Mercury has an angle of contact equal to 140^ with soda lime glas… - Vidyadip

MCQ

Mercury has an angle of contact equal to $140^{\circ}$ with soda lime glass. A narrow tube of radius $1.00 \;mm$ made of this glass is dipped in a trough containing mercury. By what amount does the mercury dip down (in $mm$) in the tube relative to the liquid surface outside? Surface tension of mercury at the temperature of the experiment is $0.465\; N m ^{-1} .$ Density of mercury $=13.6 \times 10^{3} \;kg m ^{-3}$

A
$11.96$
B
$2.34$
C
$8.24$
✓
$5.34$

✓

Answer

Correct option: D.

$5.34$

d
Angle of contact between mercury and soda lime glass, $\theta=140^{\circ}$

Radius of the narrow tube, $r=1 mm =1 \times 10^{-3} m$

Surface tension of mercury at the given temperature, $s=0.465 N m ^{-1}$

Density of mercury, $\rho=13.6 \times 10^{3} kg / m ^{3}$

Dip in the height of mercury $=h$

Acceleration due to gravity, $g=9.8 m / s ^{2}$

Surface tension is related with the angle of contact and the dip in the height as

$s=\frac{h \rho g r}{2 \cos \theta}$

$\therefore h=\frac{2 s \cos \theta}{r \rho g}$

$=\frac{2 \times 0.465 \times \cos 140}{1 \times 10^{-3} \times 13.6 \times 10^{3} \times 9.8}$

$=-0.00534 m$

$=-5.34 mm$

Here, the negative sign shows the decreasing level of mercury. Hence, the mercury level dips by $5.34\; mm$

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