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

$\begin{array}{1 1}(A)\;6 \\ (B)\;4 \\(C)\;12 \\(D)\;8 \end{array} $

1 Answer

Velocity of sound = 340 m/s
$\begin{align*}f = \frac{(2n+1) v}{4 l} ;\;\;\; n = 0.1\end{align*}$
$\begin{align*}2n + 1 &= \frac{4f \times l}{v} = \frac{4 \times 1250 \times 0.85}{340 \times 100} \\ & = 12.5\end{align*}$
$\implies 2n = 11.5$
$\therefore n = \frac{11.5}{2} = 5.7$
Maximum value of n = 5
$\therefore $ total no. of oscillation = 6
The velocity of sound in air is $340\; m/s$. is 6
Hence A is the correct answer.
answered Apr 7, 2014 by meena.p
edited Nov 27 by priyanka.c
 

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