The surface tension of soap solution is $3.5 \times 10^{-2} \mathrm{~Nm}^{-1}$. The amount of work done required to increase the radius of soap bubble from $10 \mathrm{~cm}$ to $20 \mathrm{~cm}$ is _________ $\times ~10^{-4} \mathrm{~J}$.

$(\operatorname{take} \pi=22 / 7)$

To calculate the work done to increase the radius of a soap bubble, we can use the formula: <br/><br/> $W = T \Delta A$ <br/><br/> where W is the work done, T is the surface tension, and ΔA is the change in surface area. <br/><br/> For a soap bubble, we need to consider both the inner and outer surfaces, so the surface area is doubled. The surface area of a sphere is: <br/><br/> $A = 4\pi r^2$ <br/><br/> The initial surface area of the soap bubble is: <br/><br/> $A_1 = 2 \cdot 4\pi (0.1\,\mathrm{m})^2 = 8\pi (0.1\,\mathrm{m})^2$ <br/><br/> The final surface area of the soap bubble is: <br/><br/> $A_2 = 2 \cdot 4\pi (0.2\,\mathrm{m})^2 = 8\pi (0.2\,\mathrm{m})^2$ <br/><br/> The change in surface area is: <br/><br/> $\Delta A = A_2 - A_1 = 8\pi(0.2\,\mathrm{m})^2 - 8\pi(0.1\,\mathrm{m})^2$ <br/><br/> Now, we can calculate the work done: <br/><br/> $$ W = T \Delta A = (3.5 \times 10^{-2}\,\mathrm{Nm}^{-1})[8\pi(0.2\,\mathrm{m})^2 - 8\pi(0.1\,\mathrm{m})^2] $$ <br/><br/> Using the given value of π: <br/><br/> $$ W = (3.5 \times 10^{-2}\,\mathrm{Nm}^{-1})[8(22/7)(0.2\,\mathrm{m})^2 - 8(22/7)(0.1\,\mathrm{m})^2] $$ <br/><br/> $$ W = (3.5 \times 10^{-2}\,\mathrm{Nm}^{-1})[8(22/7)(0.04\,\mathrm{m^2}) - 8(22/7)(0.01\,\mathrm{m^2})] $$ <br/><br/> $W = (3.5 \times 10^{-2}\,\mathrm{Nm}^{-1})[8(22/7)(0.03\,\mathrm{m^2})]$ <br/><br/> $$ \begin{aligned} & W=2 \times 1.32 \times 10^{-2} \\\\ &W =2 \times 132 \times 10^{-4} \mathrm{~J} \\\\ & W=264 \times 10^{-4} \mathrm{~J} \end{aligned} $$ <br/><br/> The work done to increase the radius of the soap bubble is 264 × 10⁻⁴ J.