Question

water rises to a height of 10cm in a capillary tube, and mercury falls to a depth of 3.42cm in the same capillary tube. if the density of mercury is 13.6g/cm3 and the angle of contact is 135o, the ratio of surface tension for water and mercury is

Answer 1

Let a liquid of density ρ rise a height h in a capillary of radius r

And we can suppose:

hrρg = 2Tcosθ

Where T is surface tension of that liquid and θ is angle of contact and

θw is the angle of contact of water

Let Tm and Tw be the surface tension of water and mercury.

Tm/Tw = [(3.112 * 13.6)/(10 * 1)] * [cosθw / cos(1350)]

= 5.98*cosθw

Answer 2

Explanation:

The height h through which a liquid will rise in a capillary tube of radius r is given by

h=

rρg

2Scosθ

where S is the surface tension, ρ is the density of the liquid and θ is the angle of contact.

For identical capillary tubes and two liquids ( water '1' & mercury '2')

h

2

h

1

=

rρ

1

g

2S

1

cosθ

1

×

2S

2

cosθ

2

rρ

2

g

⇒

h

2

h

1

=

ρ

1

S

1

cosθ

1

×

S

2

cosθ

2

ρ

2

⇒

S

2

S

1

=

h

2

h

1

×

cosθ

1

cosθ

2

×

ρ

2

ρ

1

⇒

S

2

S

1

=

−3.42

10

×

cos0

∘

cos135

∘

×

13.6

1

⇒

S

2

S

1

=

−3.42

10

×

1

−0.707

×

13.6

1

⇒

S

2

S

1

=0.152≈

6.5