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Q.

The frequency of tuning forks A and B are respectively 3% more and 2% less than the frequency of tuning fork C. When A and B are simultaneously excited, 5 beats per second are produced. Then the frequency of the tuning fork 'A' (in Hz) is

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a

98

b

100

c

103

d

105

answer is C.

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Detailed Solution

Let n be the frequency of fork C thennA=n+3n100=103n100  and  nB=n−2n100=98100but  nA−nB=5⇒5n100=5⇒n=100 Hz∴nA=(103)(100)100=103 Hz
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