**Solution**

Given,

Speed of sound (v) = 345 m/s

Fundamental frequency of closed pipe (f_{c}) = 220 Hz

Length of an open pipe (l_{0}) = ?

Here, given that, the wavelength of the second overtone of closed pipe (λ_{3}) = the third harmonics of an open pipe (λ'_{3})

Or, \(\frac v{\lambda_3}=\frac v{\lambda'_3}\)

Or, f3 for closed pipe = f3 for open pipe. \(\left[\because f=\frac v\lambda\right]\)

Or, \(5f_1=\frac v{\lambda'_3}\)

Or, \(\lambda'_3=\frac v{5f_1}=\frac{345}{5\times220}=0.3136\) m

Now, for open organ pipe,,

\(f_3=\frac v{\lambda'_3}=\frac{3v}{2l_0}\) [∵ λ'_{3} = 2l_{0}]

Or, \(l_0=\frac{3\lambda'_3}2=\frac{3\times0.3136}2=0.47\) m

∴ The length of an open pipe (l_{0}) = 0.47 m