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
2km/s
B
1km/s
C
0.5km/s
D
0.25km/s

Solution

## The correct option is C $$0.5\,km/s$$$$f' = \left (\dfrac{c + \nu_a}{c - \nu_a} \right )f$$where $$c$$ is the velocity of the radio wave , an electromagnetic wave , i.e.,  $$c = 3 \times 10^8\,m/s$$ and $$\nu_s$$ is velocity of aeroplane .       $$f' - f = \left [ \dfrac{c + \nu_a}{c - \nu_a} - 1 \right ]f$$ $$\Rightarrow$$             $$\Delta f = \dfrac{2 \nu_a f}{c - \nu_a}$$ Since approaching aeroplane cannot have a speed comparable to the speed of electromagnetic wave , so $$v_s \ll c .$$ $$\therefore$$                      $$\Delta f = \dfrac{2 \nu_a f}{c}$$ $$\Rightarrow$$           $$2.6 \times 10^3 = \dfrac{2\nu_A(780 \times 10^6)}{3 \times 10^8}$$                                $$\Rightarrow$$                    $$\nu_A = 0.5 \times 10^3 \,m/s$$                                 $$= 0.5\,km/s$$ Physics

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