IIT JEE Advanced 2011 Paper 1

The velocity of car is 36 km/h =10 m/s. The appeared frequency due to Doppler effect is
f = (

v+vr

v+vs

)f0
where v is the velocity of waves in the medium; vr is the velocity of the receiver relative to the medium, which is positive if the receiver is moving towards the source; vs is the velocity of the source relative to the medium; positive if the source is moving away from the receiver. Therefore, the frequency reflected from the wall is
f' = (

320

320−10

) × 8 kHz = 8.25 kHz
The frequency observed by the car driver is
f" =

320+10

320

f′ =

330

320

×8.25 kHz = 8.5 kHz