Q. In a resonance pipe the first and second resonance are obtained at depths 22.7 cm and 70.2 cm respectively. What will be the end correction?
A) 1.05 cm
B) 115.5 cm
C) 92.5 cm
D) 113.5 cm
Explanation
(a) For the end correction x,
$\frac{l_{2} + x}{l_{1} + x} = \frac{3\lambda/4}{\lambda/4} = 3$
$\Longrightarrow \ \ \ \ \ \ \ \ \ x = \frac{l_{2} - 3l_{1}}{2}$
$= \frac{70.2 - 3 \times 22.7}{2} = 1.05cm$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.