The work done in blowina siap bubble of surface tension `0.06 N,^(-1)` from 2 cm radius to 5 cm radiu is

Options

A. `0.004168 J`
B. `0.003168 J`
C. `0.003158 J`
D. `0.004158 J`

Explanation

As given, `s=0.06 Nm^(-1)`
`r_(1)=2 cm =0.02 m,r_(2)=5 cm =0.05`
Since, bubble has two surface
Initial surface areqa of the bubble `=2xx4pir_(2)^(2)`
`=2xx4pixx(0.2)^(2)`
`=32pixx10^(-4)m^(2)`
Final surface area of the bubble `=2xx4pir_(2)^(2)`
`=2xx4pixx(0.05)^(2)`
`=200pixx10^(-4)m^(2)`
So, work done `=s xx` increase in surface
`=0.06xx(200pi10^(-4)-32pixx10^(-4))`
`0.06xx168pi10^(-4)`
`0.0031168 J`