In a resonance tube experiment when the tube is filled with water up to a height of 17.0 cm from bottom, it resonates with a given tuning fork. When the water level is raised the next resonance with the same tuning fork occurs at a height of 24.5 cm. If the velocity of sound in air is 330 m/s, the tuning fork frequency is :

Waves PYQ — JEE Main 2020 · 5 Sep · Shift 1 · Q59 with Solution

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Correct answer: A

Concept used: resonance in a closed pipe

In a resonance tube experiment, the air column above water behaves like a closed organ pipe.

Resonance lengths are: $L_1 = \frac{\lambda}{4}, \quad L_2 = \frac{3\lambda}{4}, \quad L_3 = \frac{5\lambda}{4}, \dots$

Hence, the difference between two successive resonance lengths is: $L_2 - L_1 = \frac{\lambda}{2}$
Interpret the given heights

The tube is filled with water up to a height of $17.0\text{ cm}$ from the bottom for one resonance, and when water level is raised to $24.5\text{ cm}$ , the next resonance occurs.

Raising the water level decreases the air-column length.

So the difference in successive resonance lengths equals the difference in water heights: $\Delta L = 24.5 - 17.0 = 7.5\text{ cm}$

Therefore, $\frac{\lambda}{2} = 7.5\text{ cm}$ $\lambda = 15.0\text{ cm} = 0.15\text{ m}$
Use wave relation

$v = f\lambda$ $f = \frac{v}{\lambda} = \frac{330}{0.15}$

$f = 2200\text{ Hz}$
Check options
- A: $2200\text{ Hz}$ ✔️
- B: $3300\text{ Hz}$ ✘
- C: $1100\text{ Hz}$ ✘
- D: $550\text{ Hz}$ ✘

Therefore, the correct answer is A.