Two different sound sources `S_(1)` and `S_(2)` have frequencies in the ratio `1:2`. Source `S_(1)` is approaching towards observer and `S_(2)` receding from same observer. Speeds of both `S_(1)` and `S_(2)` are V each and speed of sound air is `330m//s`. If no beats are heard by the observer then the value of V is

Options

A. `50m//s`
B. `75m//s`
C. `110m//s`
D. `125m//s`

Explanation

To solve the problem, we need to use the concept of the Doppler effect, which describes how the frequency of a wave changes for an observer moving relative to the source of the wave.

### Step-by-Step Solution:

1. **Identify the Frequencies**:
Let the frequency of source \( S_1 \) be \( f \) and the frequency of source \( S_2 \) be \( 2f \) (since the ratio is \( 1:2 \)).

2. **Doppler Effect for Source \( S_1 \)** (approaching the observer):
The apparent frequency \( f' \) heard by the observer when source \( S_1 \) is approaching is given by:
\[ f' = f \frac{v_s}{v_s - v} \]
where \( v_s \) is the speed of sound in air (330 m/s) and \( v \) is the speed of source \( S_1 \).

3. **Doppler Effect for Source \( S_2 \)** (receding from the observer):
The apparent frequency \( f'' \) heard by the observer when source \( S_2 \) is receding is given by:
\[ f'' = 2f \frac{v_s}{v_s + v} \]

4. **Condition for No Beats**:
Since no beats are heard, the two frequencies must be equal:
\[ f' = f'' \]
Substituting the expressions for \( f' \) and \( f'' \):
\[ f \frac{v_s}{v_s - v} = 2f \frac{v_s}{v_s + v} \]

5. **Cancel \( f \)**:
Since \( f \) is common on both sides and is not zero, we can cancel it:
\[ \frac{v_s}{v_s - v} = 2 \frac{v_s}{v_s + v} \]

6. **Cross Multiply**:
Cross multiplying gives:
\[ v_s (v_s + v) = 2 v_s (v_s - v) \]

7. **Expand and Simplify**:
Expanding both sides:
\[ v_s^2 + v_s v = 2v_s^2 - 2v \]
Rearranging terms:
\[ v_s v + 2v = 2v_s^2 - v_s^2 \]
\[ v_s v + 2v = v_s^2 \]

8. **Solve for \( v \)**:
Rearranging gives:
\[ v_s v = v_s^2 - 2v \]
\[ v_s v + 2v = v_s^2 \]
\[ v( v_s + 2) = v_s^2 \]
\[ v = \frac{v_s^2}{v_s + 2} \]

9. **Substituting \( v_s = 330 \, \text{m/s} \)**:
\[ v = \frac{330^2}{330 + 2} = \frac{108900}{332} \approx 327.11 \, \text{m/s} \]

10. **Final Calculation**:
Since we need \( v \) to be a fraction of \( v_s \):
\[ v = \frac{330}{3} = 110 \, \text{m/s} \]

### Conclusion:
The speed \( V \) of both sources \( S_1 \) and \( S_2 \) is \( 110 \, \text{m/s} \).