A) \[90\,\,Hz\]and\[84\,\,Hz\]
B) \[100\,\,Hz\]and\[106\,\,Hz\]
C) \[96\,\,Hz\] and\[90\,\,Hz\]
D) \[206\,\,Hz\] and\[200\,\,Hz\]
Correct Answer: C
Solution :
We know that\[f=\frac{v}{2l}\] (Fundamental frequency of an open organ pipe). Number of beats heard\[=6\] \[\Rightarrow \] \[{{f}_{P}}-{{f}_{Q}}=6\] or \[\frac{v}{2(0.3)}-\frac{v}{2(0.32)}=6\] or \[v=\frac{2\times 0.3\times 0.32\times 6}{0.32-0.3}\] Now, \[{{f}_{p}}=\frac{v}{2(0.3)}=\frac{0.32\times 6}{0.02}=96\,\,Hz\] and \[{{f}_{Q}}=\frac{v}{2(0.32)}=\frac{0.3\times 6}{0.02}=90\,\,Hz\]You need to login to perform this action.
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