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

Hint: In blowing the bulb, some of the work is done for which energy is required. The relation between work done and the surface tension is used to solve this.

Formula used:
(1) Work done$ = T \times $increase in surface area
(2) Surface area of sphere $ = 4\pi {r^2}$
Where $T$ is the surface tension and r is the radius of the sphere.

Complete step by step answer:
The surface of the soap bubble possess minimum area due to surface tension. To increases the radius of soap bubble, work is to be done against the force of attraction. Which is given by
Work done, $W = T \times $increase in area of the soap bubble
Where $T$ is the surface tension.
Since, the soap bubbles is in spherical shape
So, increase in area of soap bubble
$ = 2 \times $(Final area – initial area)
$ = 2 \times \left( {4\pi r_2^2 - 4\pi r_1^2} \right)$
Where r, is the initial radius of soap bubble and ${r_2}$ is the final radius of soap bubble.
So, increase in area $ = 2 \times 4\pi \left( {r_2^2 - r_1^2} \right)$
Hence, work done $ = 2 \times T \times 4\pi \left( {r_2^2 - r_1^2} \right)$
Hence, in the question.
Surface tension, $T = 25 \times {10^{ - 3}}N/M$
Initial radius of soap bubble, ${r_1} = 10cm$
$ = 10 \times {10^{ - 2}}m$
Final radius of soap bubble, ${r_2} = 20cm$
$ = 20 \times {10^{ - 2}}m$
So, work done $ = 25 \times {10^{ - 3}} \times 2 \times 4\pi \left[ {{{\left( {20 \times {{10}^{ - 2}}} \right)}^2} - {{\left( {10 \times {{10}^{ - 2}}} \right)}^2}} \right]$
Work done $ = 20 \times {10^{ - 3}} \times 8 \times \dfrac{{22}}{7} \times \left( {400 \times {{10}^{ - 4}} - 100 \times {{10}^{ - 4}}} \right)$
$
   = 200 \times {10^{ - 3}} \times 3.14\left( {400 - 100} \right) \times {10^{ - 4}} \\
   = 200 \times 3.14 \times {10^{ - 3}} \times 300 \times {10^{ - 4}} \\
 $
$ = 18.84 \times {10^{ - 3}}$ Joules
$ = 1.88 \times {10^{ - 2}}$Joules

Note:
Here, the area of the soap bubble is taken to be $8\pi {r^2}$i.e. $2 \times 4\pi {r^2}$ instead of only $4\pi {r^2}$. This is because a soap bubble has air both inside and outside. So, it has two free surfaces. Hence, the surface area of one soap bubble $ = 2 \times $surface area of the sphere.