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A pipe of length 85 cm is closed from one end. Find the number of possible natural oscillations of aircolumn in the pipe whose frequencies lie below 1250 Hz. The velocity of sound in air is 340 m/s.   JEE Main  Solved  Paper-2014

A) 6             

B) 4

C) 12                                          

D) 8

Correct Answer: A

Solution :

\[L=n\frac{\lambda }{2}+\frac{\lambda }{4}=(2n+1)\frac{\lambda }{4}=\frac{(2n+1)v}{4f}\] \[\Rightarrow \]\[0.85=\frac{{{(2n+1)}^{v}}}{4f}\]\[\Rightarrow \]\[f=\frac{(2n+1)\times 85}{0.85\times 4}\] \[\Rightarrow \]\[f=(2n+1)100{{H}_{2}}\]\[\Rightarrow \]\[\Rightarrow \] The possible frequencies are: \[100Hz,300Hz,...1100Hz.\]