Question

What is the pressure inside the drop of mercury of radius 3.00 mm at room temperature? Surface tension of mercury at that temperature (20°C) is 4.65 x 10^-1 N m^-1. The atmospheric pressure is 1.01 x 10⁵ Pa. Also give the excess pressure inside the drop.

Answer 1

Here, Surface tension,

T = 4.65 x 10^-1 Nm^-1

Radius of drop, r = 3.00 mm = 3 x 10^-3 m

Outside pressure, p₀ = 1.01 x 10⁵ Pa

Using the relation, p = 2T/r

We have,

= {2 x 4.65 x 10^-1}/{3 x 10^-3}

= 3.1 x 10²

= 310 Nm^-2

If p₁ be the inside pressure, then

p₁ - p₀ = p

⇒ p₁ = p + p₀ = 310 + 1.01 x 10⁵

= 3.1 x 10² + 1.01 x 10⁵ Pa

= 0.0031 x 10⁵ + 1.01 x 10⁵ Pa

= 1.013 x 10⁵ Pa