If a liquid surface (density
ρ
) supports another fluid of density,
ρb
above the meniscus, then a balance of forces would result in
capillary rise h
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A liquid of density ρ and surface tension σ rises in a capillary of inner radius r to a height
h=
2σcosθ
ρgr
where θ is the contact angle made by the liquid meniscus with the capillary’s surface The liquid rises due to the forces of adhesion, cohesion, and surface tension. If adhesive force (liquid-capillary) is more than the cohesive force (liquid-liquid) then liquid rises as in case of water rise in a glass capillary. In this case, the contact angle is less than 90 deg and the meniscus is concave. If adhesive force is less than the cohesive force then liquid depresses as in case of mercury in a glass capillary. In this case, the contact angle is greater than 90 deg and the meniscus is convex.
The formula for capillary rise can be derived using pressure balance.
The capillary rise experiment is used to measure the surface tension of a liquid.
Solved Problems on Capillary Rise
Problem from IIT JEE 2014
A glass capillary tube is of the shape of a truncated cone with an apex angle
α
so that its two ends have cross sections of different radii. When dipped in water vertically, water rises in it to a height
h
, where the radius of its cross-section is
b
. If the surface tension of water is
S
, its density is
ρ
, and its contact angle with glass is
θ
, the value of
h
will be (where
g
is acceleration due to gravity), 2
S
b
ρ
g
cos
(
θ
−
α
)
2
S
b
ρ
g
cos
(
θ
+
α
)
2
S
b
ρ
g
cos
(
θ
−
α
/
2
)
2
S
b
ρ
g
cos
(
θ
+
α
/
2
)
Solution: Let
R
be the radius of the meniscus formed with a contact angle
θ
(see figure). By geometry, this radius makes an angle
θ
+
α
2
with the horizontal and,
cos
(
θ
+
α
2
)
=
b
/
R
.
Let
P
0
be the atmospheric pressure and
P
1
be the pressure just below the meniscus. Excess pressure on the concave side of meniscus of radius
R
is,
P
0
−
P
1
=
2
S
/
R
.
The hydrostatic pressure gives,
P
0
−
P
1
=
h
ρ
g
.
Eliminate
(
P
0
−
P
)
from second and third equations and substitute
R
from first equation to get,
h
=
2
S
ρ
g
R
=
2
S
b
ρ
g
cos
(
θ
+
α
2
)
.
Problem from IIT JEE 2018
A uniform capillary tube of inner radius
r
is dipped vertically into a beaker filled with water. The water rises to a height
h
in the capillary tube above the water surface in the beaker. The surface tension of water is
σ
. The angle of contact between water and the wall of the capillary tube is
θ
. Ignore the mass of water in the meniscus. Which of the following statements is (are) true?
For a given material of the capillary tube,
h
decreases with increase in
r
.
For a given material of the capillary tube,
h
is independent of
σ
.
If this experiment is performed in a lift going up with a constant acceleration, then
h
decreases.
h
is proportional to contact angle
θ
.
Solution: Consider the portion of water that rises in the capillary tube. In equilibrium, the upward force due to surface tension,
F
=
2
π
r
σ
cos
θ
, is balanced by the weight of the water,
W
=
π
r
2
h
ρ
g
, i.e.,
2
π
r
σ
cos
θ
=
π
r
2
h
ρ
g
which gives
h
=
2
σ
cos
θ
r
ρ
g
.
From this equation,
h
decreases with increase in
r
, it depends on
σ
and it is proportional to
cos
θ
. The contact angle
θ
depends on the material of the capillary tube.
Hi,This is your Answer Happy Learning.
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