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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 fork are sounded again, `6` beats per second are heard. If the frequency of fork `1` is `200 Hz`, then what was the original frequency of fork `2` ?
A. `200 Hz`
B. `202 Hz`
C. `196 Hz`
D. `204 Hz`

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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`.

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