Question

An open tube is in resonance with string (frequency of vibration of tube is $n_0$). It 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. $\frac{1}{2}$
• B. $2$
• C. $\frac{2}{3}$
• D. $\frac{3}{2}$

Answer 1

The correct option is B $2$

For open tube $n_0=\frac{v}{2l}$
For closed length available for resonance is
$l^{\prime }=l\times \frac{25}{100}=\frac{l}{4}$
$\therefore$ Fundamental frequency of water filled tube
$n=\frac{c}{4l^{\prime }}=\frac{v}{4\times \left(l/4\right)}=\frac{v}{l}=2n_0$
$\Rightarrow$ $\frac{n}{n_0}=2$