127k views
5 votes
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/c.c and the angles of contact for mercury and water are 135 degree and 0 degree respectively, the ratio of surface tension for water and mercury is ____________

User Swaprks
by
7.2k points

1 Answer

5 votes

Final Answer:

The ratio of surface tension for water to mercury is approximately 1.86.

Step-by-step explanation:

Capillary rise and fall:

Capillary rise and fall are caused by the interplay of surface tension, adhesive forces, and cohesive forces.

The height of capillary rise (h) is given by the Young-Laplace equation:

h = 2γ cos θ / (ρg r)

where:

γ is the surface tension

θ is the angle of contact

ρ is the density of the liquid

g is the acceleration due to gravity

r is the radius of the capillary tube

Setting up the ratio:

We are given the capillary rise for water (h1) and the capillary fall for mercury (h2), their respective angles of contact (θ1 and θ2), and the density of mercury (ρ2).

We need to find the ratio of the surface tension of water (γ1) to the surface tension of mercury (γ2):

γ1 / γ2 = ?

Solving the equations:

Since the experiment is performed with the same capillary tube, the radius (r) and acceleration due to gravity (g) are the same for both liquids.

We can write two equations for the capillary rise and fall of the liquids:

h1 = 2γ1 / (ρ1g r)

h2 = -2γ2 / (ρ2g r)

Divide the first equation by the second equation:

h1 / h2 = (γ1 / γ2) * (ρ2 / ρ1) * (cos θ1 / cos θ2)

Substitute the given values:

h1 = 10 cm

h2 = 3.42 cm

ρ2 = 13.6 g/cm³

ρ1 = 1 g/cm³ (density of water)

θ1 = 0°

θ2 = 135° (cos 135° = -0.707)

Calculate the ratio of surface tensions:

γ1 / γ2 = (10 cm) / (3.42 cm) * (13.6 g/cm³) / (1 g/cm³) * (cos 0°) / (cos 135°)

γ1 / γ2 ≈ 1.86

Therefore, the ratio of surface tension for water to mercury is approximately 1.86.

User MKorsch
by
7.2k points