What is the press

Total pressure inside a mercury drop is given as ‘P’,

P = Excess pressure inside the mercury drop + Atmospheric pressure

Excess pressure inside the mercury drop, P₁ =

Where,

‘S’ is the surface tension of the mercury

‘r’ is the radius of the mercury drop.

Thus,

P = P₁ + P₀

Given,

Surface tension of the mercury, S = 4.65 × 10^-1 N

Radius of the mercury drop, r = 3.0 mm = 3.0 × 10^-3 m

Atmospheric pressure, P₀ = 1.01 × 10⁵ Pa

⇒ P₁ =

⇒ P₁ = 3.1 × 10² Pa = 0.0031 × 10⁵ Pa

Therefore,

P = 0.0031 × 10⁵ Pa + 1.01 × 10⁵ Pa

⇒ P = 1.0131 × 10⁵ Pa

The pressure inside the drop of mercury is 1.0131 × 10⁵ Pa and the excess pressure inside the drop is 310 Pa.