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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 nAnB=x ....(i) -----> Wrong

nBnA=x .....(ii) -----> Correct

nB=nA+x=341+6=347Hz.


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