Question

How can you prove that capillary depression (h) in the tube of radius when inserted in mercury (sp gr. S1) above which a liquid of sp gr S2 lies is given by
h = 2σcosθ / (S1-S2)​

Answer 1

Answer:

Problem 1. Determine the capillary depression of mercury in a 4 mm ID glass

Tube. Assume surface tension as 0.45 N/m and β =115°?.

The specific weight of mercury = 13550 × 9.81 N/m3, Equating the surface force and the

pressure force, [h × γ × πD2/4] = [π × D × σ × cos β], Solving for h,

h = {4 × σ × cos β}/{γ × D} = [4 × 0.45 × cos 115]/[13550 × 9.81 × 0.004]

= – 1.431 × 10–3 m or – 1.431 mm, (depression)Problem 2. A glass tube of 8 mm ID is immersed in a liquid at 20°C. The specific

weight of the liquid is 20601 N/m 3

. The contact angle is 60°. Surface tension is 0.15 N/m.

Calculate the capillary rise and also the radius of curvature of the meniscus.

Capillary rise, h = {4 ×  × cos}/{ × D} = {4 × 0.15 × cos 60}/{20601 × 0.008}

= 1.82 × 10 -3 m or 1.82 mm.

The meniscus is a doubly curved surface with equal radius as the section is circular

P i - P o) =  × {(1/R 1) + (1/R 2)} = 2 /R

R = 2/(P - P ), (P - P ) = specific weight × h

i o i o

So, R = [2 × 0.15]/ [1.82 × 10 -3 × 2060] = 8 × 10 -3 m or 8 mm.

Problem 3. A mercury column is used to measure the atmospheric pressure. The height of column above the mercury

well surface is 762 mm. The tube is 3 mm in dia. The contact angle is 140°. Determine the true pressure in mm of mercury

if surface tension is 0.51 N/m. The space above the column may be considered as vacuum. In this case capillary depression is

involved and so the true pressure = mercury column + capillary depression.

The specific weight of mercury = 13550 × 9.81 N/m 3

, equating forces,

[h ×  ×  D 2

/4] = [ × D × × cos].

So

h = {4 ×  × cos}/{ × D}

h = (4 × 0.51) × cos 140]/[13550 × 9.81 × 0.003]

= - 3.92 × 10 -3

m or - 3.92 mm, (depression)

Hence actual pressure indicated = 762 + 3.92 = 765.92 mm of mercury.

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