The freqency of a radar is 780 M hz. The frequency of the reflected wave from an aerolane is increased by 2.6 kHz. The velocity of aeroplane is

Options

A. 0.25 km/sec
B. 0.5 km/sec
C. 1 km/sec
D. 2 km/sec

Explanation

To solve the problem, we will use the Doppler effect formula for sound waves. The formula relates the apparent frequency (f') observed by an observer moving towards a source of sound to the actual frequency (f) emitted by the source.

### Given Data:
- Actual frequency of the radar (f) = 780 MHz = \( 780 \times 10^6 \) Hz
- Increase in frequency (Δf) = 2.6 kHz = \( 2.6 \times 10^3 \) Hz
- Velocity of sound (v) = 330 m/s

### Step 1: Calculate the apparent frequency (f')
The apparent frequency (f') is given by:
\[ f' = f + \Delta f \]
Substituting the values:
\[ f' = 780 \times 10^6 \text{ Hz} + 2.6 \times 10^3 \text{ Hz} \]
\[ f' = 780000000 + 2600 \]
\[ f' = 780002600 \text{ Hz} \]

### Step 2: Use the Doppler Effect formula
The Doppler effect formula for a stationary source and a moving observer is:
\[ f' = f \left( \frac{v + v_o}{v} \right) \]
Where:
- \( v_o \) is the velocity of the observer (the airplane in this case).
- Since the source is stationary, \( v_s = 0 \).

Rearranging the formula to solve for \( v_o \):
\[ v_o = \frac{f'}{f} \cdot v - v \]

### Step 3: Substitute the values into the formula
Substituting the values we have:
\[ v_o = \frac{780002600}{780000000} \cdot 330 - 330 \]

Calculating the fraction:
\[ \frac{780002600}{780000000} \approx 1.00000333 \]

Now substituting this back into the equation:
\[ v_o = 1.00000333 \cdot 330 - 330 \]
\[ v_o \approx 330.0011 - 330 \]
\[ v_o \approx 0.0011 \cdot 330 \]
\[ v_o \approx 0.3636 \text{ m/s} \]

### Step 4: Convert to km/s
To convert the velocity from m/s to km/s:
\[ v_o \approx 0.3636 \text{ m/s} \approx 0.0003636 \text{ km/s} \]

### Final Result
The velocity of the airplane is approximately:
\[ v_o \approx 0.3636 \text{ m/s} \]