In an experiment to determine the velocity of sound in air at room temperature using a resonance tube, the first resonance is observed when the air column has a length of 20.0 cm for a tuning fork of frequency 400 Hz is used. The velocity of the sound at room temperature is 336 ms - 1. The third resonance is observed when the air column has a length of \underline\hspace2cm cm.

Waves PYQ — JEE Main 2022 · 29 Jun · Shift 2 · Q64 with Solution

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Correct answer: 104

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

For a closed pipe, resonances occur at:

L_n = \frac{(2n-1)\lambda}{4}, \quad n=1,2,3,\dots

So:

Given:

Using

\lambda = \frac{v}{f}

we get

\lambda = \frac{336}{400} = 0.84\,\text{m} = 84\,\text{cm}

Observed first resonance length is

L_1 = 20.0\,\text{cm}

But ideally,

\frac{\lambda}{4} = \frac{84}{4} = 21\,\text{cm}

The difference is due to end correction.

Let end correction be $e$ . For a closed pipe,

L_1 + e = \frac{\lambda}{4}

So,

20 + e = 21

e = 1\,\text{cm}

For third resonance,

L_3 + e = \frac{5\lambda}{4}

Substitute values:

L_3 + 1 = \frac{5\times 84}{4} = 105

L_3 = 105 - 1 = 104\,\text{cm}

The third resonance is observed when the air column length is:

\boxed{104\,\text{cm}}