In a closed organ pipe, the frequency of fundamental note is $30 H z$ . A certain amount of water is now poured in the organ pipe so that the fundamental frequency is increased to $110 H z$ . If the organ pipe has a cross – sectional area of $2 c m^{2}$ . The amount of water poured in the organ tube is (velocity of sound is 330 m/s)

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400 gm

200 gm

300 gm

600 gm

answer is B.

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Detailed Solution

$\frac{V}{4 l_{1}} = 30$
$\begin{aligned} ℓ_{1} = \frac{11}{4} \\ \frac{V}{4 ℓ_{2}} = 110 \end{aligned}$
$\begin{aligned} ℓ_{2} = \frac{3}{4} \\ Δ ℓ = 2 m \\ m = ρ (Δ V) \\ = ρ (A Δ ℓ) \\ = 1 \times 2 \times 200 \\ = 400 g \end{aligned}$