A siren placed at a railway platfere is emitting a sound of frequency 5 kHZ. A passenger sitting in a moving train A records the frequency of the siren as 5.5 kHz.
During his return journey by train B he records the frequency of the siren as 6 kHz. The ratio of the speed of train B to that of train A is

Options

A. `(242)/(252)`
B. 2
C. `(5)/(6)`
D. `(11)/(6)`

Answer

B

Explanation

Given, frequency of sound, `f_(s) `= 5 kHz
`f_(A) 5.5` kHz
frequency of siren records by moving train B,
`f_(B) = 6 ` kHz
Now, let `v_(s)` = speed of sound
`therefore` For observer A, frequency of siren records by train A is given as,
` f_(A) = f_(s) ((v_(s) + v_(A))/(v_(s)))`
or `" " 5.5 = 5 ( (v_(s) + V_(A))/(v_(s)))`
or `" " 1.1 = 1 + (v_(A))/(v_(s))`

or `" " v_(A) = 0.1 v_(S) " " ` ....(i)
`therefore` for observer B, frequency of siren records by train B is given as ,
` f_(B) = f_(s) ((v_(s) + v_(B))/(v_(s)))`
` 6 = 5 ((v_(s) + v_(B))/(v_(s))) or 6 = 5 ( 1 + (v_(B))/(v_(s)))`
`rArr " " (6)/(5) = 1 + (v_(B))/(v_(s))`
`rArr " " (6)/(5) - 1 = (v_(B))/(v_(s)) rArr ( 6 - 5)/(5) = (v_(b))/(v_(s))`

`rArr " " (1)/(5) = (v_(B))/(v_(s)) rArr (v_(B))/(v_(s)) = 2.0 `
or `" " v_(B) = 0.2 v_(s) `
Now, from Eqs. (i) and (ii) we get
or `(v_(B))/(v_(A)) = (0.2 V_(s))/(0.1 v_(s)) or (v_(B))/(v_(A)) = 2 `
So, the ratio of the speed of train B to that of train A is ` V_(B_ : V_(A) = 2 .`