Question

There is a soap film on a rectangular frame of wire of area 4 cm x 4 cm. If the area of the frame is increased to 4 cm x 5 cm, find the work done in the process.(Surface tension of soap film = 3 x 10⁻² N/m)

Answer 1

Given :

A1 = 4 × 4 cm² = 16 cm²  

A1 = 16 × 10⁻⁴m²

A2 = 4 × 5 cm² = 20 cm²  

A2 = 20 × 10⁻⁴ m²

Surfae tension of soap film T = 3 × 10⁻² N/m

W = T∆A  

∴ W = 2T∆A [A soap has two surfaces]

= 2T (A2 – A1 )

= 2T (20 × 10⁻⁴– 16 × 10⁻⁴)

= 2 × 3 × 10⁻²× 4 × 10⁻⁴  

= 6 × 4 × 10⁻⁶

= 24 × 10⁻⁶ J

∴ Work done = 2.4 × 10⁻⁵ J

Answer 2

W = 8×10^-4 T J

# Given-

A1 = 4×4 cm^2 = 16 cm^2 = 16×10^-4 m^2

A2 = 4×5 cm^2 = 20 cm^2 = 20×10^-4 m^2

# Solution-

Let's assume T be surface tension of the soap film on rectangular frame.

Work done in increasing area is given by-

W = T × 2∆A

W = T × 2 × (A2 - A1)

W = T × 2 × (20×10^-4 - 16×10^-4)

W = 8×10^-4 T J

Work done in increasing area is 8×10^-4 T J.