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Question

A U-tube contains water so that air column in each tube is 2.20 m. A tuning fork of frequency 345 Hz vibrate above one of the limbs, say limb A. An immiscible liquid C of specific gravity 0.6 is now slowly poured into the other limb B till air column in limb A resonates. Speed of sound in air =345 m/s, then at this point (Neglect the end correction),

A
air column in limb B will not resonate with the same tuning fork
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B
air column in limb B will resonate with the same tuning fork, in 2nd overtone
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C
the difference between water levels in the two limbs is 0.6 m
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D
the height of liquid C in limb B is 1.5 m
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Solution

The correct options are
A air column in limb B will not resonate with the same tuning fork
C the difference between water levels in the two limbs is 0.6 m
D the height of liquid C in limb B is 1.5 m
For 345 Hz , λ=vf =1 m.
Now the given setup is similar to organ pipe closed at one end. So the resonance occurs at
l=λ4,3λ4,5λ4,.... corresponding to fundamental, 1st, 2nd,.... overtones respectively.

We have the length of the column given as 2.20 m. This length does not match with any resonating lenghts.
So when the length of the column A reduces because of immersion of liquid into column B, we get the resonance at

7λ4= 7×0.25=1.75 m.
So the water level that rises in column A is =2.201.75=0.45 m

x×0.6=(0.90)×1
x=1.5 mDis correct.

Air column in B is 2.20+0.451.50=1.15 m.

Diffrence in level is 1.50.9=0.6 m C is correct.

For column B to resonate in 2nd overtone,
length of air column required is 5λ4=5×0.25=1.25 m.

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