A pipe closed at one end length \(0.8 \mathrm{~m}\). At its open end a \(0.5 \mathrm{mlong}\) uniform string is vibrating in its second harmonic and it resonates with the fundamental frequency of the pipe. If the tension in the wire in \(50 \mathrm{~N}\) and the speed of sound is \(320 \mathrm{~m} / \mathrm{s}\), the mass of the string used is

For fundamental harmonics of a closed end pipe, \(\frac{\lambda}{4}=L\)
Velocity of sound is in air is given by, \(v=\lambda f\)
For second harmonics of a string, \(\lambda^{\prime}=L^{\prime}\) the wavelength is the length of the string.
Speed of sound on the wave is related as, \(v_{\mathrm{S}}=\lambda^{\prime} f\).
Speed of sound is given by, \(v_{\mathrm{S}}=\sqrt{\frac{T}{\mu}}, T\) is the tension in the wire and \(\mu\) the mass per unit length of the string.
Since, the string in second harmonics is in resonance with the closed end pipe in fundamental mode,
\(\begin{aligned} & f=\frac{v}{4 L}=\frac{v_{\mathrm{S}}}{L^{\prime}}=\frac{1}{L^{\prime}} \sqrt{\frac{T}{\mu}} \\ & \Rightarrow \frac{320 \mathrm{~m} / \mathrm{s}}{4 \times 0.8 \mathrm{~m}}=\frac{1}{0.5 \mathrm{~m}} \sqrt{\frac{50 \mathrm{~N}}{\mu}} \\ & \Rightarrow \mu=\frac{1}{50} \mathrm{kgm}^{-1} \\ & \Rightarrow m=\mu L^{\prime}=\frac{0.5 \mathrm{~m}}{50 \mathrm{~kg}^{-1} \mathrm{~m}}=10 \mathrm{~g}\end{aligned}\)