wave_mechanics critical_thinking Question 8
Question: A closed organ pipe of length L and an open organ pipe contain gases of densities $ {\rho _{1}} $ and $ {\rho _{2}} $ respectively. The compressibility of gases are equal in both the pipes. Both the pipes are vibrating in their first overtone with same frequency. The length of the open organ pipe is
[IIT-JEE (Screening) 2004]
Options:
A) $ \frac{L}{3} $
B) $ \frac{4L}{3} $
C) $ \frac{4L}{3}\sqrt{\frac{{\rho _{1}}}{{\rho _{2}}}} $
D) $ \frac{4L}{3}\sqrt{\frac{{\rho _{2}}}{{\rho _{1}}}} $
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Answer:
Correct Answer: C
Solution:
Frequency of first over tone of closed pipe = Frequency of first over tone of open pipe $ \Rightarrow \frac{3v}{4L _{1}}=\frac{v}{L _{2}} $
$ \Rightarrow \frac{3}{4L _{1}}\sqrt{\frac{\gamma P}{{\rho _{1}}}} $
$ =\frac{1}{L _{2}}\sqrt{\frac{\gamma P}{{\rho _{2}}}} $
$ [ \because v=\sqrt{\frac{\gamma P}{\rho }} ] $
$ \Rightarrow L _{2}=\frac{4L _{1}}{3}\sqrt{\frac{{\rho _{1}}}{{\rho _{2}}}}=\frac{4L}{3}\sqrt{\frac{{\rho _{1}}}{{\rho _{2}}}} $
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