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JEE Main 2014 Paper

Section: Physics

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Question : 21 of 90

Marks: +1, -0

A pipe of length 85 cm is closed from one end.Find the number of possible natural oscillation of air column in the pipe whose frequencies lie below 1250 Hz.The velocity of sound in air is 340 m/s

6
4
12
8

Solution:

In fundamental mode

\frac{\lambda}{4}=0.85

\lambda=4\times 0.85

\lambda=4\times 0.85

f=\frac{v}{\lambda}=\frac{340}{4\times 0.85}

=100

Hz

∴Possible frequencies =100 Hz,300 Hz,500 hz,700Hz,900 Hz 1100 Hz below 1250 Hz

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