Correct Answer - D
Here given veolocity of he sources `v_s=0.
velocity of the observer `v_0=3m/s`
So, the apparent frequency heard by the man
`=((332+3)/332)xx256=258.3 Hz`
From the apparoching tuning fork `=f^1`
Similarly `f^11=((332-3)/332)xx256`
`=253.7 Hz`
So, beat produced by them
`=258.3-253.7=4.6Hz`