Question

Find the work done in blowing a soap bubble of surface tension 0.06Nm-1 from 2cm radius to 5cm radius.

Answer 1

hope this will help you....

Answer 2

Dear Student,

◆ Answer -

W = 3.167×10^-3 J

● Explanation -

# Given -

T = 0.06 N/m

r1 = 2 cm = 0.02 m

r2 = 5 cm = 0.05 m

# Solution -

The work done in blowing a soap bubble is -

W = T × ∆A

W = T × 8π(r2^2-r1^2)

W = 0.06 × 8×3.142 × (0.05^2 - 0.02^2)

W = 0.003167 J

W = 3.167×10^-3 J

Hence, work done in blowing a soap bubble is 3.167×10^-3 J