Surface Tension — Excess pressure, capillary rise
Properties of Matter
14
JEE Qs
8%
Hard
75
min
Master the derivations for excess pressure and capillary rise to correctly apply the formulas and understand the underlying physics and assumptions.
🧮 Key Formulas
✅ Key Points for JEE
- 1Surface tension arises from unbalanced cohesive forces at the liquid-gas interface, acting tangentially to minimize surface area.
- 2Pressure inside a curved liquid surface is always greater on the concave side relative to the convex side.
- 3The contact angle (θ) between the liquid and solid surface determines the meniscus shape (concave if θ < 90°, convex if θ > 90°) and hence capillary rise or fall.
- 4Carefully distinguish between a liquid drop (one free surface), an air bubble inside liquid (one free surface), and a soap bubble (two free surfaces) when applying excess pressure formulas.
- 5Capillary rise/fall is a balance between the upward component of surface tension force acting along the contact line and the downward weight of the liquid column.
⚠️ Common Mistakes
- ✕Confusing the formulas for excess pressure in a liquid drop (2T/R) versus a soap bubble (4T/R) by miscounting the number of free surfaces.
- ✕Incorrectly identifying which side of the curved interface has higher pressure; the pressure is always higher on the concave side.
- ✕Ignoring the contact angle (θ) or using its sine instead of cosine in the capillary rise formula, or misinterpreting its value (e.g., for mercury).
- ✕Failing to consider external atmospheric pressure when calculating absolute pressure inside bubbles/drops.
📝 Practice Questions
See allQ47.Two soap bubbles of radius 2 cm and 4 cm , respectively, are in contact with each other. The radius of curvature of the common surface, in cm , is ______ . and
Q27.A small rigid spherical ball of mass M is dropped in a long vertical tube containing glycerine. The velocity of the ball becomes constant after some time. If the density of glycerine is half of the density of the ball, then the viscous force acting on the ball will be (consider g as acceleration due to gravity) (1) 2 Mg (2) Mg (3) 3 2 Mg (4) Mg2
Q41. A tube of length L is shown in the figure. The radius of cross section at the point (1) is 2 cm and at the point (2) is 1 cm , respectively. If the velocity of water entering at point (1) is 2 m/s, then velocity of water leaving the point (2) will be (1) 4 m/s (2) 2 m/s (3) 6 m/s (4) 8 m/s
Q30.A massless spring gets elongated by amount x1 under a tension of 5 N . Its elongation is x2 under the tension of 7 N . For the elongation of (5x1 −2x2), the tension in the spring will be, (1) 39 N (2) 15 N (3) 11 N (4) 20 N
Q32.Water flows in a horizontal pipe whose one end is closed with a valve. The reading of the pressure gauge attached to the pipe is P1 . The reading of the pressure gauge falls to P2 when the valve is opened. The speed of water flowing in the pipe is proportional to (1) P1 −P2 (2) (P1 −P2)4 (3) (P1 −P2)2 (4) √P1 −P2
Q48.An air bubble of radius 1.0 mm is observed at a depth of 20 cm below the free surface of a liquid having surface tension 0.095 J/m2 and density 103 kg/m3 . The difference between pressure inside the bubble and atmospheric pressure is _____ N/m2 . (Take g = 10 m/s2 )
NCERT Chapters
- Class 11 Physics Ch 10: Mechanical Properties of Fluids