Question

The surface tension of soap solution is 3.5 × 10–2 Nm–1. The amount of work done required to increase the radius of soap bubble from 10 cm to 20 cm is________× 10–4 J. (take π = 22/7)

Answer 1

Correct Answer: 264

Given:

• Surface tension (\( S \)) = \( 3.5 \times 10^{-2} \, \text{N/m} \)
• Initial radius (\( r_i \)) = \( 10 \, \text{cm} = 0.1 \, \text{m} \)
• Final radius (\( r_f \)) = \( 20 \, \text{cm} = 0.2 \, \text{m} \)

Step 1: Formula for Work Done

The work done (\( W \)) in increasing the surface area of a soap bubble is given by:

\[ W = \Delta U = S \times \Delta A, \]

where:

• \( S \): Surface tension
• \( \Delta A \): Change in surface area.

Step 2: Calculate the Change in Surface Area

A soap bubble has two surfaces (inner and outer). The initial surface area (\( A_i \)) and final surface area (\( A_f \)) are given by:

\[ A_i = 2 \times 4\pi r_i^2, \quad A_f = 2 \times 4\pi r_f^2. \]

The change in surface area (\( \Delta A \)) is:

\[ \Delta A = A_f - A_i = 2 \times 4\pi (r_f^2 - r_i^2). \]

Substitute \( r_i = 0.1 \, \text{m} \) and \( r_f = 0.2 \, \text{m} \):

\[ \Delta A = 8\pi \left( r_f^2 - r_i^2 \right) = 8\pi \left( (0.2)^2 - (0.1)^2 \right). \]

Simplify:

\[ \Delta A = 8\pi \left( 0.04 - 0.01 \right) = 8\pi \times 0.03. \]

Step 3: Calculate the Work Done

Substitute \( \Delta A = 8\pi \times 0.03 \) and \( S = 3.5 \times 10^{-2} \, \text{N/m} \) into the formula for work done:

\[ W = S \times \Delta A = (3.5 \times 10^{-2}) \times 8 \pi \times 0.03. \]

Using \( \pi \approx \frac{22}{7} \):

\[ W = (3.5 \times 10^{-2}) \times 8 \times \frac{22}{7} \times 0.03. \]

Simplify step by step:

\[ W = 3.5 \times 8 \times 22 \times 3 \times 10^{-2} \times 10^{-2} \div 7. \]

\[ W = \frac{3.5 \times 8 \times 22 \times 3}{7} \times 10^{-4}. \]

\[ W = 264 \times 10^{-4} \, \text{J}. \]

Final Answer:

The work done is \( 264 \times 10^{-4} \, \text{J} \).