JEE Main 2022 · 28 June · evening · Numerical answer · +4 / −0
A tunning fork of frequency 340 Hz resonates in the fundamental mode with an air column of length 125 cm in a cylindrical tube closed at one end. When water is slowly poured in it, the minimum height of water required for observing resonance once again is ___________ cm.
(Velocity of sound in air is 340 ms⁻¹)
Answer and solution
Answer 50
Solution
Physics sir · 4 steps
Find the wavelength first, then step the air column down to the next resonance length below 125 cm.
- Wavelength. λ = v/f = 340/340 = 1 m = 100 cm.
L = λ/4, 3λ/4, 5λ/4, … = 25, 75, 125 cm, …
- Resonance lengths. A tube closed at one end resonates when the air column is an odd multiple of λ/4. So the 125 cm column is 5λ/4, and the next shorter column that resonates is 3λ/4 = 75 cm.
- Water fills the gap. Pouring in water shortens the air column. The water must fill the difference: 125 - 75 = 50 cm.
- So. The minimum height of water for resonance again is 50 cm.
The common slip: 25. Taking the gap between successive resonances as λ/4 gives 25 cm. In a closed tube the lengths go up in odd quarters, so the gap is λ/2 = 50 cm.
The idea to keep. In a closed tube, resonance lengths are odd multiples of λ/4, so successive resonances are λ/2 apart. Shortening the column by λ/2 gives the next resonance.
Answer from the NTA final answer key. Solution by PrepWiser.