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` grams
B. `10` grams
C. `20` grams
D. `40` grams

Answer

B

Explanation

Fundmental frequency of close organ pipe `= (V_(1))/(4l_(1))`
Second harmonic frequency of string `= (2V_(2))/(2l_(2))`
So, `(V_(1))/(4l_(1)) = (V_(2))/(l_(2)) = (320)/(4 xx 0.8) = (1)/(0.5) sqrt((50)/(mu))`
`2500 = (50)/(mu)`
`mu = (1)/(50) = (m)/(0.5)`
`m = 10 gm`.