“Formation of Drops and Bubbles” MHT-CET Class 12 Physics Crash Course | Chapter Mechanical Properties of Fluids
In this session G.S. Khairnar discussed the theory and MCQ’s of “Formation of Drops and Bubbles”. The session will be useful for all the aspirants of MHT-CET 2022. The session will be in Marathi and English.
Formation of Drops and Bubbles:-
Hello dear students we are familiar about the force due to surface tension that acts on the liquid surface. It is depends upon the density of liquid and also the surface in contact. When a liquid is placed on flat and smooth surface, it tries to occupy the spherical shape due to surface tension. Sir Laplace’s studied the formation of drops and bubbles in his law called as Laplace’s law of spherical membrane.
In this article we are going to discuss the formation of drops and bubble and the equation for pressure outside and inside free surface of liquid drop/bubble.
Let’s discuss the concept of Drops and Bubbles.……………!
Due to the surface tension free liquid drops & bubbles are spherical, if effect of gravity & air resistance are negligible.
As it is spherical in shape, the inside pressure will be greater than outside (Pi>P0) & excess pressure is (Pi-P0)
Let radius of liquid increases from r to (r + ∆r)
Initial surface area, A1 = 4 π r2
Final surface area, A2 = 4 π (r+∆r)2
= 4 π (r2+2r∆r+∆r2)
∴ A2 = (4 π r2+8 π r∆r +4 π ∆r)
As ∆r is very small, ∆r2 is neglected.
∴ Increase in surface area = dA = A2 – A1
= (4 π r2+8 π r ∆r – 4 π r2)
∴ dA = (8 π r ∆r) …………….(1)
Now, work done in increasing surface area,
∴ dW = T.dA= T.(8 π r ∆r)………(2)
By definition, we know that pressure is force acting per unit area,
Excess pressure, (Pi– P0) = Excess force / Area
∴ Excess force = Excess pressure (Pi-P0) × area
∴ dF = (Pi-P0) × 4 π r2 ……….(3)
And, Work done,
∴ dW = dF.∆r
∴ dW = (Pi-P0) × 4 π r2.∆r ……(4)
On Comparing, equations (2) & (4)
(Pi-P0) × 4 πr2. ∆r = T× (8 π r ∆r)
(Pi-P0) = ……….(5)
Equation (5) is called as Laplace’s Law of spherical membrane for drops.
For, soap bubble, there are two sides.
Change in surface area A = 2 (8 π r ∆r) = 16 π r ∆r
∴ dW = 16 π r ∆r.T
Work done, (Pi-P0) x 4 π r2.∆r= T.(16 π r ∆r)
Equation (6) is called as Laplace’s Law of spherical membrane for bubbles.
Formation of Drops and Bubbles:-
Let’s solve following MCQ’s……!
Q.1) According to Laplace’s law, the excess pressure during formation of drop is……….
Q.2) According to Laplace’s law, the excess pressure during formation of bubble is……….
Q.3) If the excess pressure in formation of mercury drop of surface tension 425 N/m is 4 atmosphere, the radius of drop should be….
a) 2 mm
b) 2 cm
c) 0.02 cm
d) 0, 2 mm
Answer
C) 0.02 cm
Q.4) Radius of soap bubble formed is 2.5 cm, excess pressure during formation of bubble is (surface tension of soap solution=0.05 N/m) ….
a) 1 Pa
b) 2 Pa
c) 3 Pa
d) 4 Pa
Answer
d) 4 Pa
Q.5) Ratio of radii of drops of mercury if the ratio of excess pressure is 1:3 ….
a) 1:3
b) 3:1
c) 1:9
d) 9:1
Answer
a) 1:3
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