Let ' $\mathrm{R}_1$ ' and ' $\mathrm{R}_2$ ' are radii of two mercury drops. A big mercury drop is formed…

Let ' $\mathrm{R}_1$ ' and ' $\mathrm{R}_2$ ' are radii of two mercury drops. A big mercury drop is formed from then under isothermal conditions. The radius of the resultant drop is
  1. $\sqrt{\mathrm{R}_1^2+\mathrm{R}_2^2}$
  2. $\left(\mathrm{R}_1^3+\mathrm{R}_2^3\right)^{\frac{1}{3}}$
  3. $\sqrt{\mathrm{R}_1^2-\mathrm{R}_2^2}$
  4. $\frac{R_1+R_2}{2}$

Solution

The volume of the bigger drop will be equal to the sum of the volumes of the smaller drops $\begin{aligned} & \frac{4}{3} \pi \mathrm{R}^3=\frac{4}{3} \pi \mathrm{R}_1^3+\frac{4}{3} \pi \mathrm{R}_2^3 \\ & \therefore \mathrm{R}=\left(\mathrm{R}_1^3+\mathrm{R}_2^3\right)^{1 / 3} \end{aligned}$

Asked in: MHT CET 2021 (20 Sep Shift 2)

Practice more Mechanical Properties of Fluids questions on Aicharya