A train approaches stationary observer, the velocity of train being `1/20` of the velocity of sound. A sharp blast is blown with the whistle of the engine at equal intervals of 1s. The interval between the successive blasts as heard by the observer

Options

A. `(1)/(20)`sec
B. `(1)/(20)`min
C. `(19)/(20)`sec
D. `(10)/(20)`min

Answer

C

Explanation

`f=(1)/(T)=(1)/(1)=1Hz`
Frequency heard by the observer
`f.=f((v)/(v-v_(s)))=1((v)/((19v)/(20)))=(20)/(19)Hz`
`t=(1)/(f.)=(19)/(20)sec`.