Answer to Question #141019 in Electric Circuits for Takam Praise

Question

A clean glass capillary tube, of internal diameter 0.04 cm, is held vertically with its lower end below the surface of clean water in a beaker, and with 10 cm of the tube above the surface. What will happen if the tube is depressed until only 5 cm of its length is above the surface? The surface tension of water is 7.2 x 10-2 Nm-1.

Answer 1

Explanations & Calculations

• The relationship between the height a particular liquid rises in a given capillary tube due to the surface tension is given by

$\qquad\qquad \begin{aligned} \small h &= \small \frac{2T\cos\theta }{r\rho g}\\ \end{aligned}$ : $\small \theta$ is the contact angle between water & glass

• And for the given situation (10cm above the water level) the capillary rise is

$\qquad\qquad \begin{aligned} \small h_1 &= \small \frac{2\times 0.072Nm^{-1}}{0.0002m \times 10^3kgm^{-3}\times 9.8ms^{-2}} \cos\theta\\ \small &= \small 0.0735\cos\theta \,\,m \\ \small&=\small 7.35\cos \theta \,\,cm \end{aligned}$

• It's obvious that the capillary rise will not change as none of the components in the equation is changed even after the tube is depressed further.

• After it is depressed only 5 cm is above the water & if $\small h_1<5cm$ no water will spill & if $\small h_1>>5cm$ water will spill.
• If the capillary rise just a very little more that 5cm water will stay in the tube until the contact angle reaches 180 degrees without spill.