The work done against surface tension in blowing a soap bubble from a radius of 5 cm to 15 cm is (surface tension of soap solution is `30 xx 10^(-3) Nm^(-1)`)

Options

A. `2.4pi`
B. `4.8pi`
C. `1.2pi`
D. `7.2pi`

Answer

B

Explanation

`{:("Before blowing"),("surface area):}}=2xx4Pr_(1)^(2)`
`{:("After blowing"),("surface area"):}}=2xx4Pr_(2)^(2)`
Work done = Difference in area `xxT`
`W=8pi(Pr_(2)^(2)-Pr_(1)^(2))T`
`=8pixx(15^(2)-5^(2))xx10^(-4)xx30xx10^(-3)`
`=8pixx[(20)xx(10)]xx30xx10^(-7)`
`=8pixx200xx30xx10^(-7)`
`=4.8pixx10^(-3)J`
`W=4.8pimJ`