Question

a) What is the pressure inside a drop of Mercury of radius 3.00mm Airtel room temperature .

b)What is the excess pressure inside the drop ?

Surface tension of Mercury at that temperature (20°C)is 4.65×10^-1 Nm^-1.The atmosphere pressure is 1.01×10^5 Pa.


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Answer 1

a) Total pressure inside a liquid drop,

P T= Pa+2T/R

where Pa= atmospheric pressure= 1.01×10^5 Pa

P T = 1.01× 10^5+ 2×4.65×10^-1/3.00×10^-3

=1.01×10^5 + 3.10×10^2

=1.01×10^5 + 0.0031 × 10^5

=1.0131×10^5 Pa= 1.01× 10^5

b)Excess pressure P = P T - P 0

= 2T/R

=3.10×10^2 Pa

=310Pa

Answer 2

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