Question

A soap bubble was slowly enlarged from a radius of 4cm to 6cm. The amount of work necessary for enlargement was 1.5 x 10^-4 joules. Calculate the surface tension of the soap bubble.​

Answer 1

`T=3*10^-3 N/m`

Step-by-step explanation:

From the question we are told that:

Radius :

`R_1=4=>0.04\\\\R_2=6=>0.06`

Work
`W=1.5 * 10^(-4)`

Generally the equation for Work done is mathematically given by

`W=TdA`

Where

`dA=A_2-A_1\\\\dA=(2 \pi r_2^2)(2 \pi r_1^2)`

`dA=8 \pi*(r_2^2-r_1^2)\\\\dA=8*3.142*(0.06^2-0.04^2)`

`dA=0.050m^2`

Therefore

`W=TdA`

`T=(1.5 * 10^(-4))/(0.05m^2)`

`T=3*10^-3 N/m`