The work done in blowing a soap bubble of surface tension `0.60 Nm^(-1)` from 2 cm radius to 5 cm radius is

Options

A. 0.004168 J
B. 0.003168 J
C. 0.003158 J
D. 0.004568 J

Answer

B

Explanation

Given, `S=0.06 Nm^(-1) r_(1)=2` cm `=0.02` m,
`r_(2)=5` cm `=0.05 m`
Since, bubble has two surfaces.
Initial surface area of the bubble `=2 xx4 pi r_(1)^(2)`
`=2xx4pixx(0.02)^(2)=32pixx10^(-4) m^(2)`
Final surface area of the bubble `=2 xx4 pi r_(2)^(2)`
`=2xx4xxpixx(0.05)^(2)`
`=200xxpi xx10^(-4) m^(2)`
So, work done = Surface tension `xx` Increase in surface area
`=0.06xx(200xxpixx10^(-4)-32pi xx10^(-4))`
`=0.06xx168 pi xx10^(-4) =0.003168 J`