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 cm², the amount of water poured in the organ tube is __________ g. (Take speed of sound in air is 330 m / s)

Solution

Physics sir · 4 steps

Water shortens the air column. Find the column length before and after, and the gap is the water.

1. Closed pipe rule. A pipe closed at one end has a node at the closed end and an antinode at the open end. In the fundamental, the air column holds a quarter wavelength, so f = v/4L. Water poured in acts as the new closed end, so only the air column length L changes.

   L = v/4f

2. Two lengths. Before: L₁ = 330/(4 × 30) = 2.75 m. After: L₂ = 330/(4 × 110) = 0.75 m. The water column is 2.75 - 0.75 = 2 m = 200 cm tall.
3. Volume to mass. Volume of water = 2 cm² × 200 cm = 400 cm³. Water has density 1 g cm⁻³, so its mass is 400 g.
4. So. The amount of water poured in is 400 g.

The common slip: 800. Using the open-pipe formula f = v/2L gives lengths of 5.5 m and 1.5 m, so a 4 m water column and 800 g. A closed pipe needs v/4L.

The idea to keep. Pouring water into a pipe only shortens its air column. Use the right end condition: for a closed pipe the fundamental is v/4L. Then the change in length times the area gives the water volume.

Answer from the NTA final answer key. Solution by PrepWiser.