The frequency of waves emitted from a radar is 750 MHz. The frequency of reflected wave from the aeroplane as observed at the radar station is incrased by `2.5` KHz. The speed of aeroplane is.

Options

A. `4 Kms^(-1)`
B. `2 Kms^(-1)`
C. `1 Kms^(-1)`
D. `0.5 Kms^(-1)`

Answer

To solve the problem, we will use the Doppler effect formula for waves. Here’s a step-by-step solution:

### Step 1: Understand the given data
- Frequency of the radar wave (f) = 750 MHz = 750 × 10^6 Hz
- Increase in frequency of the reflected wave (Δf) = 2.5 kHz = 2.5 × 10^3 Hz

### Step 2: Convert the increase in frequency to Hz
- Δf = 2.5 kHz = 2500 Hz

### Step 3: Use the Doppler Effect formula
The formula for the observed frequency (f') when the source is moving towards the observer is given by:
\[ f' = f + \Delta f \]
However, since we are interested in the change in frequency due to the motion of the airplane, we can use:
\[ \Delta f = \frac{2V_r}{\lambda} \]
Where:
- \( V_r \) = radial velocity of the airplane
- \( \lambda \) = wavelength of the radar wave

### Step 4: Calculate the wavelength (λ)
The wavelength (λ) can be calculated using the formula:
\[ \lambda = \frac{c}{f} \]
Where:
- \( c \) = speed of light = \( 3 \times 10^8 \) m/s
- \( f \) = frequency of the radar wave = \( 750 \times 10^6 \) Hz

Calculating λ:
\[ \lambda = \frac{3 \times 10^8 \text{ m/s}}{750 \times 10^6 \text{ Hz}} = 0.4 \text{ m} \]

### Step 5: Substitute values into the Doppler effect formula
Now substituting the values into the Doppler effect formula:
\[ 2500 \text{ Hz} = \frac{2V_r}{0.4 \text{ m}} \]

### Step 6: Solve for radial velocity (Vr)
Rearranging the equation to solve for \( V_r \):
\[ V_r = \frac{2500 \text{ Hz} \times 0.4 \text{ m}}{2} \]
\[ V_r = \frac{1000 \text{ m/s}}{2} = 500 \text{ m/s} \]

### Step 7: Convert to km/s
To convert the speed from m/s to km/s:
\[ V_r = \frac{500 \text{ m/s}}{1000} = 0.5 \text{ km/s} \]

### Conclusion
The speed of the airplane is **0.5 km/s**.