A) 0.5 m
B) 1.0 m
C) 1.5 m
D) 2.0 m
Correct Answer: A
Solution :
: When air is blown at the open end of a closed pipe a longitudinal wave travels in the air of the pipe from the closed end to the open end. When \[\lambda \]is wavelength and\[l\]is the length of pipe and D is the frequency of note emitted and v is the velocity of sound in air, then \[\upsilon =\frac{v}{\lambda },\](fundamental note) Given, \[\upsilon =\frac{332}{166}=2m\] But, \[\lambda =4l\] \[\therefore \] \[l=\frac{\lambda }{4}=\frac{2}{4}=0.5m\]You need to login to perform this action.
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