Question

[Ans. 14.14 m/s]
Calculate the rise of water inside a
clean glass capillary tube of radius
0.1 mm, when immersed in water of
surface tension 7 x 102 N/m. The angle
of contact between water and glass is
zero, density of water = 1000 kg/m', g
= 9.8 m/s2
[Ans. 0.142 m
An air bubble ofradius 0 2 mm is situated​

Answer 1

Given :

▪ Radius of capillary tube = 0.1mm

▪ Surface tension of water = 7×10$^{-2}$N/m

▪ Angle of contact = 0°

▪ Density of water = 1000kg/m$^3$

▪ Acceleration due to gravity = 9.8m/s$^2$

To Find :

▪ Height of water column inside the capillary tube.

Formula :

→ When a capillary tube of radius 'r' is dipped in a liquid of density $\rho$ and surface tension T, the liquid rises or falls through a distance,

$\bigstar\:\underline{\boxed{\bf{\red{H=\dfrac{2T\cos\theta}{\rho gr}}}}}$

Calculation :

$\dashrightarrow\sf\:H=\dfrac{2T\cos\theta}{\rho gr}\\ \\ \dashrightarrow\sf\:H=\dfrac{2\times 7\times 10^{-2}\times \cos0\degree}{1000\times 9.8\times 0.1\times 10^{-3}}\\ \\ \dashrightarrow\sf\:H=\dfrac{14\times 10^{-2}(1)}{0.98}\\ \\ \dashrightarrow\underline{\boxed{\bf{\purple{H=0.142m}}}}\:\orange{\bigstar}$