A) 4
B) 7
C) 2
D) 8
Correct Answer: B
Solution :
for closed pipe \[{{\operatorname{f}}_{c}}=\frac{1}{2}m{{v}^{2}}\] fundamental frequency for open pipe \[{{\operatorname{f}}_{\operatorname{o}}}=\frac{\operatorname{v}}{2{{\operatorname{l}}_{o}}}\] fundamental frequency \[{{\operatorname{f}}_{o}}-{{\operatorname{f}}_{c}}=2\] \[\frac{v}{2{{l}_{o}}}-\frac{c}{4{{l}_{o}}}=2\] \[\frac{v}{4{{l}_{o}}}=2\] ?.(1) Case-II \[{{\operatorname{l}}_{c}}=2{{l}_{o}}while\,\,lopen\,\,=\frac{{{l}_{o}}}{2}\] \[{{\operatorname{f}}_{c}}=\frac{v}{4\times 2{{l}_{o}}}=\frac{v}{8{{l}_{o}}}\Rightarrow {{f}_{o}}\frac{v}{2\times \frac{{{l}_{o}}}{2}}=\frac{v}{{{l}_{o}}}\] \[{{\operatorname{f}}_{o}}-{{f}_{c}}=x\] \[\frac{v}{{{l}_{o}}}-\frac{v}{8{{l}_{o}}}=x\] \[\frac{7v}{8{{l}_{o}}}=x\] ?.(2) X=7 by (1) and (2)
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