A) 25 cm
B) 18 cm
C) 10 cm
D) 111 cm
E) 15 cm
Correct Answer: E
Solution :
The fundamental frequency of a closed organ pipe is \[\frac{v}{4{{l}_{1}}}.\] The fundamental frequency of an open organ pipe is \[\frac{v}{4{{l}_{2}}}.\] Here, \[{{l}_{1}}\] = length of closed organ pipe and \[{{l}_{2}}\] = length of the open organ pipe when, \[{{l}_{2}}\] = 60 cm, then in accordance with question \[\frac{v}{4{{l}_{1}}}=\frac{v}{60\,cm}\] \[{{l}_{1}}=\frac{1}{4}\times 60\,cm=15\,cm\]You need to login to perform this action.
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