Mechanical Properties of Fluids

Question : 19

Total: 31

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 4.65×10–1Nm–1. The atmospheric pressure is 1.01×105Pa. Also give the excess pressure inside the drop.

Solution:

Here, radius of drop, R=3.0 mm=3.0×10−3 m
Surface tension of mercury, T=4.65×10−1 Nm−1
Pressure outside the mercury drop,
P0= atmospheric pressure =1.01×105Pa
If Piis pressure inside the drop, then excess of pressure inside the mercury drop,
Pi−P0=

2×4.65×10−1

3.0×10−3

=310 N m−2
Hence, pressure inside the mercury drop,
Pi=P0+

=1.01×105+310
=1.0131×105Pa .

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