When two tuning forks (fork $1$ and fork $2$) are sounded simultaneously, $4$ beats per second are heard. Now, some tape is attached on the prong of the fork $2$. When the tuning forks are sounded again, $6$ beats per second are heard. If the frequency of fork $1$ is $200\text{ Hz}$, then what was the original frequency of fork $2$?

Original beat frequency $=|f_1-f_2|=4\text{ Hz}$.Since $f_1=200\text{ Hz}$, $f_2$ could be $204\text{ Hz}$ or $196\text{ Hz}$.Attaching tape increases the mass of fork $2$, which decreases its frequency $f_2$.New beat frequency $=6\text{ Hz}$.Case 1: If $f_2=204\text{ Hz}$, and it decreases, the difference $|200-f_2|$ would decrease (towards $0$ at first), then eventually increase.Case 2: If $f_2=196\text{ Hz}$, and it decreases (say to $194\text{ Hz}$), the difference $|200-f_2|$ increases to $6\text{ Hz}$.This matches the observation. Thus, the original frequency was $196\text{ Hz}$.