Question

Water rises to a height of $10\text{cm}$ in a capillary tube and mercury falls to a depth of $\sqrt{2}\text{cm}$ in the same capillary tube. If the density of mercury is $\approx 14{\text{g/cm}}^3$ 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. $2:1$
• B. $1:2$
• C. $5:14$
• D. $14:5$

Answer 1

The correct option is C $5:14$

We know,
Rise in height of capillary tube
$h=\frac{2T\cos \theta }{r\rho g}$
$\Rightarrow T=\frac{h\rho rg}{2\cos \theta }$
For water:
$T_1=\frac{rg\times 10}{2\cos 0^{\circ }}=5rg$
For mercury:
$T_2=\frac{rg\times (-\sqrt{2})\times 14}{2\cos {135}^{\circ }}=14rg$

$\therefore \frac{T_1}{T_2}=\frac{5}{14}$