The surface tension of soap solution is 0.03 N/m. The amount of work done in forming a bubble of radius 5 cm is:
A) 3.77 J
B) 1.885 J
C) 0.95 × 10⁻³ J
D) 1.9 × 10⁻³ J

We are given the surface tension of a soap solution as γ=0.03 Nm\gamma = 0.03 \, \frac{N}{m}, and the radius of the bubble as r=5 cm=0.05 mr = 5 \, \text{cm} = 0.05 \, \text{m}. We are asked to calculate the amount of work done in forming the bubble.
Concept:
The work done in forming a bubble is related to the change in surface area of the soap film. When a bubble is formed, the surface area increases, and since the soap film has surface tension, the work done is the product of surface tension and the change in surface area.
For a single bubble with radius rr, the surface area AA of the bubble is:
A=4πr2A = 4 \pi r^2
Since a bubble has two surfaces (inside and outside), the total surface area AtotalA_{\text{total}} for a bubble is:
Atotal=8πr2A_{\text{total}} = 8 \pi r^2
The work done in forming the bubble is the product of the surface tension γ\gamma and the total change in surface area:
W=γ⋅ΔAW = \gamma \cdot \Delta A
where ΔA=8πr2\Delta A = 8 \pi r^2 is the total change in surface area.
Thus, the work done WW is:
W=γ⋅8πr2W = \gamma \cdot 8 \pi r^2
Substituting the given values:
γ=0.03 Nm,r=0.05 m\gamma = 0.03 \, \frac{N}{m}, \quad r = 0.05 \, \text{m} W=0.03×8π(0.05)2W = 0.03 \times 8 \pi (0.05)^2
Calculating the work done:
W=0.03×8π×0.0025=0.03×0.0628≈0.001884 JW = 0.03 \times 8 \pi \times 0.0025 = 0.03 \times 0.0628 \approx 0.001884 \, \text{J}
Thus, the work done in forming the bubble is approximately 1.885 J1.885 \, \text{J}.
Final Answer:
The amount of work done in forming the bubble is:
B) 1.885 J