Physics - Fluid Question with Solution | TestHub

Work done, $\mathrm{W}=\mathrm{S}\times \Delta \mathrm{A}$ Where, $\Delta \mathrm{A}$ is a change in surface area. Since, the radius of the big drop be $\mathrm{R}$ ${\mathrm{A}}_{\text{initial }}=4\pi {\mathrm{R}}^2$ Let the radius of the small drops be $\mathrm{r}$ ${\mathrm{A}}_{\text{final }}=8\times 4\pi {\mathrm{r}}^2$ The volume is constant $$\begin{array}{ll}\therefore & \frac{4}{3}\pi {\mathrm{R}}^3=8\times \frac{4}{3}\pi {\mathrm{r}}^3\\ \therefore & {\mathrm{R}}^3=8{\mathrm{r}}^3\\ \therefore & \mathrm{R}=2\mathrm{r}\end{array}$$ Substituting the values in the formula. $$\begin{array}{rl}& \mathrm{W}=\mathrm{S}\times \Delta \mathrm{A}\\ & \mathrm{W}=\mathrm{S}\left({\mathrm{A}}_{\text{final }}-{\mathrm{A}}_{\text{initial }}\right)\end{array}$$ $\begin{array}{rl}& \mathrm{W}=\mathrm{S}\left(8\times 4\pi {\mathrm{r}}^2-4\pi {\mathrm{R}}^2\right)\\ & \mathrm{W}=\mathrm{S}\left(32\pi {\left(\frac{\mathrm{R}}{2}\right)}^2-4\pi {\mathrm{R}}^2\right)\\ & \mathrm{W}=\mathrm{S}\left(8\pi {\mathrm{R}}^2-4\pi {\mathrm{R}}^2\right)\\ & \mathrm{W}=4\pi {\mathrm{R}}^2\mathrm{S}\end{array}$