BITSAT 2019 Slot 2 Solved Paper

Given, frequency of sound produced by sources A and B,fA=fB=400Hz
Speed of sound in air, v=340m∕s
Apparent frequency heard by the observer when source A is moving towards him,
f′A=

v

v−vS

×fA=

340

340−vS

×400
Apparent frequency heard by the observer when source B is moving away from him,
f′B=

v

v+vS

×fB=

340

340+vS

×400
Given, Beats =f′A−f′B=4

340

340−vS

×400−

340

340+vS

×400=4
⇒

34000×2vS

3402−vS2

=1

34000×2vS

3402

=1‌‌[∵vS2<<3402]
⇒ vS=1.7m∕s