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Question

# When a tuning fork of frequency 341 is sounded with another tuning fork, six beats per second are heard. When the second tuning fork is loaded with wax and sounded with the first tuning fork, the number of beats is two per second. The natural frequency of the second tuning fork is [MP PET 1989]

A

334

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B

339

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C

343

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D

347

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Solution

## The correct option is D 347 nA = Known frequency = 341 Hz, nB = ? x = 6 bps , which is decreasing (i.e.x↓) after loading ( from 6 to 1 bps ) Unknown tuning fork is loaded so nB↓ Hence nA−nB↓=x↓ ....(i) -----> Wrong nB↓−nA=x↓ .....(ii) -----> Correct ⇒nB=nA+x=341+6=347Hz.

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