Q. An open tube is in resonance with string. If tube is dipped in water, so that 75% of length of tube is inside water, then ratio of the frequency $(v_{0})$ of tube to string is
A) $v_{0}$
B) $2v_{0}$
C) $\frac{2}{3}v_{0}$
D) $\frac{3}{2}v_{0}$
Explanation
(b) When open tube is dipped in water, it becomes a tube closed at one end. Fundamental frequency for open tube is
$v_{0} = \frac{v}{2l}$
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Length available for resonance of closed tube is 0.25𝒍
$\therefore\ v_{c} = \frac{v}{4(0.25l)} = \frac{v}{2l} \times 2 = 2v_{0}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.