A pipe closed at one end has length 83 cm. The number of possible natural oscillations of air column whose frequencies lie below 100Hz are (velocity of sound in air = 332 m/s)

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A pipe closed at one end has length 83 cm. The number of possible natural oscillations of air column whose frequencies lie below 100Hz are (velocity of sound in air = 332 m/s)

$0$
$4$
$5$
$6$

Solution: The correct answer is A. 0

The frequency for a pipe closed at one end is given by:

$f = (\frac{n \times v )/}{4 L}$

v =

speed of sound =

332 m / s

L =

length of pipe =

83 c m = 0.83 m

n = 0, 1, 2\dots

$f = (\frac{1 \times 332)/(}{4 \times 0.83)}$

⟹ f = 100 H z

Hence, no natural oscillation of frequency less than

100 H z

.

i

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