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Mechanical Properties Of Fluids

Question
CBSEENPH11020009

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
equals space fraction numerator 2 straight S over denominator straight r space plus space straight P subscript straight o end fraction

equals space fraction numerator 2 space cross times space 4.65 space cross times space 10 to the power of negative 1 space end exponent over denominator left parenthesis 3 space cross times space 10 to the power of negative 3 end exponent right parenthesis end fraction plus space 1.01 space cross times 10 to the power of 5 space

equals space 1.0131 space cross times space 10 to the power of 5

equals 1.01 space cross times space 10 to the power of 5 space end exponent Pa

Excess space pressure space equals space fraction numerator 2 straight S over denominator straight r end fraction

equals fraction numerator 2 space cross times space 4.65 space cross times space 10 to the power of negative 1 end exponent over denominator left parenthesis 3 space cross times space 10 to the power of negative 3 end exponent right parenthesis end fraction space equals space 310 space Pa

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