A sounding body emitting a frequency of `150 H_(Z)` is dropped from a height. During its fall under gravity it crosses a balloon moving upwards with a constant velocity of `2 m//s` one second after it started to fall . The difference in the frequency observer by the man in balloon just before and just after crossing the body will be (velocity of sound `= 300 m//s`, `g = 10 m//s^(2))` a) 12 b) 6 c) 8 d) 4

Options

A. `12`
B. `6`
C. `8`
D. `4`

Explanation

To solve the problem, we need to calculate the difference in frequency observed by a man in the balloon just before and just after crossing the sounding body. We will use the Doppler effect formula for sound waves.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Frequency of the sounding body (source), \( f = 150 \, \text{Hz} \)
- Velocity of sound, \( V = 300 \, \text{m/s} \)
- Velocity of the balloon (observer), \( V_o = 2 \, \text{m/s} \) (upwards)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

2. **Calculate the Velocity of the Source Just Before Crossing:**
- The source is dropped from rest, so its velocity after 1 second of free fall can be calculated using the formula:
\[ V_s = g \cdot t = 10 \cdot 1 = 10 \, \text{m/s} \, \text{(downwards)} \]

3. **Calculate the Apparent Frequency Just Before Crossing (Source Approaching Observer):**
- When the source is approaching the observer, we use the formula:
\[ f' = f \cdot \frac{V + V_o}{V - V_s} \]
- Substituting the values:
\[ f' = 150 \cdot \frac{300 + 2}{300 - 10} = 150 \cdot \frac{302}{290} \]
- Calculating this gives:
\[ f' \approx 150 \cdot 1.0414 \approx 156.21 \, \text{Hz} \]

4. **Calculate the Apparent Frequency Just After Crossing (Source Receding from Observer):**
- When the source is receding from the observer, we use the formula:
\[ f'' = f \cdot \frac{V - V_o}{V + V_s} \]
- Substituting the values:
\[ f'' = 150 \cdot \frac{300 - 2}{300 + 10} = 150 \cdot \frac{298}{310} \]
- Calculating this gives:
\[ f'' \approx 150 \cdot 0.96 \approx 144.84 \, \text{Hz} \]

5. **Calculate the Difference in Frequencies:**
- The difference in frequency observed by the man in the balloon just before and just after crossing the body is:
\[ \Delta f = f' - f'' \approx 156.21 - 144.84 \approx 11.37 \, \text{Hz} \]
- Rounding this gives approximately \( 12 \, \text{Hz} \).

### Final Answer:
The difference in frequency observed by the man in the balloon just before and just after crossing the body is approximately **12 Hz**.