JEE Mains 2020 Model Paper 3 New Pattern

Let the fork at A be source S, and the fork at B be source S2. For the sound from Source S1
Apparent frequency = n1 = n(

υ−υ0

υ−υ0

)
Here = 345 + 5 = 350 ms−1 as wind and sound are in same directions.
υ0 = + 25 ms−1 , υs = 0
⇒ n1 = 500(

350−25

350−0

)
= 464.3 Hz
For the sound from source S2
υ = 345 - 5 = 340 ms−1 as wind is opposite to sound
υs = 0 ms−1 , υ0 = −25ms−1
n2 = n(

υ−υ0

υ−υs

) = 500(

υ−(−25)

υ−−0

)

500×365

340

= 536.80 Hz
n2−n1 = 536.80 - 464.30 = 72.5 Hz