The work done in increasing the size of a soap film from 10 cm × 6 cm to 10 cm × 11 cm is 3 × 10-4 joule. The surface tension of the film is:

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  1. 1.5 × 10-2 N/m
  2. 3.0 × 10-2 N/m
  3. 6.0 × 10-2 N/m
  4. 11.0 × 10-2 N/m

Answer (Detailed Solution Below)

Option 2 : 3.0 × 10-2 N/m
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Detailed Solution

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CONCEPT:

Surface Tension

Surface tension is a property of liquids that makes them acquire the least surface area possible. It is the force per unit length acting along the surface of a liquid.

The work done in increasing the surface area of a liquid film is given by the product of the surface tension and the change in surface area.

EXPLANATION:

Let's calculate the surface tension of the soap film:

The initial dimensions of the soap film are 10 cm × 6 cm. Hence, the initial area (A1) is:

A1 = 10 cm × 6 cm = 60 cm2

The final dimensions of the soap film are 10 cm × 11 cm. Hence, the final area (A2) is:

A2 = 10 cm × 11 cm = 110 cm2

The change in area (ΔA) is:

ΔA = A2 - A1 = 110 cm2 - 60 cm2 = 50 cm2

The work done (W) in increasing the size of the soap film is given as 3 × 10-4 joule.

The surface tension (γ) is calculated using the formula:

W = γ × ΔA

Rearranging the formula for γ gives:

γ = W / ΔA

Substitute the values:

γ = (3 × 10-4 joule) / (50 cm2)

Note: 1 cm2 = 10-4 m2, so 50 cm2 = 50 × 10-4 m2

γ = (3 × 10-4 joule) / (50 × 10-4 m2)

γ = (3 / 50) N/m

γ = 0.06 N/m

Therefore, the correct answer is option 2: 3.0 × 10-2 N/m.

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