A pipe open at both ends of length $1.5 \mathrm{~m}$ is dipped in water such that the second overtone of…
A pipe open at both ends of length $1.5 \mathrm{~m}$ is dipped in water such that the second overtone of vibrating air column is resonating with a tuning fork of frequency $330 \mathrm{~Hz}$. If speed of sound in air is $330 \mathrm{~m} / \mathrm{s}$ then the length of the pipe immersed in water is (Neglect end correction)
$0.35 \mathrm{~m}$
$0.25 \mathrm{~m}$
$0.55 \mathrm{~m}$
$0.45 \mathrm{~m}$
Solution
Let the length of the pipe immersed in water is $\ell$ Now it becomes closed pipe, then second overtone
$\begin{aligned}
& \frac{5 \mathrm{v}}{4(1.5-\ell)}=330 \\
& \ell=1.5-1.25=0.25 \mathrm{~m}
\end{aligned}$