A) 100 cm
B) 120 cm
C) 140 cm
D) 80 cm
Correct Answer: B
Solution :
[b] Fundamental frequency of closed organ pipe \[{{V}_{c}}=\frac{V}{4{{l}_{c}}}\] Fundamental frequency of open organ pipe \[{{V}_{0}}=\frac{V}{2{{l}_{0}}}\] Second overtone frequency of open organ pipe \[=\frac{3V}{2{{l}_{0}}}\] From question, \[\frac{V}{4{{l}_{c}}}=\frac{3V}{2{{l}_{0}}}\] \[\Rightarrow \,\,{{l}_{0}}=6{{l}_{c}}=6\times 20=120cm\]You need to login to perform this action.
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