Question

When a long glass capillary tube of radius 0.015 cm is dipped in a liquid, the liquid rises to a height of 15 cm within it. If the contact angle between the liquid and glass to close to \[{0^0}\], the surface tension of the liquid, in milliNewton \[{m^{ - 1}}\] , is [\[{\rho _{(liquid)}} = 900kg{m^{ - 3}},g = 10m{s^{ - 2}}\]] (Give answer in closes integer)

Answer 1

Hint:If a capillary is inserted in a liquid, its height will increase. This rise in liquid in the capillary is due to surface tension. The capillary rise of a liquid is determined by the surface tension, density of the liquid, viscosity of the liquid, and diameter of the capillary tube.





Formula used:
 Rise in the capillary is given as,
\[h = \dfrac{{2T\cos \theta }}{{\rho gr}}\]
Where T is surface tension
\[\theta \] is angle of contact
\[\rho \] is density
g is acceleration due to gravity
r is radius of capillary




Complete answer:
 Given radius of the capillary, r = 0.015 cm =\[0.015 \times {10^{ - 2}}m\]
Liquid rises to a height, h = 15 cm = \[15 \times {10^{ - 2}}m\]
Angle of contact, \[\theta = {0^0}\]

Figure: Rise in liquid in the capillary.
As we know that,
Rise in the capillary, \[h = \dfrac{{2T\cos \theta }}{{\rho gr}}\]
Or,
\[T = \dfrac{{h\rho gr}}{{2\cos \theta }}\]
By substituting the given values, we get

\[T = \dfrac{{(15 \times {{10}^{ - 2}}) \times 900 \times 10 \times (0.015 \times {{10}^{ - 2}})}}{{2 \times \cos {0^0}}}\] (\[\cos {0^0} = 1\] )

\[T = 1012.5 \times {10^{ - 4}}N{m^{ - 1}}\]
\[T = 101.25 \times {10^{ - 3}}N{m^{ - 1}}\]
\[T = 101{\rm{ milliNewton}}{m^{ - 1}}\]

Hence, the surface tension of the liquid, in milliNewton \[{m^{ - 1}}\]is \[T = 101{\rm{ milliNewton}}{m^{ - 1}}\]




Note: The rise of liquid in the capillary tube is due to the forces of adhesion, cohesion, and surface tension. If adhesive force (liquid-capillary) is greater than the cohesive force (liquid-liquid) then the liquid will rise in the liquid in the capillary tube.
The formula for capillary rise may be derived by balancing forces on the liquid column. The weight of the liquid is balanced by the upward force due to surface tension. This formula can also be derived using pressure balance.