### Waves And Sound Question 3

#### Question 3 - 2024 (30 Jan Shift 1)

In a closed organ pipe, the frequency of fundamental note is $30 Hz$. A certain amount of water is now poured in the organ pipe so that the fundamental frequency is increased to $110 Hz$. If the organ pipe has a cross-sectional area of $2 cm^{2}$, the amount of water poured in the organ tube is g. (Take speed of sound in air is $330 m / s$ )

## Show Answer

#### Answer: (400)

#### Solution:

$\frac{V}{4 \ell _1}=30 \Rightarrow \ell _1=\frac{11}{4} m$

$\frac{V}{4 \ell _2}=110 \Rightarrow \ell _2=\frac{3}{4} m$

$\Delta \ell=2 m$

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

$\mathbf{M}=\mathbf{4 0 0} g ;\left(\because \rho=1 g / cm^{3}\right)$