In a closed organ pipe, the frequency of fundamental note is $$30 \text{ Hz}$$. A certain amount of water is now poured in the organ pipe so that the fundamental frequency is increased to $$110 \text{ Hz}$$. If the organ pipe has a cross-sectional area of $$2 \text{ cm}^2$$, the amount of water poured in the organ tube is ________ g. (Take speed of sound in air is $$330 \text{ m s}^{-1}$$)

For a closed organ pipe, the fundamental frequency is given by:

$$f = \frac{v}{4L}$$

where $$v$$ is the speed of sound and $$L$$ is the length of the air column. Initially, the pipe produces a fundamental frequency of 30 Hz, so

$$30 = \frac{330}{4L_1}$$

Solving for $$L_1$$ gives

$$L_1 = \frac{330}{4 \times 30} = \frac{330}{120} = 2.75 \text{ m}$$

After adding water, the frequency rises to 110 Hz, which implies

$$110 = \frac{330}{4L_2}$$

and hence

$$L_2 = \frac{330}{4 \times 110} = \frac{330}{440} = 0.75 \text{ m}$$

The water fills the bottom portion of the pipe, so the height of the water column is

$$h = L_1 - L_2 = 2.75 - 0.75 = 2 \text{ m}$$

Given the cross-sectional area of the pipe is $$2 \text{ cm}^2$$, the volume of water poured in is

$$V = A \times h = 2 \text{ cm}^2 \times 2 \text{ m} = 2 \times 10^{-4} \text{ m}^2 \times 2 \text{ m} = 4 \times 10^{-4} \text{ m}^3$$

Using the density of water $$= 1000 \text{ kg/m}^3 = 1 \text{ g/cm}^3$$, the mass of water is

$$m = \rho \times V = 1000 \times 4 \times 10^{-4} = 0.4 \text{ kg} = 400 \text{ g}$$

Therefore, the mass of the water added is $$400$$ g.