Wave Mechanics Question 1126
Question: An open pipe resonates with a tuning fork of frequency 500 Hz. it is observed that two successive nodes are formed at distances 16 and 46 cm from the open end. The speed of sound in air in the pipe is
[Orissa JEE 2003]
Options:
A) 230 m/s
B) 300 m/s
C) 320 m/s
D) 360 m/s
Show Answer
Answer:
Correct Answer: B
Solution:
Distance between two consecutive nodes $ =\frac{\lambda }{2}=46-16=30 $
$ \Rightarrow \lambda =60 $ cm = 0.6m $ \
Therefore $ $ v=n\lambda =500\times 0.6=300 $
$ m/s $ .
BETA
