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JEE Main 2005 (AIEEE) · Beats

Beats question from JEE Main 2005 (AIEEE)

Q71JEE Main 2005 · AIEEEEasy

Beats

When two tuning forks (fork 11 and fork 22) are sounded simultaneously, 44 beats per second are heated. Now, some tape is attached on the prong of the fork 22 When the tuning forks are sounded again, 66 beats per second are heard. If the frequency of fork 11 is 200 Hz200\text{ Hz}, then what was the original frequency of fork 22?

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Answer: (d)

  1. 4 beats with 200 Hz, so n2=204n_2 = 204 or 196 Hz196\text{ Hz}.
  2. Tape (loading) lowers n2n_2, and the beats rise from 4 to 6.
  3. Beats rise only if n2n_2 moves away from 200, so n2<200n_2 < 200: 196→194 Hz196 \to 194\text{ Hz} gives 6 beats.
  4. So the answer is (d) 196 Hz196\text{ Hz}.
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