Physics - Fluid Question with Solution | TestHub

The drop is in equilibrium under the action of the following forces: Weight of the liquid, $\mathrm{W}=\mathrm{Mg}=\frac{4}{3}\pi {\mathrm{r}}^3\rho \mathrm{g}$ (downwards) Upthrust $=$ weight of the liquid displaced $$\therefore {\mathrm{F}}_{\mathrm{t}}=\frac{2}{3}\pi {\mathrm{r}}^3\mathrm{dg}$$ (upwards) Force due to surface tension, $\mathrm{F}=2\pi \mathrm{rT}$ (upwards) $$\begin{array}{rl}& \therefore \mathrm{Mg}=\mathrm{F}+{\mathrm{F}}_{\mathrm{t}}\\ & \therefore \mathrm{F}=\mathrm{Mg}-{\mathrm{F}}_{\mathrm{t}}\\ & \therefore 2\pi \mathrm{r}T=\frac{4}{3}\pi {\mathrm{r}}^3\rho g-\frac{2}{3}\pi {\mathrm{r}}^3\mathrm{dg}\\ & \therefore \mathrm{T}=\frac{2}{3}{\mathrm{r}}^2\rho g-\frac{1}{3}{\mathrm{r}}^2\mathrm{dg}\\ & ={\mathrm{r}}^{2}g\left(\frac{2}{3}\rho -\frac{1}{3}d\right)\\ & ={\mathrm{r}}^{2}g\left(\frac{2\rho -d}{3}\right)\\ & \therefore \mathrm{r}=\sqrt{\frac{3\mathrm{T}}{\mathrm{g}(2\rho -\mathrm{d})}}\\ & \therefore \mathrm{D}=\sqrt{\frac{12\mathrm{T}}{\mathrm{g}(2\rho -d)}}\end{array}$$