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 $13.6g/c.c.$ and the angles of contact for mercury and water are ${135}^{\circ }$ and $0^{\circ }$, respectively, the ratio of surface tension for water and mercury is

• A. $1:0.15$
• B. $1:3$
• C. $1:6.5$
• D. $1.5:1$

Answer 1

The correct option is C $1:6.5$

Height, $h=\frac{2T\cot \theta }{r\rho g}$
$\therefore$ For water, $h_w=\frac{2\times T_w\times \cos 0^{\circ }}{r\times 1\times g}$
And, for mercury,
$h_m=\frac{2\times T_m\times \cos {135}^{\circ }}{r\times 13.6\times g}$
$\therefore \frac{h_w}{h_m}=\frac{2\times T_w\times 1}{r\times 1\times g}\times \frac{r\times 13.6\times g\times \sqrt{2}}{2\times T_m\times 1}$
$[\because \cos {135}^{\circ }=-1/\sqrt{2}]$
$\Rightarrow \frac{10}{3.42}=\frac{T_w}{T_m}\times 13.6\times \sqrt{2}$
$\therefore \frac{T_w}{T_m}=\frac{10}{3.42\times 13.6\times 1.414}=\frac{1}{6.5}$
$\therefore T_w:T_m=1:6.5$.