Question

calculate work done in blowing soap bubble from 2 to 5 cm. Surface tension of solution is 6/100 n per meter

Answer 1

Surface tension of solution(S) = 6/100 N/m = 0.06 N/m
initial radius(r) = 2 cm = 0.02 m
final radius(R) = 5 cm = 0.05m

Work done = S× 4π(R² - r²)
⇒ W = 0.06×4×π(0.05² - 0.02²)
⇒ W = 0.24π(0.0025 - 0.0004) J
⇒ W = 0.00158 J

Work done is 0.00158 J

Answer 2

Surface tension of solution(S) = 6/100 N/m = 0.06 N/m

initial radius(r) = 2 cm = 0.02 m

final radius(R) = 5 cm = 0.05m

Work done = S× 4π(R² - r²)

⇒ W = 0.06×4×π(0.05² - 0.02²)

⇒ W = 0.24π(0.0025 - 0.0004) J

⇒ W = 0.00158 J

Work done is 0.00158 J