Question

What is the pressure inside the drop of mercury of radius 3.00 mm at room temerature? Surface tension of mercury at that temperature (20°c) is 4.65 x 10⁻¹ Nm⁻¹ .The atmospheric pressure is 1.01 x 10⁵ Pa. Also give the excess pressure inside the drop.

Answer 1

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Here's , Your Answer ...

Given ,

Radius of the mercury drop, r = 3.00 mm = 3 × 10–3 m .

Surface tension of mercury, S = 4.65 × 10–1 N m–1 .

Atmospheric pressure, P0 = 1.01 × 105 Pa .

We know that ,

Total pressure inside the mercury drop
= Excess pressure inside mercury + Atmospheric pressure

= 2S / r + P0

= [ 2 × 4.65 × 10-1 / (3 × 10-3) ] + 1.01 × 105

= 1.0131 × 105

Excess pressure = 2S / r

= [ 2 × 4.65 × 10-1 / (3 × 10-3) ]

= 310 Pa. Ans.

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HOPE , IT HELPS ... ✌️