Work done in increasing the size of a soap bubble from a radius of 3cm to 5cm is nearly (Surface tension of soap solution `=0.03Nm^-1`)

Options

A. ` 4 pi m J `
B. ` 0.2 pi m J `
C. `2 pi m J `
D. `0.4 pi mJ `

Answer

D

Explanation

Here, surface tension, `S=0.03Nm^(-1),r_(1)=3cm=3xx10^(-2)m,r_(2)=5cm=5xx10^(-2)m`
Since bubble has two surfaces,
Initial surface area of the bubble `=2xxpir_(1)^(2)=2xx4pixx(3xx10^(-2))^(2)=72pixx10^(-4)m^(2)`
Final surface area of the bubble `=2xx4pir_(2)^(2)=2xx4pi(5xx10^(-2))=200pixx10^(-4)m^(2)`
Increase in surface area `=200pixx10^(-4)-72pixx10^(-4)=128pixx10^(-4)`
`therefore" ""Work done"=Sxx"increase in surface area"=0.03xx128xxpixx10^(-4)=3.84pixx10^(-4)`
`4pixx10^(-4)J=0.4pimJ`