In increasing the area of a film of soap solution from 50 cm 2 to 100 cm 2 , 3 . 0 × 10 - 4 J of work is done. Calculate the value of surface tension of the soap solution.

Given: Initial area = 50 cm 2 Final area = 100 cm 2 Work done = 3 . 0 × 10 - 4 J Find: Surface Tension of the soap solution Solution: Change in area △ A = Final area - Initial area △ A = 100 cm 2 - 50 cm 2 = 50 cm 2 △ A = 50 × 10 - 4 m 2 We know, w = T 2 △ A T = w 2 △ A T = 3 · 0 × 10 - 4 2 × 50 × 10 - 4 J / m 2 T = 3 × 10 - 2 J / m 2 OR T = 3 × 10 - 2 N / m → Relation between work, surface tension and change in area.

In increasing the area of a film of soap solution from 50 cm 2 to 100 cm 2 , 3 . 0 × 10 - 4 J of work is done. Calculate the value of surface tension of the soap solution.

Given: Initial area = 50 cm 2 Final area = 100 cm 2 Work done = 3 . 0 × 10 - 4 J Find: Surface Tension of the soap solution Solution: Change in area △ A = Final area - Initial area △ A = 100 cm 2 - 50 cm 2 = 50 cm 2 △ A = 50 × 10 - 4 m 2 We know, w = T 2 △ A T = w 2 △ A T = 3 · 0 × 10 - 4 2 × 50 × 10 - 4 J / m 2 T = 3 × 10 - 2 J / m 2 OR T = 3 × 10 - 2 N / m → Relation between work, surface tension and change in area.

The surface area of a soap-bubble is 2 . 0 × 10 - 3 m 2 . How much work will be done in blowing the bubble to twice its surface area ?

Surface area of soap-bubble = 2 × 10 - 3 m Find: Work will be done in blowing (change) the bubble to twice its surface area. Solution: Initial surface area = 2 . 0 × 10 - 3 m 2 Initial surface area = 2 × 2 . 0 × 10 - 3 m 2 Change in surface area ( △ A ) = 2 × 2 . 0 × 10 - 3 - 2 . 0 × 10 - 3 = 2 . 0 × 10 - 3 m 2 We know that w = T × 2 △ A T = Surface tension of soap solution = 310 × 10 - 2 N / m ] W = 3 . 0 × 10 - 2 × 2 × 2 . 0 × 10 - 3 J w = 12 × 10 - 5 J or 1 . 2 × 10 - 4 J w = 1 . 2 × 10 - 4 J .

E M B I B E

ISC Physics Class XI Part 1>Chapter 19 - Surface Tension>NUMERICALS>Q 3

MEDIUM

11th ICSE

IMPORTANT

Earn 100

In increasing the area of a film of soap solution from $50 {cm}^{2}$ to $100 {cm}^{2}, 3.0 \times 10^{- 4} J$ of work is done. Calculate the value of surface tension of the soap solution.

Important Questions on Surface Tension

MEDIUM

11th ICSE

IMPORTANT

ISC Physics Class XI Part 1>Chapter 19 - Surface Tension>NUMERICALS>Q 4

The surface area of a soap-bubble is

2.0 \times 10^{- 3} m^{2}

. How much work will be done in blowing the bubble to twice its surface area ?

MEDIUM

11th ICSE

IMPORTANT

ISC Physics Class XI Part 1>Chapter 19 - Surface Tension>NUMERICALS>Q 5

How much work will be done in increasing the diameter of a soap bubble from

2 cm

to

5 cm

?

MEDIUM

11th ICSE

IMPORTANT

ISC Physics Class XI Part 1>Chapter 19 - Surface Tension>NUMERICALS>Q 6

A liquid drop of radius

1 mm

is broken into

1000

equal small drops. How much work will be done? Surface tension of water

= 0.07 N m^{- 1}

and

π = 22 / 7

.

MEDIUM

11th ICSE

IMPORTANT

ISC Physics Class XI Part 1>Chapter 19 - Surface Tension>NUMERICALS>Q 7

A drop of water of diameter

0.2 cm

is broken up into

27000

droplets of equal volume. How much work will be done against surface tension in the process? (Surface tension of water

= 7 \times 10^{- 2} N m^{- 1})

MEDIUM

11th ICSE

IMPORTANT

ISC Physics Class XI Part 1>Chapter 19 - Surface Tension>NUMERICALS>Q 8

A mercury drop of radius

1.0 cm

is sprayed into

10^{6}

droplets of equal size. Calculate the energy expended.

MEDIUM

11th ICSE

IMPORTANT

ISC Physics Class XI Part 1>Chapter 19 - Surface Tension>NUMERICALS>Q 9

Calculate the energy released when

1000

small water drops each of same radius

10^{- 7} m

coalesce to form one large drop. The surface tension of water is

7.0 \times 10^{- 2} {Nm}^{- 1}

.

MEDIUM

11th ICSE

IMPORTANT

ISC Physics Class XI Part 1>Chapter 19 - Surface Tension>NUMERICALS>Q 10

A big drop is formed by combining

27

small droplets of water. What will be the change in the surface energy? What will be the ratio of the surface energy of big drop to the surface energy of

27

small droplets?

MEDIUM

11th ICSE

IMPORTANT

ISC Physics Class XI Part 1>Chapter 19 - Surface Tension>NUMERICALS>Q 11

A thin wire is bent in the form of a ring of diameter

3.0 cm

. The ring is placed horizontally on the surface of soap solution and then raised up slowly. How much upward force is necessary to break the vertical film formed between the ring and the solution?