Q. An open tube is in resonance with string (frequency of vibration of tube is $n_{0}$). If tube is dipped in water so that 75% of length of tube is inside water, then the ratio of the frequency of tube to string now will be
A) 1
B) 2
C) $\frac{2}{3}$
D) $\frac{3}{2}$
Explanation
(b) For open tube, $n_{0} = \frac{v}{2l}$
For closed tube length available for resonance is
$l^{'} = l \times \frac{25}{100} = \frac{l}{4}\ \ \therefore$ Fundamental frequency of water filled tube
$n = \frac{v}{4l^{'}} = \frac{v}{4 \times (l/4)} = \frac{v}{l} = 2n_{0} \Rightarrow \frac{n}{n_{0}} = 2$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.