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

## The correct option is C 0.5 km/sf′=(c+vac−va)f where c is the velocity of the radio wave, an electro-magnetic wave, i.e., $$c = 3 \times 10^8 m/s$$ and vs is velocity of aeroplane. f′−f=[c+vac−va−1]f⇒Δf=2vafc−va Since approaching aeroplane cannot have a speed comparable to the speed of electromagnetic wave, so vs << c. ∴Δf=2vafc⇒2.6103=2vA(780×106)3×108⇒VA=0.5×103m/s=0.5km/s

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