Work done in increasing the size of a soap bubble from a radius of 3 cm to 5 cm is nearly (Surface tension of soap solution = 0.03 N m^-1 ): - Physics MCQ FIITJEE AIEEE 2011 (Set Q) 2026 | ExamTest.live

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Work done in increasing the size of a soap bubble from a radius of 3 cm to 5 cm is nearly (Surface tension of soap solution = 0.03 N m^{- 1}):

Solution & Step-by-step Explanation

For a soap bubble (two surfaces), the work done is: W = T Δ A = T \times 2 \times 4 π (r_{2}^{2} - r_{1}^{2}) W = 8 π T (r_{2}^{2} - r_{1}^{2}) Given r_{1} = 3 \times 1 0^{- 2} m, r_{2} = 5 \times 1 0^{- 2} m, and T = 0.03 N/m : W = 8 π (0.03) [(5 \times 1 0^{- 2})^{2} - (3 \times 1 0^{- 2})^{2}] W = 0.24 π (25 \times 1 0^{- 4} - 9 \times 1 0^{- 4}) W = 0.24 π (16 \times 1 0^{- 4}) W = 3.84 π \times 1 0^{- 4} J = 0.384 π mJ \approx 0.4 π mJ

Try it yourself before checking the explanation above.

Work done in increasing the size of a soap bubble from a radius of 3 cm to 5 cm is nearly (Surface tension of soap solution = 0.03 N m^{- 1}):

0.2 π mJ

2 π mJ

0.4 π mJ

4 π mJ

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