The radii of two mercury drops are $R_1$ and $R_2$. Under isothermal conditions, a single drop of radius $R$…

The radii of two mercury drops are $R_1$ and $R_2$. Under isothermal conditions, a single drop of radius $R$ is formed from them. The relations between $R, R_1$ and $R_2$ is
  1. $R^2=R_1^2+R_2^2$
  2. $R=R_1+R_2$
  3. $R=\frac{R_1+R_2}{2}$
  4. $R^3=R_1^3+R_2^3$

Solution

Total volume remains same, $\begin{aligned} & \frac{4}{3} \pi R^3=\frac{4}{3} \pi R_1^3+\frac{4}{3} \pi R_2^3 \\ & R^3=R_1^3+R_2^3 \end{aligned}$

Asked in: MHT CET 2022 (11 Aug Shift 1)

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