Two aeroplanes 'A' and 'B' are moving away from one another with a speed of `720kmph`. The frequency of the whistle emitted by 'A' is `1100Hz`. The apparent frequency of the whistle as heard by the passenger of the aeroplane 'B' is (velocity of sound in air is `350ms^(-1)`).

Options

A. `300Hz`
B. `400Hz`
C. `500Hz`
D. `600Hz`

Explanation

To find the apparent frequency of the whistle as heard by the passenger of aeroplane 'B', we will use the Doppler effect formula for sound. The formula for apparent frequency (f') when both the source and observer are moving away from each other is given by:

\[ f' = f_0 \frac{v + v_o}{v + v_s} \]

Where:
- \( f' \) = apparent frequency
- \( f_0 \) = emitted frequency (1100 Hz)
- \( v \) = velocity of sound in air (350 m/s)
- \( v_o \) = velocity of the observer (aeroplane B) (200 m/s)
- \( v_s \) = velocity of the source (aeroplane A) (200 m/s)

### Step 1: Convert the speed of the aeroplanes from km/h to m/s
The speed of the aeroplanes is given as 720 km/h. To convert this to m/s, we use the conversion factor:

\[ \text{Speed in m/s} = \text{Speed in km/h} \times \frac{1000 \text{ m}}{3600 \text{ s}} = 720 \times \frac{1000}{3600} = 200 \text{ m/s} \]

### Step 2: Substitute the values into the Doppler effect formula
Now we substitute the values into the formula:

\[ f' = 1100 \frac{350 + 200}{350 + 200} \]

### Step 3: Calculate the apparent frequency
Since both the observer and the source are moving away from each other, we have:

\[ f' = 1100 \frac{350 - 200}{350 + 200} \]

Calculating the numerator and denominator:

\[ f' = 1100 \frac{150}{550} \]

### Step 4: Simplify the fraction
Now we simplify the fraction:

\[ f' = 1100 \times \frac{15}{55} \]

### Step 5: Further simplify
\[ f' = 1100 \times \frac{3}{11} \]

### Step 6: Calculate the final value
Calculating this gives:

\[ f' = 1100 \times 0.2727 \approx 300 \text{ Hz} \]

Thus, the apparent frequency of the whistle as heard by the passenger of aeroplane 'B' is approximately **300 Hz**.