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)
Correct Answer: 264
Solution and Explanation
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 \( 264 \times 10^{-4} \, \text{J} \).
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