16 tuning forks are arranged in increasing order of frequency. Any two consecutive tuning forks when sounded together produce 8 beats per second. If the frequency of last tuning fork is twice that of first, the frequency of first tuning fork is :-

Options

A. `60`
B. `80`
C. `100`
D. `120`

Explanation

To solve the problem, we need to find the frequency of the first tuning fork given the conditions of the problem. Let's break it down step by step.

### Step 1: Define the variables
Let the frequency of the first tuning fork be \( n \). According to the problem, the frequency of the last tuning fork is twice that of the first tuning fork. Therefore, the frequency of the last tuning fork can be expressed as:
\[ \text{Frequency of last tuning fork} = 2n \]

### Step 2: Understand the beat frequency
The problem states that any two consecutive tuning forks produce 8 beats per second. This means that the difference in frequency between any two consecutive tuning forks is 8 Hz. Thus, we can express the frequencies of the tuning forks in terms of \( n \):
- Frequency of the first tuning fork: \( n \)
- Frequency of the second tuning fork: \( n + 8 \)
- Frequency of the third tuning fork: \( n + 16 \)
- ...
- Frequency of the 16th tuning fork: \( n + 120 \)

### Step 3: Set up the equation for the last tuning fork
Since the frequency of the last tuning fork (16th tuning fork) is given by:
\[ n + 120 = 2n \]
We can rearrange this equation to find \( n \):
\[ 120 = 2n - n \]
\[ 120 = n \]

### Step 4: Conclusion
Thus, the frequency of the first tuning fork is:
\[ \text{Frequency of the first tuning fork} = n = 120 \text{ Hz} \]

### Final Answer
The frequency of the first tuning fork is **120 Hz**.

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