Question

A capillary tube of radius 1 mm is kept vertical with the lower end in water . (a) Find the height of water raised in the capillary. (b) If the length of the capillary tube is half the answer of part (a), find the angle $\theta$ made by the water surface in the capillary with the wall.

Answer 1

(a) r = 1 mm = ${10}^{-3}m$

$h=\frac{2Tcos\theta }{r\rho g}$

= $\frac{2\times \left(0.076\right)}{{10}^{-3}\times 10\times 100}$

=1.52 cm

$=1.52\times {10}^{-2}m$

(b) $h^{\prime }=\frac{2Tcos\theta }{r\rho g}$

$cos\theta =\frac{h^{\prime }r\rho g}{2T}$

$cos\theta =\frac{(1.52\times {10}^{-2})\times \left({10}^{-3}\right)\times \left(10\right)\times \left(10\right)}{2\times 0.076}$

$[Because{h}^{\prime }=\left(\frac{h}{2}\right)=\frac{1.52\times {10}^{-2}}{2}]$

= $\frac{1}{2}$

So, $\theta =co{s}^{-1}\left(\frac{1}{2}\right)={60}^{\circ }$