Question

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

• A. 2 km/s
• B. 1 km/s
• C. 0.5 km/s
• D. 0.25 km/s

Answer 1

The correct option is C 0.5 km/s

$f^{\prime }=\left(\frac{c+v_a}{c-v_a}\right)f$
where c is the velocity of the radio wave, an electro-magnetic wave, i.e., \(c = 3 \times 10^8 m/s\) and $v_s$ is velocity of aeroplane.
$f^{\prime }-f=[\frac{c+v_a}{c-v_a}-1]f\Rightarrow \Delta f=\frac{2v_{a}f}{c-v_a}$
Since approaching aeroplane cannot have a speed comparable to the speed of electromagnetic wave,
so $v_s$ << c.
$\therefore \Delta f=\frac{2v_{a}f}{c}\Rightarrow 2.6{10}^3=\frac{2v_A(780\times {10}^6)}{3\times {10}^8}\Rightarrow V_A=0.5\times {10}^{3}m/s=0.5km/s$