Question

A glass capillary tube of internal radius $r=0.25\, mm$ is immersed in water. The top end of the tube projected by $2\, cm$ above the surface of the water. At what angle does the liquid meet the tube? Surface tension of water $=0.7\, N / m$.

• A. $\theta = 90^{\circ}$
• B. $\theta = 70^{\circ}$
• C. $\theta = 45^{\circ}$
• D. $\theta = 35^{\circ}$

Answer 1

Answer: (B)

Water wets glass and so the angle of contact is zero.
For full rise, neglecting the small mass in the meniscus Water wets glass and so the angle of contact is zero.
For full rise, neglecting the small mass in the meniscus
$2 \pi r T=\pi r^{2} h \rho g$
$\Rightarrow h=\frac{2 T}{r \rho g}$
[$\therefore $ water wets glass,$\theta=\theta^{\circ}$]
$=\frac{2 \times 0.07}{0.25 \times 10^{-3} \times 1000 \times 9.8}$
As the tube is only $2\, ​cm$ above the water and so, water will rise by $2\, cm$ and meet the tube at an angle such that,
$2 \pi r T \cos \theta=\pi r^{2} h'\rho g$
$\Rightarrow 2 T \cos \theta=h' r \rho g$
$\Rightarrow \cos \theta=\frac{h' r p g}{2 T}=\frac{20 \times 10^{-2} \times 0.25 \times 10^{-3} \times 1000 \times 9.8}{2 \times 0.07}$
The liquid will meet the tube at an angle, $\approx 70^{\circ}$