Question

The work done in blowing a soap bubble of diameter 3 cm is (Surface tension of soap solution = 0.035 \( \text{Nm}^{-1} \))

• A. \( 792 \, \mu\text{J} \)
• B. \( 99 \, \mu\text{J} \)
• C. \( 396 \, \mu\text{J} \)
• D. \( 198 \, \mu\text{J} \)

Hint:

Always remember that soap bubbles have two surfaces, so multiply the surface area \( 4\pi r^2 \) by 2. For liquid drops, use only \( 4\pi r^2 \).

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:

Work done in blowing a soap bubble is stored as surface potential energy. A soap bubble has two free surfaces (inner and outer). Work \( W = T \times \Delta A \). Total Area increase \( \Delta A = 2 \times (4\pi r^2 - 0) = 8\pi r^2 \).
Step 2: Key Formula or Approach:

\( W = 8 \pi r^2 T \). Given Diameter \( D = 3 \) cm \( \implies r = 1.5 \) cm = \( 1.5 \times 10^{-2} \) m. \( T = 0.035 \, \text{N/m} \).
Step 3: Detailed Explanation:

Substitute values: \( W = 8 \times \frac{22}{7} \times (1.5 \times 10^{-2})^2 \times 0.035 \) \( W = 8 \times \frac{22}{7} \times 2.25 \times 10^{-4} \times \frac{35}{1000} \) \( W = 8 \times 22 \times 2.25 \times 10^{-4} \times \frac{5}{1000} \) (Since \( 35/7 = 5 \)) \( W = 8 \times 22 \times 2.25 \times 5 \times 10^{-7} \) \( W = 176 \times 11.25 \times 10^{-7} \) Let's simplify differently: \( 8 \times 5 = 40 \). \( 40 \times 2.25 = 90 \). \( 90 \times 22 = 1980 \). \( W = 1980 \times 10^{-7} \) Joules. \( W = 198 \times 10^{-6} \) J. \( W = 198 \, \mu\text{J} \).
Step 4: Final Answer:

The work done is \( 198 \, \mu\text{J} \).