Question:

A capillary tube made of glass of radius $0.15 \mathrm{~mm}$ is dipped vertically in a beaker filled with methylene iodide (surface tension $=0.05 \mathrm{Nm}^{-1}$, density $=667 \mathrm{~kg} \mathrm{~m}^{-3}$ ) which rises to height $h$ in the tube. It is observed that the two tangents drawn from liquid-glass interfaces (from opp. sides of the capillary) make an angle of $60^{\circ}$ with one another. Then $\mathrm{h}$ is close to $\left(\mathrm{g}=10 \mathrm{~ms}^{-2}\right)$.

1. $0.137 \mathrm{~m}$

2. $0.172 \mathrm{~m}$

3. $0.087 \mathrm{~m}$

4. $0.049 \mathrm{~m}$

Correct Option: , 3

Solution:

$\mathrm{r} \rightarrow$ radius of capillary

$\mathrm{R} \rightarrow$ Radius of meniscus.

From figure, $\frac{\mathrm{r}}{\mathrm{R}}=\cos 30^{\circ}$

$\mathrm{R}=\frac{2 \mathrm{r}}{\sqrt{3}}=\frac{2 \times 0.15 \times 10^{-3}}{\sqrt{3}}$

$=\frac{0.3}{\sqrt{3}} \times 10^{-3} \mathrm{~m}$

Height of capillary

$\mathrm{h}=\frac{2 \mathrm{~T}}{\rho \mathrm{gR}}=2 \sqrt{3} \mathrm{~T}$

$h=\frac{2 \times 0.05}{667 \times 10 \times\left(\frac{0.3 \times 10^{-3}}{\sqrt{3}}\right)}$

$\mathrm{h}=0.087 \mathrm{~m}$