A cylindrical tube, open at both ends, has a fundamental frequency ${f_0}$ in air. The tube is dipped vertically into water such that half of its length is inside water. The fundamental frequency of the air column now is
AIEEE 2012,AIPMT 1997, Medium
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The fundamental frequency of open tube
$v_{0}=\frac{v}{2 l_{0}}$ $...(i)$
That of closed pipe
$v_{c}=\frac{v}{4 l_{c}}$ $...(ii)$
According to the problem $l_{c}=\frac{l_{0}}{2}$
Thus $v_{c}=\frac{v}{l_{0} / 2} \Rightarrow v_{c} \frac{v}{2 l} \ldots$$ (iii)$
From equations $(i)$ and $(iii)$
$v_{0}=v_{c}$
Thus, $v_{c}=f \quad\left(\because v_{0}=f \text { is given }\right)$
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