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Vidaara.orgClass 11 · Physics
CodeVID-P11-09-SUR-01
Surface Tension — Assignment
Chapter: Mechanical Properties of Fluids
Topic: Surface Tension
Maximum Marks: 30
Time: 60 minutes
Name: ____________________ Roll No.: __________ Date: ____________

General Instructions

  • All questions are compulsory.
  • Section A carries 1 mark each, Section B 2 marks, Section C 3 marks and Section D 5 marks.
  • Show all working for Sections B, C and D. Only final answers are given at the end — for full solutions, raise your doubts with your teacher.
Section A — Multiple Choice Questions 5 × 1 = 5 marks
1.
A free liquid drop tends to take a spherical shape because surface tension minimises its:
  • A.volume
  • B.surface area
  • C.density
  • D.pressure
2.
Surface energy per unit area is numerically equal to:
  • A.the angle of contact
  • B.the surface tension
  • C.the density
  • D.twice the surface tension
3.
The angle of contact for mercury on glass is:
  • A.zero
  • B.acute
  • C.obtuse
  • D.exactly $90^\circ$
4.
Capillary rise is greater in a tube that is:
  • A.wider
  • B.narrower
  • C.longer
  • D.shorter
5.
The excess pressure inside a liquid drop of radius $R$ is:
  • A.$\frac{T}{R}$
  • B.$\frac{2T}{R}$
  • C.$\frac{4T}{R}$
  • D.$\frac{2T}{R^2}$
Section B — Short Answer (2 marks) 3 × 2 = 6 marks
6.
Define surface tension and write its SI unit.
7.
Distinguish between cohesive and adhesive forces.
8.
Find the excess pressure in a water drop of radius $2\ \text{mm}$. (Take $T=0.072\ \text{N/m}$.)
Section C — Short Answer (3 marks) 2 × 3 = 6 marks
9.
Derive the expression for capillary rise $h=\frac{2T\cos\theta}{\rho g r}$.
10.
Why is the excess pressure in a soap bubble twice that in a liquid drop of the same radius?
Section D — Long Answer (5 marks) 1 × 5 = 5 marks
11.
Explain surface tension on the molecular level and define surface energy. Derive the formula for capillary rise and state why water rises but mercury is depressed in a glass tube.

Answer Key

Section A — Multiple Choice Questions
  1. (B) surface area
  2. (B) the surface tension
  3. (C) obtuse
  4. (B) narrower
  5. (B) $\frac{2T}{R}$
Section B — Short Answer (2 marks)
  1. Force per unit length acting along a liquid surface, $T=\frac{F}{L}$; SI unit N/m.
  2. Cohesive forces act between molecules of the same substance; adhesive forces act between molecules of two different substances.
  3. $P=\frac{2T}{R}=\frac{2\times0.072}{2\times10^{-3}}=72\ \text{Pa}$.
Section C — Short Answer (3 marks)
  1. The upward force from surface tension is $2\pi r T\cos\theta$; this supports the weight of the column $\pi r^2 h\rho g$. Equating gives $h=\frac{2T\cos\theta}{\rho g r}$.
  2. A soap bubble has two liquid–air surfaces (inner and outer) while a drop has only one, so the excess pressure $\frac{4T}{R}$ is double the drop's $\frac{2T}{R}$.
Section D — Long Answer (5 marks)
  1. Surface molecules have fewer neighbours, so a net inward pull makes the surface contract; surface energy is the work done per unit increase of area, equal to $T$. Balancing surface-tension force $2\pi r T\cos\theta$ against the column weight $\pi r^2 h\rho g$ gives $h=\frac{2T\cos\theta}{\rho g r}$. For water on glass $\theta$ is acute ($\cos\theta>0$) so it rises; for mercury $\theta$ is obtuse ($\cos\theta<0$) so it is depressed.
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