Question

A liquid of density $\rho$ and surface tension $\sigma$ rises to a height $h$ in a capillary tube of diameter $d$. The weight of the liquid in the capillary tube is:
(Consider liquid to be perfectly wetting in nature.)

• A. $2\pi \sigma h$
• B. $\frac{2\pi \sigma h^2}{d}$
• C. $\pi \sigma d$
• D. $\frac{\pi \sigma d^2\rho }{h}$

Answer 1

The correct option is C $\pi \sigma d$

Let the height to which the liquid rises in the capillary tube be $h$.
Cross-sectional area of the capillary tube is $A=\pi r^2$
The force due to surface tension pulling the liquid column vertically upwards is,
$F=\left(2\pi r\right)\times \sigma \times \cos \theta ...\left(ii\right)$
Since liquid is perfectly wetting,
$\theta =0^{\circ },\cos \theta =1$
Here, $2\pi r=$ length of circumference on which force due to surface tension will be acting.

For vertical equilibrium of liquid column,
$F=W$
$\Rightarrow 2\pi r\sigma =W$
$(\because d=2r)$
$\Rightarrow W=2\pi \sigma r=\pi \sigma d$.
Hence the correct choice is $\left(c\right)$.