Question

Surface tension of a soap bubble is $2.0 \times 10^{–2} Nm^{–1}$. Work done to increase the radius of soap bubble from 3.5 cm to 7 cm will be: [Take $\pi=\frac{22}{7}$]

• A. 5.76 × 10–4 J
• B. 0.72 × 10–4 J
• C. 9.24 × 10–4 J
• D. 18.48 × 10–4 J

Answer 1

Answer: (D)

18.48 × 10–4 J

Answer 2

The correct answer is 18.48 × 10–4 J