Correct Answer - D
As the source is moving away from the listetner hence frequency observed by listerner is
`f_1=(v)/(v+v_S)f=(340)/(340+2)xx512`
`=(340)/(342)xx512=509Hz`
The frequency reflected from wall (we can assume an observer at rest) is
`f_2=(v)/(v-v_S)xxf`
`=(340)/(338)xx512=515Hz`
Therefore beats heard by observer `(L)` is `515-509=6`.