The frequency of a radar is 780 MHz. After getting reflected from an approaching aeroplane, the apparent frequency is more than the actual frequency by 2.6 kHz. The aeroplane has a speed of

Options

A. `2(km)/(s)`
B. `1(km)/(s)`
C. `0.5(km)/(s)`
D. `0.25(km)/(s)`

Explanation

To solve the problem, we will use the Doppler effect formula for sound waves. The scenario involves a radar emitting waves that reflect off an approaching airplane. We need to find the speed of the airplane based on the given frequency and the change in apparent frequency.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Frequency of the radar (f) = 780 MHz = \(780 \times 10^6\) Hz
- Change in apparent frequency (\(f' - f\)) = 2.6 kHz = \(2.6 \times 10^3\) Hz
- Speed of electromagnetic waves (c) = \(3 \times 10^8\) m/s

2. **Doppler Effect Formula:**
The formula for the apparent frequency when the source is stationary and the observer is moving towards the source is:
\[ f' = f \left( \frac{c + v_a}{c} \right) \]
where \(v_a\) is the speed of the observer (the airplane in this case).

3. **Calculate the Apparent Frequency After Reflection:**
After the wave reflects off the airplane, the airplane acts as a source moving towards the radar. The frequency detected after reflection (\(f''\)) can be calculated using:
\[ f'' = f' \left( \frac{c}{c - v_a} \right) \]

4. **Express the Change in Frequency:**
The change in frequency is given by:
\[ f'' - f = 2.6 \times 10^3 \text{ Hz} \]
Substituting the expressions for \(f''\) and \(f'\):
\[ f'' - f = f \left( \frac{c + v_a}{c} \right) \left( \frac{c}{c - v_a} \right) - f \]

5. **Simplify the Equation:**
Rearranging the equation gives:
\[ f'' - f = f \left( \frac{(c + v_a)c - (c)(c - v_a)}{c(c - v_a)} \right) \]
Simplifying further, we find:
\[ f'' - f = f \left( \frac{2 v_a}{c - v_a} \right) \]

6. **Substituting Known Values:**
Set the change in frequency equal to the given value:
\[ 2.6 \times 10^3 = 780 \times 10^6 \left( \frac{2 v_a}{c - v_a} \right) \]

7. **Solving for \(v_a\):**
Rearranging gives:
\[ v_a = \frac{(2.6 \times 10^3)(c - v_a)}{2(780 \times 10^6)} \]
Since \(c\) is much larger than \(v_a\), we can approximate \(c - v_a \approx c\):
\[ v_a \approx \frac{(2.6 \times 10^3)(c)}{2(780 \times 10^6)} \]

8. **Calculating the Speed of the Airplane:**
Substituting \(c = 3 \times 10^8\):
\[ v_a \approx \frac{(2.6 \times 10^3)(3 \times 10^8)}{2(780 \times 10^6)} \]
\[ v_a \approx \frac{7.8 \times 10^{11}}{1560 \times 10^6} = 500 \text{ m/s} \]

### Final Answer:
The speed of the airplane is approximately **500 m/s**.