Neet Solved Paper 2015 Question 25
Question: The fundamental frequency of a closed organ pipe of length 20 cm is equal to the second overtone of an organ pipe open at both the ends. The length of organ pipe open at both the ends is
Options:
A) 80 cm
B) 100 cm
C) 120 cm
D) 140 cm
Show Answer
Answer:
Correct Answer: C
Solution:
The fundamental frequencies of closed and open organ pipe are given as
$ v _{c}=\frac{v}{4l} $ $ v _{o}=\frac{v}{2l’} $ Given the second
overtone (i.e. third harmonic) of open pipe is equal to the fundamental frequency of closed pipe i.e.
$ 3v _{o}=v _{c} $
$ \Rightarrow 3\frac{v}{2l’}=\frac{v}{4l} $
$ \Rightarrow \ 1’=6l=6\times 20=120cm $
