The work done in increasing the volume of a soap bubble of radius R and surface tension T by 700% will be

Correct Option: 24πR²T

Step-by-step Explanation:

Let the initial radius of the soap bubble be R and its surface tension be T.

Since a soap bubble has two free surfaces (an inner surface and an outer surface), the initial total surface area is given by:
$A_1=2\times 4\pi R^2=8\pi R^2$
The initial volume of the bubble is:
$V_1=\frac{4}{3}\pi R^3$

According to the problem, the volume of the bubble is increased by 700%.
Thus, the final volume $V_2$ becomes:
$V_2=V_1+700\%\text{of}{V}_1$
$V_2=V_1+7V_1=8V_1$

Let $R^{\prime }$ be the new radius of the bubble. We can write the relation for the new volume as:
$\frac{4}{3}\pi {\left(R^{\prime }\right)}^3=8\times \frac{4}{3}\pi R^3$
Taking the cube root on both sides, we get:
$R^{\prime }=2R$

Now, the final surface area of the bubble $A_2$ is:
$A_2=2\times 4\pi {\left(R^{\prime }\right)}^2$
Substituting $R^{\prime }=2R$ into the equation:
$A_2=8\pi {\left(2R\right)}^2=8\pi \left(4R^2\right)=32\pi R^2$

The increase in the surface area ($\Delta A$) is:
$\Delta A=A_2-A_1$
$\Delta A=32\pi R^2-8\pi R^2=24\pi R^2$

The work done ($W$) in increasing the volume of the soap bubble is equal to the surface tension multiplied by the increase in surface area:
$W=T\times \Delta A$
$W=24\pi R^{2}T$