JEE Advanced 2020 Full Test 3 Paper 1

As source (horn of bus) is approaching stationary wall (say, listener), therefore, apparent frequency striking the wall is
v′=‌

vv

v−vs

.....(1)
Sound of this frequency will be reflected by the wall (now, source). The passenger is the listener moving towards source. Therefore, frequency heard by the listener
v′′=‌

(v+vL)v′

v

using (1)
v′′=‌

v+vL

v

×‌

vv

v−vs

=‌

(v+vL)v

v−vs

=‌

(330+5)×200

330−5

=‌

335

325

×200
v′′=206Hz
∴ Beat frequency =(v′′−v)
=206−200=6Hz