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CIA 3 PRESENTATION
ON
PROBLEMS BASED ON
SURFACE TENSION
VAIBHAV SANJAY PATIL
DIV : - SEME-A
ROLL NO. : - 04
GUIDED BY : -
PROF. D. R. SATPUTE
SURFACE TENSION
surface tension is defined as the force acting over the
surface of liquid per unit length of the surface
perpendicular to the force.
SI Unit is N/m
PROBLEM 1
• Find the surface tension in a soap bubble of 40 mm diameter, when
the inside pressure is 2.5 N/m2 above atmospheric pressure.
GIVEN : -
Dia. Of bubble = D = 40 mm = 4 x 10-3 m
Pressure = P = 2.5 N/m2
SOLUTION : -
For soap bubble…
P = 8 x σ / D
2.5 = (8 x σ) / (4 x 10-3)
σ = (2.5)(4 x 10-3)
___________
8
σ = 0.0125 N/m
PROBLEM 2
The surface tension of water in contact with air at 20ºC is 0.0725 N/m.
The pressure inside a droplet of water is to be 0.02 N/cm2 grater than
the outside pressure. Calculate the diameter of the droplet of water.
GIVEN :-
Surface tension = σ = 0.0725 N/m
Pressure = P = 0.02 N/Cm2 = 0.02 x 104 N/m2
SOLUTION :-
P = 4 x σ / D
0.02 x 104 = 4 x 0.0725 / D
D = 4 x 0.0725 / 0.02 x 104
D = 0.00145 m
THANK YOU

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problems on surface tension

  • 1. CIA 3 PRESENTATION ON PROBLEMS BASED ON SURFACE TENSION VAIBHAV SANJAY PATIL DIV : - SEME-A ROLL NO. : - 04 GUIDED BY : - PROF. D. R. SATPUTE
  • 2. SURFACE TENSION surface tension is defined as the force acting over the surface of liquid per unit length of the surface perpendicular to the force. SI Unit is N/m
  • 3. PROBLEM 1 • Find the surface tension in a soap bubble of 40 mm diameter, when the inside pressure is 2.5 N/m2 above atmospheric pressure. GIVEN : - Dia. Of bubble = D = 40 mm = 4 x 10-3 m Pressure = P = 2.5 N/m2 SOLUTION : - For soap bubble… P = 8 x σ / D 2.5 = (8 x σ) / (4 x 10-3) σ = (2.5)(4 x 10-3) ___________ 8 σ = 0.0125 N/m
  • 4. PROBLEM 2 The surface tension of water in contact with air at 20ºC is 0.0725 N/m. The pressure inside a droplet of water is to be 0.02 N/cm2 grater than the outside pressure. Calculate the diameter of the droplet of water. GIVEN :- Surface tension = σ = 0.0725 N/m Pressure = P = 0.02 N/Cm2 = 0.02 x 104 N/m2 SOLUTION :- P = 4 x σ / D 0.02 x 104 = 4 x 0.0725 / D D = 4 x 0.0725 / 0.02 x 104 D = 0.00145 m