## Capillary and a rod

There is a narrow rod of diameter of 1.5 mm mounted coaxially into a glass capillary of inner diameter 2.0 mm.

Determine the capillary water elevation in this arrangement.

• #### Hint

This task can be solved similarly to the way the task Capillary was solved — by means of gravity and surface force equality.

• #### Notation

 d1 = 2.0 mm = 2.0·10−3 m inner circumference of the capillary d2 = 1.5 mm = 1.5·10−3 m circumference of the rod h = ? elevation difference due to capillary elevation Tabulated values: σ = 73 mN m−1 = 73·10−3 N m−1 surface tension of water ρ = 1 000 kg m−3 water density g = 9.81 m s−2 gravitational acceleration
• #### Analysis

We will solve this task in the same way we derived in the task Capillary height of a water column in a simple capillary.

That means that we will assume the water wets both the rod and the capillary completely. Force caused by the surface tension of water is acting on the glass and points downwards. This means that according to Newton’s third law, the glass acts upon water with a force of the same magnitude, but opposite direction — that is, upward. Then, there is also gravitational force, which is pointing down. When water level is at still, both of these forces need to be equal. And from this equality we will determine the water elevation.

• #### Solution

There are two forces acting upon water: gravitational force and the “force of glass”, which is a reaction to water surface force.

For magnitude of gravitational force, the following applies:

$F_G\,=\,mg\,=\,V \varrho g\,.$

Now we see that it is necessary to determine the volume of water between the capillary and the rod. That is equal to the difference between the volume of capillary with height h, and the volume of the rod with the same height. In both cases there are cylinders, so

$V\,=\, \pi \frac{d_1^2}{4} \,h\,-\, \pi \frac{d_2^2}{4} \,h$ $V\,=\, \pi \frac{d_1^2\,-\,d_2^2} {4} \,h.$

When we input the volume to the formula for magnitude of gravitational force, we get

$F_G\,=\,mg\,=\,\pi \frac{d_1^2\,-\,d_2^2} {4}\, h \varrho g\,.$

Now let’s determine the magnitude of surface force. This force is acting both around the inner circumference of the tube and the circumference of the rod, so following must apply:

$F\,=\, \sigma \, ( o_1\,+\,o_2 ) \,=\,\sigma \, \left( \pi d_1\,+\,\pi d_2 \right)$ $F\,=\, \sigma \pi \, \left( d_1\,+\, d_2 \right).$

Both forces must equal.

$F_G\,=\, F$ $\,\pi \frac{d_1^2\,-\,d_2^2} {4}\, h \varrho g\,=\, \sigma \pi \, \left( d_1\,+\, d_2 \right)$

From the last equation we express desired elevation h.

$h\,=\, \frac{4\sigma \, \left( d_1\,+\, d_2 \right) }{\left( d_1^2\,-\,d_2^2\right) \varrho g } \,=\, \frac{4\sigma \, \left( d_1\,+\, d_2 \right) }{\left( d_1\,+\, d_2 \right)\left( d_1\,-\,d_2\right) \varrho g }$ $h\,=\, \frac{4\sigma }{\left( d_1\,-\,d_2\right) \varrho g }$

Finally, we input given values into our resulting expression:

$h\,=\, \frac{4\,\cdot\,73\cdot{10^{-3}} }{\left(2\cdot{10^{-3}} \,-\,1{,}5\cdot{10^{-3}} \right)\,\cdot\, 1000 \,\cdot\, 9{,}81 }\,\mathrm{m}\,\dot=\,0.0595\,\mathrm{m}$ $h\,\dot=\,6\,\mathrm{cm}$