16.The frequency of a stretched uniform wire of certain length is in resonance with the fundamental frequency of closed tubes.If length of wire is decreased by 0.5m,it is in resonance with first overtone of closed pipe.The intiallength of wire is ------

Let the initial length of the wire is l. and L is the length of the tube.
Frequency of the wire in the first case,
$f_1=\frac{1}{2l}\sqrt{\frac{T}{\mu }}$
wavelength of the sound wave,
${\lambda }_1=4L$
Frequency of the wire in the 2nd case,
$f_2=\frac{1}{2\left(l-0.5\right)}\sqrt{\frac{T}{\mu }}$
wavelength of the sound wave,
${\lambda }_1=\frac{4L}{3}$
Equating the velocity of sound in both the cases,
$$\begin{gathered} f_1{\lambda }_1=f_2{\lambda }_2 \\ \Rightarrow \frac{1}{2l}\sqrt{\frac{T}{\mu }}\times 4L=\frac{1}{2\left(l-0.5\right)}\sqrt{\frac{T}{\mu }}\times \frac{4L}{3} \\ \Rightarrow l=3\left(l-0.5\right) \\ \Rightarrow l=0.75 \end{gathered}$$
The initial length of the wire is 0.75 m.