# 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.

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