Eight spherical rain drops of the same mass and radius are falling down with a terminal speed of $6 \mathrm{~cm}-\mathrm{s}^{-1}$. If they coalesce to form one big drop, what will be the terminal speed of bigger drop? (Neglect the buoyancy of the air)

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Aicharya · Accepted Answer

Let now radius of big drop is $R$. Then, $\begin{aligned} \frac{4}{3} \pi R^3 & =\frac{4}{3} \times \pi r^3 \cdot 8 \\ R & =2 r \end{aligned}$ where $r$ is radius of small drops. Now, terminal velocity of drop in liquid. $v_e=\frac{2}{9} \times \frac{r^2}{\eta}(\rho-\sigma) g$ where $\eta$ is coefficient of viscosity and $\rho$ is density of drop $\sigma$ is density of liquid. Terminal speed drop is $6 \mathrm{~cm} \mathrm{~s}^{-1}$

Let terminal velocity becomes $v^{\prime}$ after coalesce, then

Dividing Eq. (i) by Eq. (ii), we get $\begin{aligned} \frac{6}{v^{\prime}}=\frac{\frac{2}{9} \frac{r^2}{\eta}(\rho-\sigma) g}{\frac{2}{9} \frac{R^2}{\eta}(\rho-\sigma) g} \\ \text { or } \quad \frac{6}{v^{\prime}}=\frac{r^2}{(2 r)^2} \\ \text { or } \quad v^{\prime}=24 \mathrm{~cm} \mathrm{~s}^{-1} \\ \end{aligned}$

Eight spherical rain drops of the same mass and radius are falling down with a terminal speed of $6…

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