Question

A long thin walled capillary tube of mass $M=10\text{gm}$ and radius $r=7\text{mm}$ is partially immersed in a liquid of surface tension $T=0.75\text{N/s}$ . The angle of contact between the liquid and the tube wall is ${60}^{\circ }$. The amount of force needed to hold the tube vertically is $N$.
(Neglect buoyant force on the tube.)

Answer 1

given,
$M=10\times {10}^{-3}kg$
$r=7\times {10}^{-3}m$
$T=0.75N{m}^{-1}$
$\theta ={60}^0$

$F_{ext}=T\cos \theta \times 2\pi r+T\cos \theta \times 2\pi r+Mg$,
(this is the sum of force on the inner wall +sum of foces on the outer wall+weight of fluid acting )
$F_{ext}=4\pi rT\cos \theta +Mg$
$F_{ext}=4\frac{22}{7}\times 7\times {10}^{-3}\times 0.75\times \frac{1}{2}+10\times {10}^{-3}\times 10$
$F_{ext}=0.133N$
This much force is required to hold the tube verticaly