Question

If a 5 cm long capillary tube with 0.1 mm internal diameter and open at both ends is slightly dipped in water having surface tension 75 dyne $c{m}^{-1}$, state whether

(i) water will rise halfway in the capillary

(ii) Water will rise up to the upper end of the capillary

(iii) Water will overflow out of the upper end of the capillary

Explain your answer

Answer 1

Given that surface tension of water T = 75 dyne/cm
Radius $r=\frac{0.1}{2}mm=0.05mm=0.005cm$
density $\rho =1gm/c{m}^3$ angle of contact $\theta =0^{\circ }$
Let h be the height to which water rise in the capillary tube Then
$h=\frac{2T\cos \theta }{r\rho g}=\frac{2\times 75\times \cos 0^{\circ }}{0.005\times 1\times 981}cm=30.58cm$
But length of capillary tube h' = 5 cm
(i) Because $h>\frac{h^{\prime }}{2}$ therefore the first possibility does not exist
(ii) Because the tube is of insufficient length therefore the water will rise upto the upper end of the tube
(iii) The water will not overflow out of the upper end of the capillary It will rise only upto the upper end of the capillary
The liquid meniscus will adjust its radius of curvature R' in such way that
R'h' = Rh $[\because hR=\frac{2T}{\rho g}=constant]$
where R is the radius of curvature that the liquid meniscus would possess if the capillary tube were of sufficient length
$\therefore$ $R^{\prime }=\frac{Rh}{h^{\prime }}=\frac{rh}{h^{\prime }}[\because R=\frac{r}{\cos \theta }=\frac{r}{\cos 0^{\circ }}=r]=\frac{0.005\times 30.58}{5}=0.0306cm$