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Excess pressure inside soap bubble of radius 2 cm (γ = 0.025 N/m):
A2.5 Pa
B1.25 Pa
C10 Pa
D5 Pa (= 4γ/r = 4 × 0.025/0.02)
Answer & Solution
Correct answer: D. 5 Pa (= 4γ/r = 4 × 0.025/0.02)
Soap bubble has 2 surfaces (inner + outer): ΔP = 4γ/r = 4 × 0.025 / 0.02 = 5 Pa.
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