A) 80
B) 90
C) 95
D) 85.5
Correct Answer: C
Solution :
For the closed organ pipe the fundamental frequency is given by \[{{n}_{1}}=\frac{v}{4l}\] So, \[{{n}_{1}}=\frac{v}{4{{l}_{1}}}\] ?(i) and \[{{n}_{2}}=\frac{v}{4{{l}_{2}}}\] ?(ii) From Eqs.(i) and (ii) or \[{{n}_{1}}-{{n}_{2}}=4=\frac{v}{4}\left( \frac{1}{{{l}_{1}}}-\frac{1}{{{l}_{2}}} \right)\] or \[4=\frac{300}{4}\left( \frac{1}{{{l}_{1}}}-\frac{1}{1} \right)(\because {{l}_{2}}=1\,m\,given)\] or \[\frac{300}{4{{l}_{1}}}=4+\frac{300}{4}\] or \[\frac{75}{{{l}_{1}}}=4+75=79\] or \[{{l}_{1}}=\frac{75}{79}=0.949\,m\approx 95\,cm\]
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