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
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(b) For open tube, ${n_0} = \frac{v}{{2l}}$
For closed tube length available for resonance is
$l' = l \times \frac{{25}}{{100}} = \frac{l}{4}$
Fundamental frequency of water filled tube
$ \therefore$ $\,n = \frac{v}{{4l'}} = \frac{v}{{4 \times (l/4)}} = \frac{v}{l} = 2{n_0}$
==> $\frac{n}{{{n_0}}} = 2$
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