Question

A narrow tube of radius $1 \, mm$ made out of soda lime glass is dipped in a tank containing mercury. The angle of contact of mercury with the surface of the glass is $140^{0}$ and surface tension of mercury at the temperature of the experiment is $0.465 \, N \, m^{- 1}$ . Mercury dips down in the tube through a depth of : (Density of mercury $=13.6\times 10^{3} \, kg \, m^{- 3}$ )

• A. $2.34 \, mm$
• B. $5.34 \, mm$
• C. $6.25 \, mm$
• D. $1.44 \, mm$

Answer 1

Answer: (B)

Here, $\theta =140^\circ ,r=1\times 10^{- 3}m$
$S=0.465 \, N \, m^{- 1},\rho =13.6\times 10^{3}kg \, m^{- 3}$
As $h=\frac{2 S cos \theta }{r \rho g}=\frac{2 \times 0.465 \times \left(\right. - cos ⁡ 40 ^\circ \left.\right)}{\left(10\right)^{- 3} \times 13.6 \times \left(10\right)^{3} \times 9.8 \, }$
$\left(\because cos 140 ^\circ = - cos ⁡ 40 ^\circ \right)$
$=-5.34\times 10^{- 3}m=-5.34 \, mm$
Note: Negative value of $h$ shows that the mercury level is depressed in the tube