Question

Work done in this increasing the radius of a soap bubble from 4cm to 5cm is ____ joule (given surface tension of soap water to be 25×10^-3 N/m)

Answer 1

The work done in increasing the radius is 0.5654 x 10^-3 J

Explanation:

Given data:

Surface tension of soap water "T" = 25×10^-3 N/m

Initial radius "ri" = 4 cm = 4 x 10^-2 m

Final radius "rf" = 5 cm = 5 x 10^-2 m

Work done "W" = T x ΔA

                         = T x 8π( rf^2−ri^2 )

                         = 25×10^-3 x 8 x 3.14 x 9 x 10^-4

                        = 0.5654 x 10^-3 J

Thus the work done in increasing the radius is 0.5654 x 10^-3 J

Also learn more

Explain different temperature scales

https://brainly.in/question/1038150

Answer 2

Explanation:

here is your answer!

approximation has been taken = 0.5654×10–³J