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. `0.4 pi mj`
B. `4 pi mj`
C. `0.2 pi mj`
D. `2 pi mj`

Answer

A

Explanation

Here `S=0.03 N//m, r_(1)=3cm=3xx10^(-2)m`, `r_(2)=5 cm=5xx10^(-2)m`
Work done (W) in increasing the size of soap bubble = surface tension x increase in area of both the free surface of the bubble
`W =Sxx(4pi R_(2)^(2)-4pi R_(1)^(2)) =8pi S[R_(2)^(2)-R_(1)^(2)]`
`=8pi(0.03) [(5xx10^(-2))-(3xx10^(-2))^(2)]`
`=3.84xxpixx10^(-4)j=4pixx10^(-4)j= =.4pi mj`