Work done in increasing the size of a soap bubble from a radius of 3 cm to 5 cm is nearly (Surface tension of soap solution = 0.03 Nm^-1)

Calculation:

Given,

R₁ = 3 cm = 3 × 10⁻² m

R₂ = 5 cm = 5 × 10⁻² m

T = 0.03 N/m (Surface tension)

Original surface area = 2 × 4πR₁

For second bubble = 2 × 4πR₂

Work done = Surface tension × extension in area

Work done = T × ΔA

Substitute the values:

Work done = 0.03 × 2[4πR₂² - 4πR₁²]

Work done = 0.03 × 8π[(5)² - (3)²] × 10⁻⁴

Work done = 0.03 × 8π × 16 × 10⁻⁴

Work done = 0.384π × 10⁻³ J

Hence, the work done is approximately 0.4π mJ.