MCQ
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
  • A
    $3{f_0}/4$
  • ${f_0}$
  • C
    ${f_0}/2$
  • D
    $2{f_0}$

Answer

Correct option: B.
${f_0}$
b
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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