Question

A cylindrical tube, open at both ends, has a fundamental frequency $v$. The tube is dipped vertically in water so that half of its length is inside the water. The new fundamental frequency is

• A. $v/4$
• B. $v/2$
• C. $v$
• D. $2v$

Answer 1

The correct option is A $v/4$

For open pipe in air,

Fundamental frequency, $\nu =\frac{V}{2l}$

where, $l=\text{l}engthofthetube$

For a pipe closed on one end (dipped in water),

Fundamental frequency, ${\nu }^{{}^{\prime }}=\frac{V}{4l^{{}^{\prime }}}$

$\because l^{{}^{\prime }}=\frac{l}{2}$

$\therefore {\nu }^{{}^{\prime }}=\frac{V}{4\times \frac{l}{2}}$

$\therefore {\nu }^{{}^{\prime }}=\frac{V}{2l}=\nu$

Answer: Option $\left(c\right)$