The fundamental frequency of a closed organ pipe of length 20 cm is equal to the second overtone of an organ pipe open at both the ends. The length of organ pipe open at both the ends is :

120 cm

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140 cm

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80 cm

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100 cm

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Solution

The correct option is B 120 cm
$\frac{V}{4 L_{c l o s e d}} = \frac{3 V}{2 L_{o p e n}}$
$\Rightarrow \frac{1}{4 \times 20} = \frac{3}{2 L_{c l o s e d}}$
$\Rightarrow L_{c l o s e d} = 120 c m$

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The fundamental frequency of a closed organ pipe of length 20 cm is equal to the second overtone of an organ pipe open at both the ends. The length of organ pipe open at both the ends is :

The correct option is B 120 cmV4Lclosed=3V2Lopen⇒14×20=32Lclosed⇒Lclosed=120cm

The correct option is B 120 cm
$\frac{V}{4 L_{c l o s e d}} = \frac{3 V}{2 L_{o p e n}}$
$\Rightarrow \frac{1}{4 \times 20} = \frac{3}{2 L_{c l o s e d}}$
$\Rightarrow L_{c l o s e d} = 120 c m$