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Two tuning forks A and B give 4 beats/s when sounded together. If the fork B is loaded with wax 6 beats/s are heard. If the frequency of fork A is 320 Hz, then the natural frequency of the tuning fork B will beA. 320B. 316C. 312D. 326 |
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Answer» Correct Answer - B `n_(B) = n_(A) +- 4` `= 320 +- 4 = 324 or 315 Hz` When `n_(B)` is loaded with wax and sounded with the fork `n_(A)` produces 6 b/s `:. n_(B) = n_(A) +-6` `= 320 +- 6 = 326 or 314 Hz` Before loading frequency of tuning fork increases by 2 `:. n_(B) = 316 Hz` |
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