Step-by-step Solution:

1. Find the fundamental frequency of the closed pipe. A hollow pipe closed at one end forms a standing wave with a node at the closed end and an antinode at the open end. For the fundamental frequency (first harmonic), the length of the pipe $L_p$ is equal to one-quarter of the wavelength $λ_p$. $L_p = \frac{\lambda_p}{4}$ The frequency $f_p$ is related to the speed of sound $v_{sound}$ and wavelength $λ_p$ by $v_{sound} = f_p \lambda_p$. Substituting $λ_p = 4 L_p$, we get the fundamental frequency of the closed pipe: $f_p = \frac{v_{sound}}{4 L_p}$ Given: $L_p = 0.8$ m and $v_{sound} = 320$ m/s. $f_p = \frac{320}{4 \times 0.8} = \frac{320}{3.2} = 100 \text{ Hz}$

2. Determine the frequency of the string. The problem states that the string resonates with the fundamental frequency of the pipe. This means the frequency of the string's vibration $f_s$ is equal to the fundamental frequency of the pipe $f_p$. $f_s = f_p = 100 \text{ Hz}$

3. Relate the string's frequency to its properties. The string is vibrating in its second harmonic (n=2). The frequency of the n-th harmonic of a string fixed at both ends is given by: $f_s = \frac{n}{2L_s} v_s$ where $L_s$ is the length of the string and $v_s$ is the speed of the wave on the string. For the second harmonic (n=2): $f_s = \frac{2}{2L_s} v_s = \frac{v_s}{L_s}$ Given: $f_s = 100$ Hz and $L_s = 0.5$ m. We can find the speed of the wave on the string, $v_s$: $100 = \frac{v_s}{0.5}$ $v_s = 100 \times 0.5 = 50 \text{ m/s}$

4. Calculate the linear mass density of the string. The speed of a wave on a string is also given by the formula: $v_s = \sqrt{\frac{T}{\mu}}$ where T is the tension in the string and μ is the linear mass density (mass per unit length). Squaring both sides: $v_s^2 = \frac{T}{\mu}$ Rearranging to solve for μ: $\mu = \frac{T}{v_s^2}$ Given: T = 50 N and we calculated $v_s = 50$ m/s. $\mu = \frac{50}{50^2} = \frac{50}{2500} = \frac{1}{50} \text{ kg/m}$

5. Calculate the total mass of the string. The linear mass density μ is defined as the total mass of the string $m_s$ divided by its length $L_s$. $\mu = \frac{m_s}{L_s}$ So, the mass of the string is: $m_s = \mu \times L_s$ Substituting the values we found: $m_s = \frac{1}{50} \text{ kg/m} \times 0.5 \text{ m} = \frac{0.5}{50} = \frac{1}{100} = 0.01 \text{ kg}$

6. Convert the mass to grams. Since the options are in grams, we convert the mass from kilograms to grams. $m_s = 0.01 \text{ kg} \times \frac{1000 \text{ g}}{1 \text{ kg}} = 10 \text{ g}$

Comparing our result with the options: (A) 5 grams (B) 10 grams (C) 20 grams (D) 40 grams

The calculated mass of the string is 10 grams, which corresponds to option B.