Wave Mechanics Question 1120
Question: If the length of a closed organ pipe is 1.5 m and velocity of sound is 330 m/s, then the frequency for the second note is
[CBSE PMT 2002]
Options:
A) 220 Hz
B) 165 Hz
C) 110 Hz
D) 55 Hz
Show Answer
Answer:
Correct Answer: B
Solution:
For closed pipe second note = $ \frac{3v}{4l}=\frac{3\times 330}{4\times 1.5}=165 $
$ Hz $ .
BETA
