A soap bubble (surface tension`= 30 xx 10^(-3) N.m^(-1)`) has radius 2 cm. The work done in doubling the radius is :

Options

A. 0
B. `1.1305xx10^(-4)J`
C. `12.261xx10^(-4)J`
D. `4.403xx10^(-4)J`

Answer

D

Explanation

Work done = Surface tension `xx` Increase in area.
For soap bubble increase in area
`=2[4pi(2r)^(2)-4pir^(2)]=24pir^(2)`
`therefore` Work done = T`xx24pir^(2)`
`=30xx10^(-3)xx24xx22/7xx(2xx10^(-2))^(2)`
`=4.043xx10^(-4)J`
Hence correct choice is (d).