Question

A 0.4-mm-diameter glass tube is inserted into water at 20∘C in a cup. The surface tension of water at 20∘C is σs=0.073N/m. The contact angle can be taken as zero degrees. The capillary rise of water in the tube is (a) 2.9 cm (b) 7.4 cm (c) 5.1 cm (d) 9.3 cm (e) 14.0 cm

Answer 1

Answer:

(b) 7.4cm

Explanation:

Given the following variables:

the radius of a glass capillary tube as r,

the coefficient of surface tension of the liquid in the glass as σ,

the density of the liquid as ρ,

the angle of contact between the liquid and the walls of the tube as θ and;

the capillary rise or height to which the liquid rises in the tube as h.

The variables are related by the following relation;

h = (2 x σ cos θ) / (r x ρ x g)           -------------------------------(i)

Where;

g = acceleration due to gravity.

Therefore from the question;

diameter of tube = 0.4mm

=> radius, r  = (0.4 / 2)mm = 0.2mm = 0.0002m

θ = contact angle = 0°

σ = surface tension coefficient =  σs = 0.073N/m

ρ = density of liquid (water) = 1000kg/m³  [a constant]

Taking g = 9.81m/s², substitute these values into equation (i) to calculate the capillary rise, h, as follows;

h = (2 x 0.073 cos 0) / (0.0002 x 1000 x 9.81)

h = (2 x 0.073 x 1) / (1.962)

h = 0.074m

Convert the result to cm by multiplying by 100 as follows;

0.074m x 100 = 7.4cm

Therefore the capillary rise of water in the tube is 7.4cm

Answer 2

Answer:

7.4 cm

Explanation:

The capillary rise of the water in the tube can be calculated by the formula

h = 2Tcosθ / rρg

where T, surface tension = 0.073N/m , r radius = 0.4 mm/2 = 0.2 mm x 1 m / 1000 mm = 0.2 × 10⁻³m, ρ density of water = 1000 kg/m³ and g acceleration due to gravity = 9.81 m/s²

h = 2 × 0.073N/m cos0 / (0.2 × 10⁻³m ×  1000 kg/m³ × 9.81 m/s²) = 0.0744 m = 7.44 cm