A) 324 Hz
B) 320 Hz
C) 316 Hz
D) 314 Hz
Correct Answer: A
Solution :
\[{{n}_{A}}=?,\,\,{{n}_{B}}=Known\,frequency\,\,=\,\,320\,Hz\,\,\] \[\operatorname{x} =4 bps\], which remains same after filing. Unknown fork A is filed so \[{{n}_{A}}\uparrow \] Hence \[{{n}_{A}}\uparrow -\,{{n}_{B}}=x\,\,\to \,\,Wrong\] \[{{n}_{B}}-\,\,{{n}_{A}}\,\uparrow \,\,=\,\,x\,\,\to \,\,Correct\] \[\Rightarrow \,\,\,{{n}_{A}}={{n}_{B}}\,-\,\,x=320-4\,\,=\,\,316\,\,Hz\] This is the frequency before filing. But in question frequency after filing is asked which must be greater than 316 Hz, such that it produces 4 beats per sec. Hence it is 324 Hz.
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