QJEEAdv18Ph1671: JEE Advanced 2018 Physics - Class 11 JEEAdv18Ph1671 Solution: In an experiment to measure the speed of sound by a resonating…

To determine which statements are true, we'll analyze the experiment step by step, using the theory of resonating air columns in a closed tube (closed at one end and open at the other).

Fundamentals of Resonance in a Closed Tube

For a tube closed at one end, the resonant lengths occur at:

$L_n = \left( \frac{2n - 1}{4} \right) \lambda + e$

where:

$L_n$ is the length of the air column for the $n^\text{th}$ resonance,

$\lambda$ is the wavelength of the sound wave,

$e$ is the end correction,

$n = 1, 2, 3, \dots$

Successive resonances (like those at $L_1$ and $L_2$ ) can be used to eliminate the end correction and find the wavelength.

Calculations

Step 1: Calculate the Wavelength ( $\lambda$ )

Given:

First resonance length ( $L_1$ ) = 50.7 cm

Second resonance length ( $L_2$ ) = 83.9 cm

The difference between successive resonances in a closed tube is:

$L_2 - L_1 = \frac{\lambda}{2}$

Solving for $\lambda$ :

$\lambda = 2(L_2 - L_1) = 2(83.9\, \text{cm} - 50.7\, \text{cm}) = 2(33.2\, \text{cm}) = 66.4\, \text{cm}$

Conclusion: The wavelength of the sound wave is 66.4 cm.

Option C is true.

Step 2: Calculate the Speed of Sound ( $v$ )

Using the relationship $v = f\lambda$ :

Frequency ( $f$ ) = 500 Hz

Wavelength ( $\lambda$ ) = 66.4 cm = 0.664 m

$v = f\lambda = 500\, \text{Hz} \times 0.664\, \text{m} = 332\, \text{m/s}$

Conclusion: The speed of sound determined from the experiment is 332 m/s.

Option A is true.

Step 3: Calculate the End Correction ( $e$ )

Assuming the first resonance corresponds to the second harmonic ( $n = 2$ ):

$L_1 = \left( \frac{2n - 1}{4} \right) \lambda + e$

For $n = 2$ :

$L_1 = \left( \frac{3}{4} \right) \lambda + e$

Solving for $e$ :

$e = L_1 - \left( \frac{3}{4} \right) \lambda = 50.7\, \text{cm} - \left( \frac{3}{4} \times 66.4\, \text{cm} \right)$

$e = 50.7\, \text{cm} - 49.8\, \text{cm} = 0.9\, \text{cm}$

Conclusion: The end correction is 0.9 cm.

Option B is true.

Step 4: Determine if $L_1$ is the Fundamental Harmonic

Assuming $L_1$ corresponds to the fundamental harmonic ( $n = 1$ ):

$L_1 = \left( \frac{1}{4} \right) \lambda + e$

$e = L_1 - \left( \frac{1}{4} \right) \lambda = 50.7\, \text{cm} - 16.6\, \text{cm} = 34.1\, \text{cm}$

An end correction of 34.1 cm is unreasonably large for typical experiments, indicating that $L_1$ does not correspond to the fundamental harmonic.

Conclusion: The resonance at 50.7 cm does not correspond to the fundamental harmonic.

Option D is false.