In a closed organ pipe, the frequency of fundamental note is 30Hz . A certain amount of water is now poured in the organ pipe so that the fundamental frequency is increased to 110Hz . If the organ pipe has a

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Fundamental frequency for closed organ pipe in the first part:

$\dfrac{V}{4{l}_{1}} = 30 ⇒ {l}_{1} = \dfrac{11}{4}m$

Similarly, for second part:

$\dfrac{V}{4{l}_{2}} = 110 ⇒ {l}_{2} = \dfrac{3}{4}m$

Therefore, change in length $\Deltal = 2m$

Change in volume $= A \Deltal = 400{cm}^{3}$

$M = 400g; \left(∵ ρ = 1g{cm}^{- 3}\right)$

In a closed organ pipe, the frequency of fundamental note is 30Hz ....

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