1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

# For a closed organ pipe three successive resonance frequencies are observed at 425 Hz, 595 Hz and 765 Hz respectively. If the speed of sound in air is 340 m/s, then the length of the pipe is :

A
2.0 m
No worries! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today!
B
0.4 m
No worries! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today!
C
1.0 m
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
0.2 m
No worries! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today!
Open in App
Solution

## The correct option is B 1.0 mGiven: v = 340m/s, Ratio of 3 frequencies = 425 : 595 : 765Solution:Since frequencies are in odd number ratio , the pipe has to be a closed pipe.Ratio of 3 frequencies = 425 : 595 : 765 = 5 : 7 : 9 So fundamental frequency =f=4255=85Hz For fundamental frequency l=v4f=3404×85=1mHence C is the correct option

Suggest Corrections
0
Join BYJU'S Learning Program
Related Videos
Beats
PHYSICS
Watch in App
Explore more
Join BYJU'S Learning Program