`A, B`and `C` are three tuning forks. Frequency of `A` is `350 H_(Z)` . Beats produced by `A` and `B` are `5//s` and by `B` and `C` are `4//s` . When a wax is put on `A` beat frequency between `A` and `B` is `2 H_(Z)` and between `a` and `C` is `6 H_(z)` . Then, frequency of `B` and `C` respectively , are

Options

A. `355 H_(Z)`, `349 h_(Z)`
B. `345 H_(Z)` , `341H_(Z)`
C. `355 H_(Z)` , `341 H_(Z)`
D. `345 H_(Z)`, `349H_(Z)`

Answer

A, B, C, D

Explanation

By putting wax on `A` Its frequency will decrease. Bit beat frequency between `A` and `B` is also decreasing.
:. `f_(A) gt f_(B)`
or `f_(A) - f_(B) = 5 H_(Z)` …(i)
:. `f_(B) = f_(A) - 5 = 345 H_(Z)`
Now, `f_(B) ~ f_(c) = 4 H_(Z)`
`f_(C)` is either `341 H_(Z)` or `349 H_(Z)` .
If it is `341 H_(Z)` , then frequency with `A` will be `9 H_(Z)` .
If it is `349 H_(Z)` , then beat frequency will be `1 H_(Z)` .
If wax is loaded on `A`, its frequency will decrease.
To produced `6 beats//s` with `C` it should become either `347 H_(Z)` if `(f_(C) = 341 H_(Z))` or it should become
`342 H_(Z)` if `(f_(c) = 349 H_(Z))`.
If it become `347 H_(Z)`, then only it produced `2 beats//s` with `B`, which is given in the question.
`:. f_(c) = 341 H_(Z)`