Physics - Fluid Question with Solution | TestHub

Balancing the forces on drop 

Buoyant force $+$ surface tension $=\mathrm{mg}$

$\sigma \frac{V}{2}\mathrm{g}+2\pi \mathrm{RT}=\rho Vg$

$2\pi \mathrm{RT}=\frac{\left(2\rho -\sigma \right)}{2}·\frac{4}{3}\pi {\mathrm{R}}^3 g; \left[\text{As }V=\frac{4}{3}\pi {\mathrm{R}}^3\right]$

$\Rightarrow R^3=\frac{3\mathrm{T}}{\left(2\rho -\sigma \right)\mathrm{g}}\Rightarrow R=\sqrt{\frac{3\times 7.5\times {10}^{-2} \mathrm{N}-{\mathrm{m}}^{-1}}{\left(2\rho -\sigma \right)\times 10}}$

$R=\frac{3}{20\sqrt{\left(2\rho -\sigma \right)}} \mathrm{m}=\frac{15}{\sqrt{2\rho -\sigma }} \mathrm{cm}$