Question

A long glass capillary tube is dipped in water. It is known that water wets glass. The water level rises by $h$ in the tube. The tube is now pushed down so that only length $h/2$ is outside the water surface. The angle of contact at the water surface at the upper end of the tube will be

• A. ${\tan }_2^{-1}$
• B. ${60}^o$
• C. ${30}^o$
• D. ${15}^o$

Answer 1

The correct option is B ${60}^o$

Given,

Rises the level of water in the tube $=h$.

Again, length outside the water $=\frac{h}{2}$

Here, $h=\frac{2s\cos 0^o}{rpg}=\frac{2s}{rpg}$ ...(i)
According to question,
$\frac{h}{2}=\frac{2s\cos {\theta }^{\prime }}{rpg}$ ...(ii)
Dividing eq. (ii) by (i) we get,
$\frac{1}{2}=\cos {\theta }^{\prime }$
or ${\theta }^{\prime }={60}^o$

So , The angle of contact at the water surface at the upper end of the tube will be ${\theta }^{\prime }=60.$