`A, B`and `C` are three tuning forks. Frequency of `A` is `350 H_(Z)` . Beats produced by `A` and `B` are `5//s` and by `B` and `C` are `4//s` . When a wax is put on `A` beat frequency between `A` and `B` is `2 H_(Z)` and between `a` and `C` is `6 H_(z)` . Then, frequency of `B` and `C` respectively , are

Options

A. `(a)355 H_(Z)`, `349 h_(Z)`
B. `(b)345 H_(Z)` , `341H_(Z)`
C. `(c)355 H_(Z)` , `341 H_(Z)`
D. `(d)345 H_(Z)`, `349H_(Z)`

Explanation

To solve the problem step by step, we will analyze the information given and apply the concept of beats.

### Step 1: Understand the given frequencies and beat frequencies
- The frequency of tuning fork A (FA) is given as 350 Hz.
- The beat frequency between A and B is 5 Hz.
- The beat frequency between B and C is 4 Hz.
- After putting wax on A, the new frequency of A (FA') will be less than 350 Hz, and the new beat frequency between A and B is 2 Hz.
- The beat frequency between A and C after putting wax is 6 Hz.

### Step 2: Set up equations based on the beat frequencies
1. For the first case (before putting wax):
- The beat frequency between A and B is given by:
\[ |FA - FB| = 5 \]
- This gives us two possibilities:
\[ FA - FB = 5 \quad \text{(1)} \]
or
\[ FB - FA = 5 \quad \text{(2)} \]

2. For the second case (after putting wax):
- The new frequency of A is FA', and the beat frequency between A and B is:
\[ |FA' - FB| = 2 \]
- This gives us:
\[ FA' - FB = 2 \quad \text{(3)} \]
- Since FA' < FA, we can conclude:
\[ FB - FA' = 2 \quad \text{(4)} \]

3. For the beat frequency between B and C:
- We have:
\[ |FB - FC| = 4 \]
- This gives us:
\[ FB - FC = 4 \quad \text{(5)} \]
- or
\[ FC - FB = 4 \quad \text{(6)} \]

### Step 3: Solve for FB using equations
From equation (1):
\[ FA - FB = 5 \]
Substituting FA = 350 Hz:
\[ 350 - FB = 5 \]
\[ FB = 350 - 5 = 345 \text{ Hz} \]

### Step 4: Calculate FA' using FB
Using equation (4):
\[ FB - FA' = 2 \]
Substituting FB = 345 Hz:
\[ 345 - FA' = 2 \]
\[ FA' = 345 - 2 = 343 \text{ Hz} \]

### Step 5: Solve for FC using FB
Using equation (5):
\[ FB - FC = 4 \]
Substituting FB = 345 Hz:
\[ 345 - FC = 4 \]
\[ FC = 345 - 4 = 341 \text{ Hz} \]

### Summary of Frequencies
- Frequency of B (FB) = 345 Hz
- Frequency of C (FC) = 341 Hz

### Final Answer
The frequencies of B and C are:
- Frequency of B = 345 Hz
- Frequency of C = 341 Hz