A radar sends a radio signal of frequency `9 xx 10^3Hz` towards an aircraft approaching the radar. If the reflected wave shows a frequency shift of `3 xx 10^3Hz` the speed with which the aircraft is approaching the radar in ms-1 (velocity of the rador signal is `3 xx 10^8 ms^(-1)` )

Options

A. 150
B. 100
C. 50
D. 25

Answer

To solve the problem of finding the speed of the aircraft approaching the radar, we will use the concept of the Doppler effect. Here’s a step-by-step solution:

### Step 1: Understand the Doppler Effect
The Doppler effect describes the change in frequency of a wave in relation to an observer moving relative to the wave source. In this case, the radar is the source, and the aircraft is the observer moving towards the source.

### Step 2: Identify Given Values
- Frequency of the radar signal (fs) = \(9 \times 10^3 \, \text{Hz}\)
- Frequency shift (Δf) = \(3 \times 10^3 \, \text{Hz}\)
- Speed of the radar signal (v) = \(3 \times 10^8 \, \text{m/s}\)

### Step 3: Calculate the Observed Frequency
The observed frequency (f0) when the aircraft approaches the radar can be calculated as:
\[ f_0 = f_s + \Delta f \]
Substituting the values:
\[ f_0 = 9 \times 10^3 + 3 \times 10^3 = 12 \times 10^3 \, \text{Hz} \]

### Step 4: Apply the Doppler Effect Formula
For a stationary source and a moving observer, the Doppler effect formula is:
\[ f_0 = f_s \left( \frac{v + v_0}{v} \right) \]
Where:
- \(v_0\) = speed of the observer (aircraft)
- \(v\) = speed of the wave (radar signal)

### Step 5: Substitute Known Values into the Formula
Substituting the known values into the formula:
\[ 12 \times 10^3 = 9 \times 10^3 \left( \frac{3 \times 10^8 + v_0}{3 \times 10^8} \right) \]

### Step 6: Simplify the Equation
To simplify, we can multiply both sides by \(3 \times 10^8\):
\[ 12 \times 10^3 \times 3 \times 10^8 = 9 \times 10^3 (3 \times 10^8 + v_0) \]
This simplifies to:
\[ 36 \times 10^{11} = 27 \times 10^{11} + 9 \times 10^3 v_0 \]

### Step 7: Solve for \(v_0\)
Rearranging the equation gives:
\[ 36 \times 10^{11} - 27 \times 10^{11} = 9 \times 10^3 v_0 \]
\[ 9 \times 10^{11} = 9 \times 10^3 v_0 \]
Dividing both sides by \(9 \times 10^3\):
\[ v_0 = \frac{9 \times 10^{11}}{9 \times 10^3} = 10^8 \, \text{m/s} \]

### Step 8: Conclusion
The speed of the aircraft approaching the radar is:
\[ v_0 = 1 \times 10^8 \, \text{m/s} \]