How much work will be done in increasing the diameter of a soap bubble from 2cm to 5cm? Surface tension solution is `3.0xx10^(-2)N//m.`

Options

A. `3.2 xx 10^(-4) J`
B. `3.9 xx 10^(-4) J`
C. `4.2 xx 10^(-4) J`
D. `4.7 xx 10^(-4) J`

Answer

B

Explanation

Given, `d_(1) = 2 cm rArr r_(1) = 1 cm`
`d_(2) = 5 cm rArr r_(2) = 2.5 cm`
`S = 3.0 xx 10^(-2) Nm^(-1)`
`W = ?`
Soap bubbles has two surfaces, Hence,
`W = S Delta A`
Here, `Delta A = 2[4pi{(2.5 xx 10^(-2))^(2) - (1.0 xx 10^(-2))^(2)}]`
`= 132 xx 10^(-2) m^(2)`
`:. W = (3.0 xx 10^(-2)) (1.32 xx 10^(-2))J`
`= 3.96 xx 10^(-4) J`