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Q.

Two tuning forks with natural frequencies 510 Hz each move relative to a stationary observer. One fork moves away from observer  while the other moves towards him at the same speed. The observer hears beats of frequency 6Hz. Then the speed of tuning fork is (velocity of sound in air = 340 m/s)

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a

1.5 m/s

b

2m/s

c

3 m/s

d

1 m/s

answer is B.

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Detailed Solution

When source is approaching a stationary observer,  apparent frequency n1= vv-vsn and when it is going away n2=vv+vsngiven n1-n2=6=2v vsnv2-vs2=2 vsnv    as v2>>vs2  vs=6×v2n=6×3402×510=2m/s
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