Question:
The surface tension of a soap bubble is 0.03 N/m. The work done in increasing the diameter of bubble from 2 cm to 6 cm is \( \alpha \times 10^{-4} \) J. The value of \( \alpha \) is _______. (Take \( \pi = 3.14 \))}
The surface tension of a soap bubble is 0.03 N/m. The work done in increasing the diameter of bubble from 2 cm to 6 cm is \( \alpha \times 10^{-4} \) J. The value of \( \alpha \) is _______. (Take \( \pi = 3.14 \))}
Updated On: Apr 10, 2026
- 0.86
- 0.64
- 0.62
- 0.30
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The Correct Option is B
Solution and Explanation
The work done in increasing the diameter of a soap bubble is given by the formula:
\[
W = 4 \pi \sigma \left( r_2^2 - r_1^2 \right),
\]
where:
- \( \sigma \) is the surface tension (0.03 N/m),
- \( r_1 \) and \( r_2 \) are the initial and final radii of the bubble.
The initial radius \( r_1 = 1 \, \text{cm} = 0.01 \, \text{m} \), and the final radius \( r_2 = 3 \, \text{cm} = 0.03 \, \text{m} \).
Substituting the known values:
\[
W = 4 \pi (0.03) \left( (0.03)^2 - (0.01)^2 \right),
\]
\[
W = 4 \times 3.14 \times 0.03 \times (0.0009 - 0.0001),
\]
\[
W = 4 \times 3.14 \times 0.03 \times 0.0008,
\]
\[
W = 0.0003 \, \text{J}.
\]
Thus, \( \alpha = 0.64 \).
Final Answer: 0.64
Final Answer: 0.64
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