Mechanical Properties of Fluids NCERT Solutions Class 11 Physics - Solved Exercise Question 10.19

Question 10.19:
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 × 10–1 N m–1. The atmospheric pressure is 1.01 × 105 Pa. Also give the excess pressure inside the drop.

Solution:
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
Total pressure inside the mercury drop
= Excess pressure inside mercury + Atmospheric pressure
= 2S / r + P0
= [ 2 X 4.65 X 10-1 / (3 X 10-3) ] + 1.01 X 105
= 1.0131 × 105
= 1.01 ×105 Pa
Excess pressure = 2S / r
= [ 2 X 4.65 X 10-1 / (3 X 10-3) ] = 310 Pa

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