CGPET 2019 Solved Paper

Given, density of liquid drop $=\rho$ Density of floating half immersed in liquid $=\rho _{0}$ Surface tension of liquid $=s$ Now, balancing the forces acting on the drops floating $\therefore w=F_{b}+F_{s}$ where, $w$ is the weight of the liquid drop, $F_{b}$ is the Buoyant force acting on the drop and $F_{S}$ is the surface tension acting on drop. Then, $w=F_{b}+F_{s}$ or $\frac{4}{3} \pi r^{3} \rho g = \frac{2}{3} \pi r^{3} \rho_{0} g + s \times 2 \pi r$ or $\frac{4}{3} \pi r^{3} \rho g - \frac{2}{3} \pi r^{3} \rho_{0} g = s \times 2 \pi r$ or $r^{3}\left[\frac{4}{3} \pi \rho g - \frac{2}{3} \pi \rho_{0} g\right] = s \times 2 \pi r$ or $r^{2} \frac{[4 \pi \rho g - 2 \pi \rho_{0} g]}{3} = s \times 2 \pi$ or $r^{2} \frac{2 \pi g [2 p - \rho_{0}]}{3} = s \times 2 \pi$ or $r^{2} \frac{g(2 p - p_{0})}{3} = s$ or $r^{2} = \frac{3 s}{g(2 p - p_{0})}$ or $r = \sqrt{\frac{3 s}{g(2 \rho - \rho_{0})}}$. So, the radius of drop is $r = \sqrt{\frac{3 s}{g(2 \rho - \rho_{0})}}$.