Question

Water rises to a height of $10\text{cm}$ in a certain capillary tube. Another identical tube when dipped in mercury, the level of mercury is depressed by $3.42\text{cm}$. Density of mercury is $13.6{\text{gm/cm}}^3$. The angle of contact for water in contact with glass is $0^{\circ }$ and mercury in contact with glass is ${135}^{\circ }$. Then the ratio of surface tension of water to that of mercury is:
(use $\frac{0.0342}{\sqrt{2}}=0.024$)

• A. $0.33$
• B. $0.25$
• C. $0.75$
• D. $0.15$

Answer 1

The correct option is D $0.15$

For Water:
Rise in height, $h_1=10\text{cm}=0.1\text{m}$
Density of water, ${\rho }_1=1000{\text{kg/m}}^3$
Angle of contact, ${\theta }_1=0^{\circ }$
radius of tube$=r$
Surface Tension$=T_1$
$\because h=\frac{2Tcos\theta }{\rho gr}...\left(i\right)$

$\Rightarrow h_1=\frac{2T_1\cos {\theta }_1}{{\rho }_{1}gr}$
Or, $T_1=\frac{h_1{\rho }_{1}gr}{2\cos {\theta }_1}$

$T_1=\frac{0.1\times 1000\times 10\times r}{2\times 1}$
$\therefore T_1=500r...\left(ii\right)$

$\Rightarrow$
For Mercury
Depression in level, $h_2=-0.0342\text{m}$
($-ve\text{sign}$ as height is depressed)
${\rho }_2=13.6{\text{gm/cm}}^3=13600{\text{kg/m}}^3$
${\theta }_2={135}^{\circ }$
Radius of tube $=r$

Similarly from Eq $\left(i\right)$,
$\Rightarrow T_2=\frac{-0.0342\times 13600\times 10\times r}{2\cos {135}^{\circ }}$

$T_2=\frac{-0.0342\times 136000\times r}{2\times (-\frac{1}{\sqrt{2}})}$
$\therefore T_2=3264r...\left(iii\right)$

$\Rightarrow \frac{T_1}{T_2}=\frac{500r}{3264r}$
$\therefore \frac{T_1}{T_2}=0.153$