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)`

Answer

B

Explanation

`f'=((c+c_a)/(c-v_a))f`
where c is the velocity of the radio wave, an elecromagnetic wave, i.e,.`c=3xx10^8(m)/(s)` and `v_S` is velocity of aeroplane.
`f'-f=[(c+v_a)/(c-v_a)-1]f`
`impliestrianglef=(2v_0f)/(c-v_a)`
since approaching aeroplane cannot have a speed comparable to the speed of electromagnetic so `v_Sltltc`.
`trianglef=(2v_af)/(c )`
`implies2.6xx10^3=(2V_A(780xx10^6))/(3xx10^8)`
`impliesV_A=0.5xx10^3(m)/(s)`
`=0.5(km)/(s)`