A hollow pipe of length $0.8 m$ is closed at one end. At its open end a $0.5 m$ long uniform string is vibrating in its second harmonic and it resonates with the fundamental frequency of the pipe. If the tension in the wire is $50 \mathrm{N}$ and the speed of sound is $320 m{s}^{-1},$ the mass of the string is 

Let l be the length of the string and L the length of the pipe. The frequency of the string vibrating in the second harmonic is 

$\mathrm{v}=\frac{2}{2l}\sqrt{\frac{T}{\mu }};\mu =$ mass per unit length of the string 

The fundamental frequency of the closed pipe is 

$v^{\mathrm{\prime }}=\frac{v}{4L};v=$speed of sound

For resonance, $v=v\text{'}$, i.e.,

$\frac{1}{l}\sqrt{\frac{T}{\mu }}=\frac{\mathrm{v}}{4L}$

Substituting the values of $l,T,\mathrm{V}\text{ and }L\text{, }$ we get $\mu =\frac{1}{50}{\mathrm{kgm}}^{-1}$

Mass of string $=\mu l=\frac{1}{50}\times 0.5={10}^{-2}\mathrm{kg}=10\text{ grams. }$ So the correct choice is (2).