Q. Estimate the capillary rise or depression in a tube of 3 mm diameter when tube is immersed in (a) Water (b) Mercury. Surface tensions for water and mercury at 20º C are 0.075 N/m & 0.52 N/m respectively.
Q. Estimate the capillary rise or depression in a tube of 3 mm diameter when tube is immersed in (a) Water (b) Mercury. Surface tensions for water and mercury at 20º C are 0.075 N/m & 0.52 N/m respectively.
SOLUTION:
Capillary rise or depression h
= (2σ * cosθ)/(wr) a) In Case of Water :
Surface tension σ = 0.075N / m
Contact angle θ = 0°
Specific weight ω = 9810N / (m ^ 3)
Radius of tube r = 3/2 = 1.5mm = 1.5 * 10 ^ – 3m
h = (2 * 0.075(cos 0°))/(9810 * 1.5 * 10 ^ – 3)
h = 0.0102m
Capillary rise h = 0.0102m
= 10.2mm (b) In Case of Mercury :
σ = 0.52N / m
θ = 130°
ω = Specific gravity of mercury × Specific weight of water
= (13.6 * 9810)
= – 3.34 ×10 ^ – 3m
-ve sign indicates that there is capillary depression.
Capillary depression = 3.34 × 10-³ m = 3.34 mm.
Q. What should be the diameter of a droplet of water, is the pressure inside is to hbe 0.0018 kg(f) / c * m ^ 2 greater than the out side? Given the value of surface tension of water in contact with air at 20° C as 0.0075 kg (f)/m.
SOLUTION:
The internal pressure intensity ‘P’ in excess of the outside pressure is given by equation
P = (4σ )/d
d= 4σ/ P
σ= 0.0075kf / m
P = 0.0018kg(f) / cm²
d = 4 * 0.0075/100 * 1/0.0018
= 0.167cm = 1.67mm
Q. Estimate the capillary rise or depression in a tube of 2 mm diameter when tube is immersed in (a) Water (b) Mercury. Surface tensions for water and mercury at 30º C are 0.085 N/m & 0.62 N/m respectively.
SOLUTION:
Capillary rise or depression h
= (2σ * cosθ)/(wr) a) In Case of Water :
Surface tension σ = 0.085N / m
Contact angle θ = 0°
Specific weight ω = 9810N / (m ^ 3)
Radius of tube r = 2/2 = 1mm = 1 * 10-³m
h = (2 * 0.085(cos 0°))/(9810 * 1 * 10 ^ – 3)
h = 0.017m
Capillary rise h = 0.017m
= 17mm (b) In Case of Mercury :
σ = 0.62N / m
θ = 130°
ω = Specific gravity of mercury × Specific weight of water
= (13.6 * 9810)
