Question

In a closed organ pipe, the frequency of the 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 as 330 m/s)

Answer 1

Correct Answer: 400

Step 1: Calculate the Initial Length of Air Column:

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

f = V / 4ℓ₁

where f = 30 Hz and V = 330 m/s.

Solving for ℓ₁:

ℓ₁ = V / 4 × f = 330 / (4 × 30) = 330 / 120 = 11 / 4 m = 2.75 m

Step 2: Calculate the New Length of Air Column:

When the frequency increases to 110 Hz:

f' = V / 4ℓ₂

Solving for ℓ₂:

ℓ₂ = V / 4 × f' = 330 / (4 × 110) = 330 / 440 = 3 / 4 m = 0.75 m

Step 3: Determine the Change in Length:

Δℓ = ℓ₁ - ℓ₂ = 2.75 - 0.75 = 2 m

Step 4: Calculate the Volume of Water Added:

The volume of water added corresponds to the volume of the air column displaced:

Change in volume = A × Δℓ = 2 cm² × 200 cm = 400 cm³

Step 5: Convert Volume to Mass:

Given that the density of water ρ = 1 g/cm³:

M = ρ × Volume = 1 × 400 = 400 g