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.
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Concept :-
Excess pressure inside a liquid is given by ∆P = 2S/R
where,
S is surface tension of liquid and R is radius of drop.
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Radius of drop (R) = 3 mm = 3 × 10^-3 m
surface tension of mercury (S) = 4.65 × 10^-1 N/m = 0.465 N/m
Atmospheric pressure (Po) = 1.01 × 10^5 Pa
pressure inside the drop = Atmospheric pressure+ Excess pressure inside the mercury drop .
= Po + 2S/R
= 1.01 × 10^5 + 2×0.465 /3 × 10^-3
= 1.01 × 10^5 + 0.0031× 10^5
= 1.013 × 10^5 Pa
Excess pressure inside the drop (∆P) = 2S/R = 2×0.465/3 × 10^-3
= 310 Pa
Excess pressure inside a liquid is given by ∆P = 2S/R
where,
S is surface tension of liquid and R is radius of drop.
______________________________________________________
Radius of drop (R) = 3 mm = 3 × 10^-3 m
surface tension of mercury (S) = 4.65 × 10^-1 N/m = 0.465 N/m
Atmospheric pressure (Po) = 1.01 × 10^5 Pa
pressure inside the drop = Atmospheric pressure+ Excess pressure inside the mercury drop .
= Po + 2S/R
= 1.01 × 10^5 + 2×0.465 /3 × 10^-3
= 1.01 × 10^5 + 0.0031× 10^5
= 1.013 × 10^5 Pa
Excess pressure inside the drop (∆P) = 2S/R = 2×0.465/3 × 10^-3
= 310 Pa
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