wave_mechanics organ_pipe_vibration_of_air_column Question 52
Question: In a resonance pipe the first and second resonances are obtained at depths 22.7 cm and 70.2 cm respectively. What will be the end correction
[J & K CET 2005]
Options:
A) 1.05 cm
B) 115.5 cm
C) 92.5 cm
D) 113.5 cm
Show Answer
Answer:
Correct Answer: A
Solution:
For end correction x, $ \frac{l _{2}+x}{l _{1}+x}=\frac{3\lambda /4}{\lambda /4}=3 $
$ x=\frac{l _{2}-3l _{1}}{2}=\frac{70.2-3\times 22.7}{2}=1.05cm $
BETA
