A hollow pipe of length 0.8 m is closed at one end. At its open end...

The fundamental frequency of a closed organ pipe of length $L_p$ is given by $f_p = \dfrac{v}{4 L_p}$.

Substituting the given values, $v = 320$ m/s and $L_p = 0.8$ m:

$f_p = \dfrac{320}{4 \times 0.8} = \dfrac{320}{3.2} = 100$ Hz

The frequency of the fourth harmonic of a string of length $L_s$ fixed at both ends is given by $f_s = \dfrac{4}{2 L_s} \sqrt{\dfrac{T}{\mu}}$, where $T$ is the tension and $\mu$ is the linear mass density.

Given that the string resonates with the fundamental frequency of the pipe, $f_s = f_p = 100$ Hz.

Substituting $L_s = 0.5$ m and $T = 50$ N:

$100 = \dfrac{4}{2 \times 0.5} \sqrt{\dfrac{50}{\mu}}$

$100 = 4 \sqrt{\dfrac{50}{\mu}}$

$25 = \sqrt{\dfrac{50}{\mu}}$

Squaring both sides, we get:

$625 = \dfrac{50}{\mu} \Rightarrow \mu = \dfrac{50}{625} = 0.08$ kg/m

The mass of the string $m$ is the product of its linear mass density and length:

$m = \mu L_s = 0.08 \times 0.5 = 0.04$ kg

Converting to grams, $m = 40$ g.

Answer: $40$ g