Question

If a 5 cm long capillary tube with 0.1 mm internal diameter, open at both ends, is slightly dipped in water having surface tension 75 dyne/cm, state whether:
water will overflow out of the upper end of capillary. Explain your answer.

Answer 1

Given the surface tension of water,

T = 75 dyne/cm

Radius r = 0.1/2 mm = 0.05 mm = 0.005 mm

$density\rho =1gm/c{m}^3$, angle of contact, $\theta =0^{\circ }$

Let h be the height to which water rise in capillary tube

Then

$h=\frac{2Tcos\theta }{rpg}=\frac{2\times 75\times cos{0}^{\circ }}{0.005\times 1\times 981}cm=30.58cm$

But the length of capillary tube, h = 5 cm

c. The water will not overflow out of the upper end of the capillary. It will rise only up to the upper end of the capillary

The liquid meniscus will adjust its radius of curvature R in a such a way that

$R^1{h}^1=Rh[hR=\frac{2T}{\rho g}=constant]$

Where R is the radius of curvature liquid meniscus would possess if the capillary tube were of sufficient length

$R^1=\frac{Rh}{h}=\frac{rh}{h^1}[R=\frac{r}{cos\theta }=\frac{r}{cos{0}^{\circ }}=r]$

$=\frac{0.005\times 30.58}{5}=0.0306cm$