Question

A U-tube contains water so that air column in each tube is $2.20m$. 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 $=345m/s$, then at this point (Neglect the end correction),

• A. air column in limb B will not resonate with the same tuning fork
• B. air column in limb B will resonate with the same tuning fork, in $2^{nd}$ overtone
• 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 $m$

Answer 1

Answer: (A), (C), (D)

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 $Hz$ , $\lambda =\frac{v}{f}=1m$.
Now the given setup is similar to organ pipe closed at one end. So the resonance occurs at
$l=\frac{\lambda }{4},\frac{3\lambda }{4},\frac{5\lambda }{4},....$ corresponding to fundamental, 1st, 2nd,.... overtones respectively.

We have the length of the column given as $2.20m$. 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

$\frac{7\lambda }{4}=$$7\times 0.25=1.75m$.
So the water level that rises in column A is $=2.20-1.75=0.45m$

$x\times 0.6=(0.90)\times 1$
$x=1.5m\to D\to$is correct.

$\therefore$ Air column in B is $2.20+0.45-1.50=1.15m$.

$\Rightarrow$Diffrence in level is $1.5-0.9=0.6m$ $\Rightarrow C\to$is correct.

For column B to resonate in $2^{nd}$ overtone,
length of air column required is $\frac{5\lambda }{4}=5\times 0.25=1.25m$.