Question

Water rises to a height of $10cm$ in a capillary tube and mercury falls to a depth of $3.42cm$ in the same capillary tube. If the density of mercury is $13.6kg/m^3$ 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)

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