Question

Water rises to a height of $10\, cm$ in a capillary tube and mercury falls to a depth of $3.42\, cm$ in the same capillary tube. If the density of mercury is $13.6\, kg / m ^{3}$ and angle of contact is $135^{\circ}$, the ratio of surface tension for water and mercury is (angle of contact for water and glass is $0^{\circ}$ ).

• A. $1: 0.5$
• B. $1: 3$
• C. $1: 6.5$
• D. $1.5: 1$

Answer 1

Answer: (C)

$h=\frac{2 T \cos \theta}{r \rho g}$
For water,
$10\, cm =\frac{2 \times T_{w} \times \cos 0^{\circ}}{r \times 1 \times g}$ ...(1)
$\{T_{w}-$ Surface tension of water
For mercury,
$-3.42 cm =\frac{2 \times T_{M} \times \cos 135^{\circ}}{r \times 13.6 \times g}$ ...(2)
$\{T_{M}-$ Surface tension of mercury
Dividing $Eq ^{ n }(1)$ by (2)
$\frac{10}{-3.42}=\frac{2 \times T_{w} \times 1 \times r \times 13.6 \times g}{r \times 1 \times g \times 2 \times T_{M} \times \frac{-1}{\sqrt{2}}}$
$\Rightarrow \frac{10}{3.42}=\sqrt{2} \times 13.6 \times \frac{T_{w}}{T_{M}}$
$\Rightarrow \frac{10}{3.42 \times 1.41 \times 13.6}=\frac{T_{W}}{T_{M}}$
$\Rightarrow \frac{1}{6.5}=\frac{T_{W}}{T_{M}}$