A) \[\frac{v}{4l}\]
B) \[\frac{v}{2l}\]
C) \[\frac{v}{4{{l}^{2}}}\Delta l\]
D) \[\frac{v}{2{{l}^{2}}}\Delta l\]
Correct Answer: C
Solution :
Fundamental frequency of closed organ pipe \[n=\frac{v}{4l}\] Here, \[{{n}_{1}}=\frac{v}{4l}\] and \[{{n}_{2}}=\frac{v}{4(l+\Delta l)}\] Number of beats per second\[={{n}_{1}}-{{n}_{2}}\] \[=\frac{v}{4l}-\frac{v}{4(l+\Delta l)}\] \[=\frac{v}{4}\left[ \frac{l+\Delta l-l}{l(l+\Delta l)} \right]\] \[=\frac{v}{4}\frac{\Delta l}{{{l}^{2}}+l\Delta l}\] Neglecting the term\[l\Delta l\]. Number of beats per second \[=\frac{v}{4{{l}^{2}}}\Delta l\]
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