JEE Main 2024 Physics question
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)
Answer: 400
Solution
f = v/4ℓ: ℓ₁ = 330/(4 × 30) = 11/4 m and ℓ₂ = 330/(4 × 110) = 3/4 m. The water column is ℓ₁ − ℓ₂
See the full step-by-step solution
Create a free account to read the complete working, and practise questions like this in a timed test.
Create a free account