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 afer crossing the body will be (velocity of sound `= 300 m//s`, `g = 10 m//s^(2))`

Options

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

Answer

To solve the problem step by step, we will calculate the apparent frequency observed by the man in the balloon just before and just after crossing the sounding body.

### Step 1: Determine the velocity of the sounding body just before crossing the balloon
The sounding body is dropped from a height and falls under gravity. The velocity of the sounding body just before crossing the balloon after 1 second can be calculated using the formula:

\[ v = g \cdot t \]

Where:
- \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity)
- \( t = 1 \, \text{s} \)

Calculating this gives:

\[ v = 10 \cdot 1 = 10 \, \text{m/s} \]

### Step 2: Calculate the apparent frequency just before crossing the balloon
The apparent frequency \( f' \) observed by the observer in the balloon just before crossing the body can be calculated using the Doppler effect formula:

\[ f' = f \cdot \frac{v + v_o}{v - v_s} \]

Where:
- \( f = 150 \, \text{Hz} \) (frequency of the source)
- \( v = 300 \, \text{m/s} \) (velocity of sound)
- \( v_o = 2 \, \text{m/s} \) (velocity of the observer moving upwards)
- \( v_s = 10 \, \text{m/s} \) (velocity of the source moving downwards)

Substituting the values:

\[ f' = 150 \cdot \frac{300 + 2}{300 - 10} \]

Calculating this gives:

\[ f' = 150 \cdot \frac{302}{290} \]

### Step 3: Calculate the apparent frequency just after crossing the balloon
After crossing, the source is moving downwards at 10 m/s and the observer is moving upwards at 2 m/s. The apparent frequency \( f'' \) can be calculated using:

\[ f'' = f \cdot \frac{v - v_o}{v + v_s} \]

Substituting the values:

\[ f'' = 150 \cdot \frac{300 - 2}{300 + 10} \]

Calculating this gives:

\[ f'' = 150 \cdot \frac{298}{310} \]

### Step 4: Calculate the difference in frequencies
Now, we find the difference in the frequencies observed just before and just after crossing:

\[ \Delta f = f' - f'' \]

Substituting the expressions we derived:

\[ \Delta f = 150 \cdot \frac{302}{290} - 150 \cdot \frac{298}{310} \]

Factoring out 150 gives:

\[ \Delta f = 150 \left( \frac{302}{290} - \frac{298}{310} \right) \]

### Step 5: Simplify and calculate the final difference
To combine the fractions, we need a common denominator:

\[ \Delta f = 150 \left( \frac{302 \cdot 310 - 298 \cdot 290}{290 \cdot 310} \right) \]

Calculating the numerator:

\[ 302 \cdot 310 = 93620 \]
\[ 298 \cdot 290 = 86420 \]
\[ 93620 - 86420 = 7200 \]

Thus:

\[ \Delta f = 150 \cdot \frac{7200}{290 \cdot 310} \]

Calculating the denominator:

\[ 290 \cdot 310 = 89900 \]

Finally:

\[ \Delta f = 150 \cdot \frac{7200}{89900} \]

Calculating this gives:

\[ \Delta f \approx 12 \, \text{Hz} \]

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