Ten tuning forks are arranged in increasing order of frequency is such a way that any two nearest tuning forks produce `4 beast//sec`. The highest freqeuncy is twice of the lowest. Possible highest and the lowest frequencies are

Options

A. 80 and 40
B. 100 and 50
C. 44 and 22
D. 72 and 36

Answer

To solve the problem, we will follow these steps:

1. **Understanding the Problem**: We have 10 tuning forks arranged in increasing order of frequency. The difference in frequency between any two nearest tuning forks produces 4 beats per second. The highest frequency is twice the lowest frequency.

2. **Defining Variables**: Let the lowest frequency be \( f_1 \). Then, the frequencies of the tuning forks can be expressed as:
- \( f_1, f_1 + 4, f_1 + 8, f_1 + 12, f_1 + 16, f_1 + 20, f_1 + 24, f_1 + 28, f_1 + 32, f_1 + 36 \)

3. **Identifying the Highest Frequency**: The highest frequency, which is the 10th tuning fork, can be expressed as:
- \( f_{10} = f_1 + 36 \)

4. **Using the Given Condition**: According to the problem, the highest frequency is twice the lowest frequency:
- \( f_{10} = 2f_1 \)

5. **Setting Up the Equation**: We can now set up the equation using the expressions for \( f_{10} \):
- \( f_1 + 36 = 2f_1 \)

6. **Solving the Equation**: Rearranging the equation gives:
- \( 36 = 2f_1 - f_1 \)
- \( 36 = f_1 \)

7. **Finding the Highest Frequency**: Now that we have \( f_1 \), we can find the highest frequency:
- \( f_{10} = f_1 + 36 = 36 + 36 = 72 \)

8. **Final Answer**: The lowest frequency is \( 36 \) Hz and the highest frequency is \( 72 \) Hz.

Thus, the possible highest and lowest frequencies are:
- Lowest Frequency: \( 36 \) Hz
- Highest Frequency: \( 72 \) Hz