Find the work done in blowing a soap bubble of surface tension `0.06 Nm^(-1)` from 2 cm radius to 5 cm radius.

Options

A. 3.1mJ
B. 1.25 mJ
C. 2.51 mJ
D. 4.55 mJ

Answer

A

Explanation

Here, `S=0.6 N m^(-1), r_(1)=2 cm = 2 xx 10^(-2) m`,
`r_(2)=5 cm =5 xx 10^(-2)` m
Since bubble has two surfaces, initial surfaces area of he bubble `=2 xx 4pir_(1)^(2)=2xx4pixx(2xx10^(-2))^(2)`
`=32pi xx 10^(-4) m^(2)`
Final surface area of the bubble
`2xx4pir_(2)^(2)=2xx4pi(5xx10^(-2))^(2)`
`=200 pi xx 10^(-4) m^(2)`
Increase in surface area=`200 pi xx10^(-4) -32 pi xx 10^(-4)`
`=168pixx10^(-4) m^(2)`
`therefore` Work done=surface tension `xx` increase in surface area
`=0.06xx168 pi xx 10^(-4)`
`=3.12xx10^(-3) J =3.12 mJ`