Let ${ }^{\prime} \mathrm{R}_{1}$ ' and ${ }^{\prime} \mathrm{R}_{2}$ ' are radii of two mercury drops. A…

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

Solution

The Volume of the bigger drop is equal to the sum of the volumes of the smaller drops. $\begin{aligned} & \frac{4}{3} \pi R^{3}=\frac{4}{3} \pi R_{1}^{3}+\frac{4}{3} \pi R_{2}^{3} \\ \therefore \quad & R=\sqrt[3]{R_{1}^{3}+R_{2}^{3}} \end{aligned}$

Asked in: MHT CET 2020 (14 Oct Shift 2)

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