Question

obtain an expression for the height to which a liquid of density and surface tension will rise in a capillary tube of radius r depends on it gravitational acceleration given that h inversely proportional to 1/r by dimensional analysis​

Answer 1

Given that,

Radius of tube = r

Height of tube = h

Let us consider a liquid whose angle of contact with the glass capillary tube is θ. We have a concave surface.

In the figure,

T is the surface tension of the liquid and r is the radius of tube and R is the radius of concave surface of liquid.

ρ is the density of liquid.

We need to calculate the pressure difference between two sides of top surface

According to figure,

$P_{a}-P_{o}=\dfrac{2T}{R}$

Here, $R\cos\theta=r$

$P_{a}-P_{o}=\dfrac{2T\cos\theta}{r}$....(I)

Considering point A and B they must be at same pressure

$P_{o}+h\rho g=P_{a}$

$P_{a}-P_{o}=h\rho g$....(II)

We need to calculate the height of the capillary

Using equation (I) and (II)

$\dfrac{2T\cos\theta}{r}=h\rho g$

$h=\dfrac{2T\cos\theta}{\rho g r}$

So, The height of capillary is inversely proportional to r.

This is proved.

The height of a liquid column is given by,

$h=\dfrac{2T\cos\theta}{\rho g r}$

Where, h = height

T = surface tension

g = acceleration due to gravity

r = radius  

$\theta$ = contact angle

$\rho$ = density of liquid

We need to proof the height of tube is inversely proportional to radius

Using dimension formula in right hand side

$\dfrac{2T\cos\theta}{\rho g r}$

Here, 2 and cosθ are constant

$=\dfrac{T}{\rho g r}$

Put the dimension formula

$=\dfrac{[MT^{-2}]}{[ML^{-3}]\times[LT^{-2}]\times[L]}$

Here, M shows the mass.

T shows the time.

L shows the length

$=[L]$

Right hand side = Left hand side

Hence, This is required answer.

Answer 2

Explanation:

Radius of tube = r

Height of tube = h

Let us consider a liquid whose angle of contact with the glass capillary tube is θ. We have a concave surface.

In the figure,

T is the surface tension of the liquid and r is the radius of tube and R is the radius of concave surface of liquid.

ρ is the density of liquid.

We need to calculate the pressure difference between two sides of top surface

According to figure,

P_{a}-P_{o}=\dfrac{2T}{R}P

a

−P

o

=

R

2T

Here, R\cos\theta=rRcosθ=r

P_{a}-P_{o}=\dfrac{2T\cos\theta}{r}P

a

−P

o

=

r

2Tcosθ

....(I)

Considering point A and B they must be at same pressure

P_{o}+h\rho g=P_{a}P

o

+hρg=P

a

P_{a}-P_{o}=h\rho gP

a

−P

o

=hρg ....(II)

We need to calculate the height of the capillary

Using equation (I) and (II)

\dfrac{2T\cos\theta}{r}=h\rho g

r

2Tcosθ

=hρg

h=\dfrac{2T\cos\theta}{\rho g r}h=

ρgr

2Tcosθ

So, The height of capillary is inversely proportional to r.

This is proved.

The height of a liquid column is given by,

h=\dfrac{2T\cos\theta}{\rho g r}h=

ρgr

2Tcosθ

Where, h = height

T = surface tension

g = acceleration due to gravity

r = radius

\thetaθ = contact angle

\rhoρ = density of liquid

We need to proof the height of tube is inversely proportional to radius

Using dimension formula in right hand side

\dfrac{2T\cos\theta}{\rho g r}

ρgr

2Tcosθ

Here, 2 and cosθ are constant

=\dfrac{T}{\rho g r}=

ρgr

T

Put the dimension formula

=\dfrac{[MT^{-2}]}{[ML^{-3}]\times[LT^{-2}]\times[L]}=

[ML

−3

]×[LT

−2

]×[L]

[MT

−2

]

Here, M shows the mass.

T shows the time.

L shows the length

=[L]=[L]

Right hand side = Left hand side

Hence, This is required answer.