Two wires are in unison. If the tension in one of the wires is increased by 2%, 5 beats are produced per second. The initial frequency of each wire is

n∝T⇒Δnn=ΔT2TIf tension increases by 2%, then frequency must increases by 1%.If initial frequency n1 = n then final frequency n2 – n1 = 5⇒101100n−n=5⇒n=500Hz.Short trick : If you can remember then apply following formula to solve such type of problems.Initial frequency of each wire (n)=(Number of beats heard per sec)×200(per centage change in tension of the wire)Here  n=5×2002=500Hz