The amount of work done in blowing a soap bubble such that its diameter increases from d to D is (T=surface tension of the solution)

Options

A. `pi (D^(2) - d^(2))S`
B. `2pi (D^(2) - d^(2))S`
C. `4pi (D^(2) - d^(2))S`
D. `8pi (D^(2) - d^(2))S`

Answer

B

Explanation

Change in surface area `=2 xx4pi [(D//2)^(2)-(d//2)^(2)]`
`=2pi (D^(2)-d^(2))`
`:.` Work done = Surface tension `xx` change in area
`=2pi S (D^(2)-d^(2))`