1.

The freuquency 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)` IsA. `98`B. `100`C. `103`D. `105`

Answer» Correct Answer - C
Let `n` the frequency of fork `C` then
`n_A=n+(3n)/(100)=(103n)/(100)` and `n_B=n-(2n)/(100)=(98)/(100)`
but `n_A-n_B=5implies(5n)/(100)=5impliesn=100Hz`
`:. n_A=((103)(100))/(100)=103Hz`


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