A source of sound producing wavelength 50 cm is moving away from a stationary observer with `(1/5)^(th)` speed of sound. Then what is the wavelength of sound received by the observer?

Options

A. 55 cm
B. 40 cm
C. 60 cm
D. 70 cm

Explanation

To solve the problem, we need to find the wavelength of sound received by a stationary observer when the source of sound is moving away from the observer.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Wavelength of the source, \( \lambda = 50 \) cm.
- Speed of sound, \( V \) (we will keep it as a variable for now).
- Speed of the source, \( V_s = \frac{1}{5} V \) (the source is moving away from the observer).

2. **Determine the Effective Speed of Sound:**
- Since the source is moving away from the observer, the effective speed of sound relative to the observer is given by:
\[ V + V_s = V + \frac{1}{5} V = \frac{5}{5} V + \frac{1}{5} V = \frac{6}{5} V \]

3. **Use the Formula for Wavelength Received by the Observer:**
- The formula to calculate the wavelength received by the observer when the source is moving away is:
\[ \lambda' = \frac{V + V_s}{V} \cdot \lambda \]
- Substituting the values we have:
\[ \lambda' = \frac{\frac{6}{5} V}{V} \cdot 50 \text{ cm} \]

4. **Simplify the Expression:**
- The \( V \) cancels out:
\[ \lambda' = \frac{6}{5} \cdot 50 \text{ cm} \]
- Now, calculate \( \lambda' \):
\[ \lambda' = \frac{300}{5} \text{ cm} = 60 \text{ cm} \]

5. **Conclusion:**
- The wavelength of sound received by the observer is \( 60 \) cm.

### Final Answer:
The wavelength of sound received by the observer is **60 cm**.