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A man is watching two trains, one leavin...

A man is watching two trains, one leaving and the other coming in with equal speed of 4 m/s. If they sound their whistles, each of frequency 240 Hz, the number of beats heard by the man (velocity of sound in air is `320(m)/(s)`) will be equal to

A

3

B

4

C

5

D

6

Text Solution

Verified by Experts

The correct Answer is:
D
The frequency of the whistle heard by the main for the incoming train is `n_(i) = ((v)/(v - v_(s)))n` and for the outgoing train, `n_(o) = ((v)/(v + v_(s)))n`
`:. n_(i) - n_(o) = nv [(1)/(v - v_(s)) - (1)/(v + v_(s))]`
`= nv [(v + v_(s) - v + v_(s))/((v - v_(s)) (v + v_(s)))]`
`= (2nv_(s))/((v - v_(s)) (v + v_(s)))`
`= (2 xx 240 xx 320 xx 4)/((320 - 4) (320 + 4))`
`= (320 xx 120)/(79 xx 81) div 6`
`:.` No of beats/sec = 6
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