Water rises in a glass capillary tube to a height of 8.4cm. What is the diameter of the capillarytube?

The height to which a liquid will rise in a capillary tube is determined by several factors, as shown in the following equation:

\(h = \frac{{2T\cos \theta }}{{r\rho g}}\)

In this equation, \(h\) is the height of the liquid inside the capillary tube relative to the surface of the liquid outside the tube, \(T\) is the surface tension of the liquid, \(\theta \) is the contact angle between the liquid and the tube, \(r\) is the radius of the tube, \(\rho \) is the density of the liquid, and \(g\) is the acceleration due to gravity, \(9.8\;{\rm{m}}/{{\rm{s}}^2}\). When the tube is made of a material to which the liquid molecules are strongly attracted, they will spread out completely on the surface, which corresponds to a contact angle of \({0^\circ }\). This is the situation for water rising in a glass tube.

We know that

\(g = 9.8\;{\rm{m}}/{{\rm{s}}^2}\)

\({\rm{d}}\)of water \( = 997\;{\rm{kg}}/{{\rm{m}}^3}\)

\(\cos \phi .\quad = 1\) (angle of contact \(b/w\)

water and glass is \({0^\circ }\), i.e. \(\phi = 0\) )

\({\rm{T}}\)of water. \( = 72{\rm{mN}}/{\rm{m}} = 0.072\;{\rm{N}}/{\rm{m}}\)

\({\rm{h}}(\)given \() = 8.4\;{\rm{cm}} = 0.084\;{\rm{m}}\)

Putting values in the equation,

\(\begin{aligned}{}0.084 &= \frac{{(2 \times 0.072 \times 1)}}{{(r \times 997 \times 9.8)}}\\r &= 0.000175\;{\rm{m}}\end{aligned}\)

i.e. \(r = 0.0175\;{\rm{cm}}\)

Diameter \( = 2 \times \) radius

Diameter \( = 2 \times 0.0175\;{\rm{cm}}\)

Diameter \( = 0.035\;{\rm{cm}}\)