question_answer

What is minimum length of a tube, open at both ends, that resonates with tuning fork of frequency 350 Hz? (velocity of sound in air = 350 m/s]

A)  50 cm        

B)  100 cm

C)  75cm        

D)  25cm

Correct Answer: A

Solution :

 The distance between two consecutive antinodes is equal to half the wavelength. If \[l\]be the length of pipe and \[\lambda \] the wavelength, then frequency (n) is \[n=\frac{v}{\lambda }=\frac{v}{2\,l}\] Given, n = 350 Hz, v = 350 m/s. \[\therefore \] \[350=\frac{350}{2\,l}\]