A) 10 cm
B) 8 cm
C) 12 cm
D) 5 cm
Correct Answer: D
Solution :
Third overtone of an open organ pipe\[{{n}_{1}}=\frac{4v}{2{{l}_{1}}}\] Second overtone of closed pipe\[{{n}_{2}}=\frac{5v}{4{{l}_{2}}}\] As\[{{n}_{1}}\]and\[{{n}_{2}}\]are in resonance. \[{{n}_{1}}={{n}_{2}}=n\] \[\frac{4v}{2{{l}_{1}}}=\frac{5v}{4{{l}_{2}}}\] \[\frac{4v}{2\times 8}=\frac{5v}{4{{l}_{2}}}\] \[\Rightarrow \] \[{{l}_{2}}=5\,cm\]
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