A hollow pipe of length \(0.8 \mathrm{~m}\) is closed at one end. At its open end, a \(0.8 \mathrm{~m}\) long uniform string is vibrating in its second harmonic and it resonates with the fundamental frequency of pipe. If the tension in the sting is \(50 \mathrm{~N}\) and speed of sound in air is \(320 \mathrm{~m} / \mathrm{s}\), the mass of the strings is

For a closed organ pipe, the fundamental frequency is given \(f=\frac{v}{4 L}\), where \(v=320 \mathrm{~m} / \mathrm{s}\) is the velocity of sound in the medium of organ pipe and \(L=0.8 \mathrm{~m}\) being the length of pipe.
Now, we are given second harmonic frequency of wire is equal and in resonance with fundamental frequency of pipe, thus;
\(f=\frac{v}{4 L}=\frac{1}{l} \sqrt{\frac{T}{\mu}}\)
Putting, \(T=50 \mathrm{~N}\) and \(l=0.8 \mathrm{~m}\)
\(\begin{aligned} & \frac{320}{4(0.8)}=\frac{1}{0.5} \sqrt{\frac{50}{\mu}} \\ & \Rightarrow \sqrt{\frac{\mu}{50}}=\frac{1}{50}\end{aligned}\)
The length of string \(=0.5 \mathrm{~m}\)
Thus, mass of string \(=0.02 \times 0.5=0.01 \mathrm{~kg}=10 \mathrm{~g}\)