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 $13.6g/c{m}^3$ and the angle of contact is ${135}^o$, the ratio of surface tension for water and mercury is

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

Answer 1

The correct option is C $1:6.5$

Let liquid density be $\rho ,$ rise in capillary height $h$ and tube radius $r.$

Balancing pressures on either side,

$hr\rho g$ = $2Tcos\theta$ where $T=$ surface tension, $\theta =$ angle of contact

$\Rightarrow \frac{T_M}{T_W}=\frac{3.112\times 13.6}{10\times 1}\times \frac{\cos {\theta }_w}{\cos {135}^{\circ }}$

where $T_M=$ surface tension in mercury

$T_W=$ surface tension in water

$\frac{T_M}{T_W}=\frac{10}{65}=1:6.5$