Question

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

• A. $5.34 \,mm$
• B. $2.35\, mm$
• C. $6.25\, mm$
• D. $1.44\, mm$

Answer 1

Answer: (A)

Here, $\theta=140^{\circ}, r=1\times10^{-3}\,m $
$S=0.465 \,N \, m^{-1}, \rho=13.6\times10^{3}\, kg \, m^{-3}$
As $h=\frac{2S \, cos\,\theta}{r\rho g}=\frac{2\times0.465\times\left(-cos\,40^{\circ}\right)}{10^{-3}\times13.6\times10^{3}\times9.8}$
$\left(\because cos\,140^{\circ}=-cos\,40^{\circ}\right)$
$= -5.34 \times 10^{-3} \,m = -5.34 \,mm$
Note: Negative value of $h$ shows that the mercury level is depressed in the tube