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Two trains A and B simultaneously start moving along parallel tracks from a station along same direction. A starts with constant acceleration 2 m/s2 from rest while B with the same acceleration but with initial velocity of 40 m/s. Twenty seconds after the start, passenger of A hears whistle of B. If frequency of whistle is 1194 Hz and velocity of sound in air is 322 m/s, then the frequency observed by the passenger is

A
1186 Hz
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B
1086 Hz
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C
986 Hz
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D
886 Hz
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Solution

The correct option is B 1086 Hz

fapp=f0(v±v0v±vs)

PR=0+12.2.202=400 m

PQ=t2

PS=40+122t2

QS=QR+RS

PSPQ=PRPQ+xsound(20t)
So,
(40t+t2)t2=400t2+322(20t)
40t=400t2+6440322t
t2+362t6840=0t=18

fapp=f0(v±v0v±vs)v020 s,vst=18 s.

v0=0+2×20=40m/svs=u+at=40+2×18=76m/s

fapp=1194(322+40322+76)
fapp=1086 Hz

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