A) 98
B) 100
C) 103
D) 105
Correct Answer: C
Solution :
Let n be the frequency of fork C then \[{{n}_{A}}=n+\frac{3n}{100}=\frac{103n}{100}\] and \[{{n}_{B}}=n-\frac{2n}{100}=\frac{98}{100}\] but \[{{n}_{A}}-{{n}_{B}}=5\]Þ \[\frac{5n}{100}=5\]Þ \[n=100\,Hz\] \ \[{{n}_{A}}=\frac{(103)(100)}{100}=103\,Hz\]You need to login to perform this action.
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