A hollow pipe of length `0.8m` 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 N` and the speed of sound is `320 ms^(-1)`, the mass of the string is

Options

A. 5 g
B. 10 g
C. 20 g
D. 40 g

Answer

b

Explanation

`2(v_(1)/(2l_(1)))=v_(2)/(4l_(2))`
`therefore sqrt(T//mu)/l_(1)=320/(4l_(2))`
(`mu=` mass per unit length of wire)
or `sqrt(50//mu)/(0.5)= 320/(4xx0.8)`
Solving we get, `mu+0.02 kg //m^(-1)=20gm^(-1)`
`therefore` Mass of string `=20 gm^(-1)xx0.5m=10g`