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A whistle of frequency 500 Hz tied to the end of a string of length 1.2 m revolves at 400 rev/min. A listener standing some distance away in the plane of rotation of whistle hears frequencies in the range (speed of sound = 340 m/s)

a
436 to 586
b
426 to 574
c
426 to 584
d
436 to 674

detailed solution

Correct option is A

The linear velocity of Whistle vS=rω=1.2×2π40060=50 m/sWhen Whistle approaches the listener, heard frequency will be maximum and when listener recedes away, heard frequency will be minimumSo, nmax=n vv−vs=500 340290=586Hznmin= n vv+vs=500 340390=436Hz

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