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Question

# When two tuning forks (fork 1 and fork 2) are sounded simultaneously, 4 beats per second are heard. Now, some tape is attached on the prong of the fork 2. When the tuning forks are sounded again, 6 beats per second are heard. If the frequency of fork 1 is 200Hz, then what was the original frequency of fork 2?

A
202Hz
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B
200Hz
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C
204Hz
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D
196Hz
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Solution

## The correct option is D 196HzFrequency of fork 1=200Hz=n0No. of beats heard when fork 2 is sounded with fork 1=Δn=4Now we know that if on loading (attaching tape) an unknown fork, the beat frequency increases (from 4 to 6 in this case) then the frequency of the unknown fork 2 is given by, n=n0−Δn=200−4=196Hz

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