Two tuning forks when sounded together produced 4 beats/sec. The frequency of one fork is 256. The number of beats heard increases when the fork of frequency 256 is loaded with wax. The frequency of the other fork is

Suppose two tuning forks are named A and B with frequencies nA=256 Hz (known), nB = ? (unknown), and beat frequency x = 4 bps.Frequency of unknown tuning fork may benB=256+4=260 Hz or =256−4=252 HzIt is given that on sounding waxed fork A (fork of frequency 256 Hz) and fork B, number of beats (beat frequency) increases. It means that with decrease in frequency of A, the difference in new frequency of A and the frequency of B has increased. This is possible only when the frequency of A while decreasing is moving away from the frequency of B.This is possible only if nB = 260 Hz. Alternate method : It is given  nA=256 Hz, nB=? and x = 4 bpsAlso after loading A (i.e. nA ↓), beat frequency (i.e. x) increases (↑).Apply these informations in two possibilities to known the frequency of unknown tuning fork.         nA ↓– nB = x↑        ... (i)          nB  – nA ↓= x↑       ... (ii)It is obvious that equation (i) is wrong (ii) is correct so nB = nA + x = 256 + 4 = 260 Hz.