Correct Answer - C
The frequency of fork `2` is `= 200 +- 4 = 196` or `204 Hz`
Since, on attaching the tape the prong of fork `2`, its frequency decrease, but now the number of beats per second, is `6` i.e., the frequency difference now increases. It is possible only when before attaching the tape, the frequency of fork `2` is less thanthe frequency of tunin fork `1`. Hence, the frequency of fork `2 = 196 Hz`.