Question

A glass capillary tube of internal radius $r=0.25\text{mm}$ is immersed in water. The top end of the tube projects by $2\text{cm}$ above the surface of water. At what angle does the liquid meet the ends of the tube?
[surface tension of water = $0.07\text{N}{\text{m}}^{-1}$]

• A. ${60}^{\circ }$
• B. $0^{\circ }$
• C. ${70}^{\circ }$
• D. ${180}^{\circ }$

Answer 1

The correct option is C ${70}^{\circ }$

Water wets the glass, so the angle of contact is zero.
Neglecting the small mass of the meniscus, for full rise of the liquid, we can say that
$2\pi rT=\pi r^{2}h\rho g$
or $h=\frac{2T}{r\rho g}$
From the data given in the question,
$h=\frac{2\times 0.07}{0.25\times {10}^{-3}\times 1000\times 9.8}$
$=0.057=5.7\text{cm}$
But given that, tube extends only $2\text{cm}$ above the water and so water will rise by $2\text{cm}$ and meet the tube at an angle such that
$\frac{2T}{R}=h^{\prime }\rho g$
But, $\cos \alpha =\frac{r}{R}$
Thus we get,
$2T\cos \alpha =r{h}^{\prime }\rho g$
$\Rightarrow \cos \alpha =\frac{h^{\prime }r\rho g}{2T}$

From the data given in the question,
$\cos \alpha =\frac{2\times {10}^{-2}\times 0.25\times {10}^{-3}\times 1000\times 9.8}{2\times 0.07}$
$\Rightarrow \alpha \approx {70}^{\circ }$
Thus. option (c) is the correct answer