Q. A cylindrical tube open at both ends, has a fundamental frequency f in air. The tube is dipped vertically in water so that half of it is in water. The fundamental frequency of air column is now
A) f/2
B) f
C) 3f/4
D) 2f
Explanation
(b) $f_{open} = \frac{v}{2\iota_{open}}$
$f_{closed} = \frac{v}{4\iota_{closed}} = \frac{v}{\frac{4\iota_{open}}{2}}$
$\left( As\ \iota_{closed} = \frac{\iota_{open}}{2} \right)$
$= \frac{v}{2f_{open}} = f_{open}\ i.e.,$
Frequency remains unchanged.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.