 Solution

ID: past-exam-of-the-mphil-in-micro-and-nanotechnology-enterprise-course-of-the-university-of-cambridge/2019/paper-1/section-b/10/ii/solution

2 π r γ cos θ

against the weight

ρ g π r^{2} h

of the liquid column. The capillary action is

h = \frac{2 γ c o s θ}{ρ g r} = \frac{4 γ c o s θ}{ρ g D} .

With perfect wetting,

cos θ = 1

D = 1 0^{- 3} m

ρ = 1000 kg m^{- 3}

and

g = 9.81 m s^{- 2}

, $h = 2.94 \times 1 0^{- 2} m = 29.4 mm$ . The diameter, not the radius, is

1 mm

. This neglects the small meniscus-volume correction, gas-density buoyancy and contact-angle hysteresis.

 New to topics? Read the docs here!