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JEE Class main Answered

Solve Q70 in the photo (HC VERMA )
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Asked by Vidushi412 | 23 Mar, 2019, 10:50: PM
answered-by-expert Expert Answer
frequency of tuning fork = no = 256 Hz
 
speed of sound = v = 332 m/s;  speed of listner = vl = 3 m/s
when the listener is running towards one tuning fork  the frequency heard by him = begin mathsize 12px style n subscript 0 open square brackets 1 space plus space v subscript l over v close square brackets end style = 256[ 1 + (3/332) ] = 258.3 Hz
when the listener is running away from other tuning fork  the frequency heard by him = begin mathsize 12px style n subscript 0 open square brackets 1 space minus space v subscript l over v close square brackets end style = 256[ 1 - (3/332) ] = 253.7 Hz
Hence bets frequency = 258.3 - 253.7 = 4.6 Hz
Answered by Thiyagarajan K | 24 Mar, 2019, 10:48: AM
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