A tuning fork A produces 4 beats/ s with tuning fork, B of fequency 256 Hz. When the fork A is filled beats are found to occurs at shorter intervals, then the original frequency will be

Options

A. 260 Hz
B. 252 Hz
C. 256 Hz
D. 258 Hz

Answer

C

Explanation

As tuning fork A produces 4 beats with tuning fork B of frequency `upsilon_(B)(=256Hz)`, the frequency of `A(upsilon_(A))` will be `upsilon_(A)=upsilon_(B)pm4=256pm4,i.e.,upsilon_(A)=252Hzor260Hz`
Now on filing due to decrease in inertia frequency of A will increase and occurrence of beats at shorter duration means increase in beat frequency, so if
`upsilon_(A)=252Hz`, then
`256-upsilon_(A)=4Hz`
and so with increase in `upsilon_(A)` beat frequency will decrease.
If `upsilon_(A)=260Hz`, then
`upsilon_(A)-256=4Hz`
and so with increase in `upsilon_(A)` beat frequency will increase and as on filing beat frequency increses the frequency of A before filing was 260 Hz.