Questions
Two closed organ pipes of length 100 cm and 101 cm 16 beats in 20 sec. When each pipe is sounded in its fundamental mode calculate the velocity of sound
detailed solution
Correct option is C
Number of beats per second,n=1620=45⇒n=n1−n2=v41l1−1l2⇒45=v411−11.01=0.01v4×1.01v=16×1015=323.2 ms−1Talk to our academic expert!
Similar Questions
A tube of certain diameter and of length 48 cm is open at both ends. Its fundamental frequency of resonance is found to be 320 Hz. The velocity of sound in air is 320 m/s. One end of the tube is now closed, considering the effect of end correction, calculate the lowest frequency of resonance for the tube (in Hz).
799 666 8865
support@infinitylearn.com
6th Floor, NCC Building, Durgamma Cheruvu Road, Vittal Rao Nagar, HITEC City, Hyderabad, Telangana 500081.
free study material
JEE Mock Tests
JEE study guide
JEE Revision Notes
JEE Important Questions
JEE Sample Papers
JEE Previous Year's Papers
NEET previous year’s papers
NEET important questions
NEET sample papers
NEET revision notes
NEET study guide
NEET mock tests
