Question

. A capillary tube of radius 0.5 mm is dipped vertically in a liquid of surface tension 0.04 N/m and relative density 0.8 g/cc. Calculate the height of capillary rise, if the angle of contact is 10°. (Given g = 9.8m/s²)

Answer 1

r = 0.5mm = 0.5 × 10⁻³m

= 5 × 10⁻⁴ m

T = 0.04 N/m

ρ = 0.8 gm/cc

= 0.8x 10 ⁻³/10 ⁻⁶  = 800 kg/m3  

θ = 10⁰

g = 9.8 m/s²

T = rh ρ g /2 cos θ  

∴ h = 2T cos θ/ r g ρ

∴ h = 2x0.04 cos10/ 5 x10⁻⁴x 800x 9.8  

h = 2x 0.04x 0.9848/ 5x 800x 9.8 10⁻⁴

h = 2.01 × 10⁻² m

∴ The height of the capillary rise is 2.01 × 10⁻² m

Answer 2

Answer:

r = 0.5mm = 0.5 × 10⁻³m

= 5 × 10⁻⁴ m

T = 0.04 N/m

ρ = 0.8 gm/cc

= 0.8x 10 ⁻³/10 ⁻⁶  = 800 kg/m3  

θ = 10⁰

g = 9.8 m/s²

T = rh ρ g /2 cos θ  

∴ h = 2T cos θ/ r g ρ

∴ h = 2x0.04 cos10/ 5 x10⁻⁴x 800x 9.8  

h = 2x 0.04x 0.9848/ 5x 800x 9.8 10⁻⁴

h = 2.01 × 10⁻² m

∴ The height of the capillary rise is 2.01 × 10⁻² m