Question

A glass capillary tube of inner diameter $0.28\text{mm}$ is lowered vertically into water in a vessel. The pressure to be applied on the water in the capillary tube so that water level in the tube is same as that in the vessel $\left({\text{in N/m}}^2\right)$ is
(Surface tension of water $=0.07\text{N/m}$ and atmospheric pressure $={10}^5{\text{N/m}}^2)$

• A. ${10}^3$
• B. $99\times {10}^3$
• C. $100\times {10}^3$
• D. $101\times {10}^3$

Answer 1

The correct option is D $101\times {10}^3$

Diameter of tube$d=0.28\text{mm}=0.28\times {10}^{-3}\text{m}$
$r=0.14\times {10}^{-3}\text{m}$
Surface tension, $T=0.07\text{N/m}$
Atm. pressure = ${10}^5{\text{N/m}}^2$
Now, pressure due to $h$ height column of water in tube $=\rho gh$
And by formula of capillary rise, $h=\frac{2T\cos \theta }{r\rho g}$
For water $\theta =0^{\circ }$
So, we get
$P=\rho gh=\frac{2T}{r}=\frac{2\times 0.07}{0.14\times {10}^{-3}}={10}^3{\text{N/m}}^2$
So, pressure due to height $h$ is $={10}^3{\text{N/m}}^2$
$\therefore$ Total pressure required to be applied $=P+$ Atmospheric pressure
$={10}^3+{10}^5$
$=101\times {10}^3{\text{N/m}}^2$