Question

A liquid of density $\rho$ and surface tension S rises in a capillary tube upto its equilibrium position. Let $\theta$ is the angle of contact between the liquid and capacity tube. Choose the correct option(s) regarding capillary rise.

• A. The magnitude of work done by surface tension force during complete rise is $\frac{4\pi S^2{\cos }^2\theta }{\rho g}$
• B. The work done by gravity force during complete rise is $-\frac{2\pi S^2{\cos }^2\theta }{\rho g}$
• C. The heat liberates during complete rise is $\frac{2\pi S^2{\cos }^2\theta }{\rho g}$
• D. The heat liberates during complete rise is $\frac{6\pi S^2{\cos }^2\theta }{\rho g}$

Answer 1

Answer: (A), (B), (C)

The magnitude of work done by surface tension force during complete rise is $\frac{4\pi S^2{\cos }^2\theta }{\rho g}$
The work done by gravity force during complete rise is $-\frac{2\pi S^2{\cos }^2\theta }{\rho g}$
The heat liberates during complete rise is $\frac{2\pi S^2{\cos }^2\theta }{\rho g}$

Let the height upto which the liquid rises in the capillary be $h$

$⟹$ $h=\frac{2Scos\theta }{\rho gr}$ where $r$ is the radius of the capillary tube

Total upward force due too surface tension $F_u=\left(Scos\theta \right)2\pi r$

$\therefore$ Work done by surface tension $W_{st}=F_{u}h$

$⟹$ $W_{st}=\frac{2Scos\theta }{\rho gr}\times \left(Scos\theta \right)2\pi r=\frac{4\pi S^{2}co{s}^2\theta }{\rho g}$

Mass of the liquid rose in the capillary $m=\rho \left(\pi r^{2}h\right)$

Also the centre of mass of the liquid rose by $\frac{h}{2}$.

$\therefore$ Work done by gravity force $W_g=-mg\frac{h}{2}$

OR $W_g=-\rho \left(\pi r^{2}h\right)\times \frac{h}{2}=-\rho \pi r^{2}g\frac{2Scos\theta }{\rho gr}\times (\frac{Scos\theta }{\rho gr})$

$⟹$ $W_g=-\frac{2\pi S^{2}co{s}^2\theta }{\rho g}$

Heat liberated $H=W_{st}+W_g=\frac{4\pi S^{2}co{s}^2\theta }{\rho g}-\frac{2\pi S^{2}co{s}^2\theta }{\rho g}=\frac{2\pi S^{2}co{s}^2\theta }{\rho g}$