Question

Calculate the work done against surface tension in blowing a soap bubble from a radius of 10 cm to 20 cm, if the surface tension of soap solution in $25\times {10}^{-3}$ N/m.

Answer 1

$T=25\times {10}^{-3}N/m$

Initial radius$=10㎝$

Final radius$=20㎝$

Final surface area $=△A=2\times (4\pi r_f^2-4\pi r_i^2)$

Work done$=T\times △A$

$△A=2\times [4\pi {\left(\frac{20}{100}\right)}^2-4\pi {\left(\frac{10}{100}\right)}^2]$

$=\frac{2\times 4\pi }{{\left(10\right)}^2}[4-1]=\frac{24\pi }{{\left(10\right)}^2}$

$W=T\times △A$

$W=\frac{25\times {10}^{-3}\times 24\pi }{{\left(10\right)}^2}$

$W=1885.71\times {10}^{-5}$

$W=0.018857J$