If $l_{1}$ and $l_{2}$ are the lengths of air column for the first and second resonance when a tuning fork of frequency n is sounded on a resonance tube, then the distance of the displacement antinode from the top end of the resonance tube is :

2 (l_{2} - l_{2})

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\frac{1}{2} (2 l_{1} - l_{2})

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\frac{1_{2} - 3 l_{1}}{2}

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\frac{1_{2} - 1_{1}}{2}

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Solution

The correct option is C $\frac{1_{2} - 3 l_{1}}{2}$
If $l_{1} = \frac{λ}{4}$ , and $l_{2} = \frac{3 λ}{4}$ , and according to the image, displacement anti node from the top should be zero,
therefore,

distance $= (l_{2} - 3 l_{1}) / 2 = 0$ .

answer C is correct.

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If l1 and l2 are the lengths of air column for the first and second resonance when a tuning fork of frequency n is sounded on a resonance tube, then the distance of the displacement antinode from the top end of the resonance tube is :

The correct option is C 12−3l12If l1=λ4, and l2=3λ4, and according to the image, displacement anti node from the top should be zero,therefore,distance =(l2−3l1)/2=0.answer C is correct.

If $l_{1}$ and $l_{2}$ are the lengths of air column for the first and second resonance when a tuning fork of frequency n is sounded on a resonance tube, then the distance of the displacement antinode from the top end of the resonance tube is :

The correct option is C $\frac{1_{2} - 3 l_{1}}{2}$
If $l_{1} = \frac{λ}{4}$ , and $l_{2} = \frac{3 λ}{4}$ , and according to the image, displacement anti node from the top should be zero,
therefore,

distance $= (l_{2} - 3 l_{1}) / 2 = 0$ .

answer C is correct.