Question

Water rises in a capillary tube to a certain height such that the upward force due to surface tension is balanced by $75\times {10}^{-4}\text{N}$ force due to the weight of the liquid. If the surface tension of water is $6\times {10}^{-2}{\text{Nm}}^{-1}$, the inner circumference of the capillary must be
(Assume angle of contact=$0$)

• A. $1.25\times {10}^{-2}\text{m}$
• B. $0.50\times {10}^{-2}\text{m}$
• C. $6.5\times {10}^{-2}\text{m}$
• D. $12.5\times {10}^{-2}\text{m}$

Answer 1

The correct option is D $12.5\times {10}^{-2}\text{m}$

Let $T$ be the surface tension, $\theta$ the angle of contact and $R$ be the radius of capillary tube.

The vertical component of the surface tension force will balance the weight of the liquid column.
Therefore,
$Tcos\theta \times 2\pi R=mg$
Since angle of contact, $\theta =0^{\circ }$
$T\times 2\pi R=mg$
$2\pi R=\frac{mg}{T}=\frac{75\times {10}^{-4}}{6\times {10}^{-2}}$
$\Rightarrow 2\pi R=12.5\times {10}^{-2}m$
Hence the circumference of the capillary tube is $12.5\times {10}^{-2}m$