the standard fork:
\[ f_B = x - 0.03x = 0.97x \]

### Step 2: Set up the equation for beats
- The number of beats per second produced by two tuning forks is given by the absolute difference of their frequencies:
\[ f_A - f_B = 6 \text{ beats per second} \]

Substituting the expressions for \( f_A \) and \( f_B \):
\[ 1.03x - 0.97x = 6 \]

### Step 3: Simplify the equation
- Simplifying the left side:
\[ 1.03x - 0.97x = 0.06x \]
Thus, we have:
\[ 0.06x = 6 \]

### Step 4: Solve for \( x \)
- To find \( x \), divide both sides by 0.06:
\[ x = \frac{6}{0.06} = 100 \]

### Conclusion
- The frequency of the standard fork is:
\[ \boxed{100 \text{ Hz}} \]