JEE Main & Advanced
Sample Paper
JEE Main - Mock Test - 7
question_answer
A cylindrical tube open at both the ends has a fundamental frequency of 390 Hz in air. If 1/4th of the tube is immersed vertically in water the fundamental frequency of air column is
A)260 Hz
B)130 Hz
C)390 Hz
D)520 Hz.
Correct Answer:
A
Solution :
[a]: A cylindrical tube open at both the ends, its fundamental frequency is where v is the velocity of sound in air. \[\upsilon =\frac{v}{\lambda }=\frac{v}{2L}=390Hz\] (i)
If a (1/4)th of cylindrical tube is immersed in a water, it will become a closed pipe of length three-fourth that of an open pipe as shown in figure (ii). Therefore, its fundamental frequency is \[\upsilon '=\frac{v}{\lambda '}=\frac{v}{3L}\] \[\upsilon '=\frac{v}{3L}=\frac{2}{3}\left( \frac{v}{2L} \right)=\frac{2}{3}\times 390Hz\](using (i)) = 260 Hz.