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

Explanation

To solve the problem, we need to find the possible highest and lowest frequencies of the tuning forks, given the conditions in the question.

### Step-by-Step Solution:

1. **Understanding the Arrangement of Tuning Forks:**
- We have 10 tuning forks arranged in increasing order of frequency.
- Since there are 10 forks, there will be 9 intervals between them.

2. **Understanding Beats:**
- It is given that any two nearest tuning forks produce 4 beats per second. This means the difference in frequency between any two nearest tuning forks is 4 Hz.

3. **Setting Up the Frequencies:**
- Let the lowest frequency be \( n_1 \).
- The highest frequency, according to the problem, is twice the lowest frequency, so we can express it as \( n_2 = 2n_1 \).

4. **Relating Frequencies with Beats:**
- The difference between the highest and lowest frequency can be expressed as:
\[ n_2 - n_1 = 9 \times 4 \]
- This is because there are 9 intervals, each contributing a difference of 4 Hz.

5. **Substituting the Expression for \( n_2 \):**
- Substitute \( n_2 = 2n_1 \) into the equation:
\[ 2n_1 - n_1 = 36 \]
- This simplifies to:
\[ n_1 = 36 \]

6. **Finding the Highest Frequency:**
- Now, substitute \( n_1 \) back to find \( n_2 \):
\[ n_2 = 2n_1 = 2 \times 36 = 72 \]

7. **Conclusion:**
- Therefore, the lowest frequency \( n_1 \) is 36 Hz and the highest frequency \( n_2 \) is 72 Hz.

### Final Answer:
- The possible lowest frequency is 36 Hz and the highest frequency is 72 Hz.