A hollow pipe of length \(0.8 \mathrm{~m}\) is closed at one end. At its open end a \(0.5 \mathrm{~m}\) long uniform string is vibrating in its second harmonic and it resonates with the fundamentals frequency of the pipe if the tension in the wire is \(50 \mathrm{~N}\) and the speed of sound is $320 \mathrm{~ms}^{-1}$, what is the mass of the string. (A) \(20 \mathrm{~g}\) (B) \(5 \mathrm{~g}\) (C) \(40 \mathrm{~g}\) (D) \(10 \mathrm{~g}\)

First, let's find the fundamental frequency of the pipe. A closed pipe has a fundamental frequency that depends on its length and the speed of sound. The formula for the fundamental frequency of a closed pipe is: \(f_p = \frac{v}{4L}\) Where \(f_p\) is the fundamental frequency of the closed pipe, \(v\) is the speed of sound, and \(L\) is the length of the pipe. In our case, the length of the pipe is \(0.8\mathrm{~m}\) and the speed of sound is \(320\mathrm{~ms}^{-1}\). Thus, the fundamental frequency of the pipe is: \(f_p = \frac{320\mathrm{~ms}^{-1}}{4(0.8\mathrm{~m})} = 100\mathrm{~Hz}\)

Now, we can find the frequency of the second harmonic of the string. Since the string is vibrating in its second harmonic, its frequency is twice its fundamental frequency. Given that the pipe and the string are in resonance, the frequency of the second harmonic of the string must be equal to the fundamental frequency of the pipe. Therefore, the frequency of the second harmonic of the string is: \(f_s = 100\mathrm{~Hz}\)

Next, we can calculate the speed of the wave on the string. We know that the frequency and wavelength of a wave are related by the formula: \(v = f\lambda\) Where \(v\) is the speed of the wave, \(f\) is the frequency, and \(\lambda\) is the wavelength. Since the string is vibrating in its second harmonic, the wavelength of the wave is equal to twice the length of the string: \(\lambda = 2L\) Thus, the speed of the wave on the string is: \(v = f_s\lambda = 100\mathrm{~Hz} \cdot 2(0.5\mathrm{~m}) = 100\mathrm{~ms}^{-1}\)

Now, we can use the formula for the speed of a wave on a string to find the mass of the string. The formula for the speed of a wave on a string is: \(v = \sqrt{\frac{T}{\mu}}\) Where \(v\) is the speed of the wave on the string, \(T\) is the tension in the string, and \(\mu\) is the linear mass density of the string, which is equal to the mass of the string divided by its length: \(\mu = \frac{m}{L}\) We're looking for the mass of the string, so we need to rearrange the formula for \(m\): \(m = \mu L = \frac{T L}{v^2}\) Given the tension in the string (\(50\mathrm{~N}\)), the length of the string (\(0.5\mathrm{~m}\)), and the speed of the wave on the string (\(100\mathrm{~ms}^{-1}\)), we can find the mass of the string: \(m = \frac{50\mathrm{~N}(0.5\mathrm{~m})}{(100\mathrm{~ms}^{-1})^2} = \frac{25}{10000}\mathrm{~kg} = 0.0025\mathrm{~kg}\) Finally, we can convert the mass of the string into grams: \(m = 0.0025\mathrm{~kg} \cdot 1000 = 2.5\mathrm{~g}\) This value is closest to (B) \(5\mathrm{~g}\) from the given options, and as such, we can conclude that the mass of the string is approximately 5 grams.