A source of sonic oscillation with frequency `n_0=600Hz` moves away and at right angles to a wall with velocity `u=30(m)/(s)`. A stationary reciever is located on the line of source in succession wall`rarr`source`rarr`receiver. If velocity of osund propagation is `v=330(m)/(s)`, then
Q. The beat frequency recorded by the receiver is

Options

A. (a)`110Hz`
B. (b)`210Hz`
C. (c)`150Hz`
D. (d)`220Hz`

Explanation

To solve the problem, we need to calculate the beat frequency recorded by the receiver when a source of sound moves away from a wall and a stationary receiver is positioned in line with the source and wall.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Frequency of the source, \( n_0 = 600 \, \text{Hz} \)
- Velocity of the source, \( u = 30 \, \text{m/s} \)
- Velocity of sound, \( v = 330 \, \text{m/s} \)

2. **Calculate the Frequency of the Direct Wave (\( n_d \)):**
The frequency of the direct wave can be calculated using the formula:
\[ n_d = n_0 \times \frac{v}{v - u} \]
Substituting the known values:
\[ n_d = 600 \times \frac{330}{330 - 30} = 600 \times \frac{330}{300} \]
Simplifying this:
\[ n_d = 600 \times 1.1 = 660 \, \text{Hz} \]

3. **Calculate the Frequency of the Reflected Sound (\( n_r \)):**
The frequency of the reflected sound can be calculated using the formula:
\[ n_r = n_0 \times \frac{v}{v + u} \]
Substituting the known values:
\[ n_r = 600 \times \frac{330}{330 + 30} = 600 \times \frac{330}{360} \]
Simplifying this:
\[ n_r = 600 \times \frac{11}{12} = 550 \, \text{Hz} \]

4. **Calculate the Beat Frequency:**
The beat frequency (\( f_b \)) is the absolute difference between the frequencies of the direct wave and the reflected sound:
\[ f_b = |n_d - n_r| = |660 - 550| = 110 \, \text{Hz} \]

5. **Final Answer:**
The beat frequency recorded by the receiver is \( 110 \, \text{Hz} \).

### Summary:
The beat frequency recorded by the receiver is **110 Hz**.