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**Hint:**As the two drops of water coalesce to form a bigger drop of water, the volume of the bigger drop will be the same as the sum of the volume of the two smaller drops so that the volume before and after the coalescing remains constant. The terminal velocity of a drop is known to be proportional to the square of the radius of the drop.

**Formula used:**

The volume of a drop is given by, $V = \dfrac{4}{3}\pi {r^3}$ where $r$ is the radius of the drop.

**Complete step by step answer:**

Step 1: Apply the conservation of the volume of the drops before and after coalescing.

Let $r$ be the radius of each smaller drop of water and $R$ be the radius of the bigger drop of water formed.

Let $V = \dfrac{4}{3}\pi {r^3}$ be the volume of each smaller drop before coalescing and $V' = \dfrac{4}{3}\pi {R^3}$ ------- (1) be the volume of the bigger drop formed after the coalescing of the smaller drops.

Then the volume of the two drops before coalescing to form the bigger drop will be

$2V = 2 \times \left( {\dfrac{4}{3}\pi {r^3}} \right)$ -------- (2)

Since the volume is conserved during the coalescing, we can equate equations (1) and (2) to get, $\dfrac{4}{3}\pi {R^3} = 2 \times \left( {\dfrac{4}{3}\pi {r^3}} \right)$

$ \Rightarrow R = {2^{\left( {\dfrac{1}{3}} \right)}}r$

$ \Rightarrow \dfrac{R}{r} = {2^{\left( {\dfrac{1}{3}} \right)}}$ --------- (3)

Equation (3) indicates the relation between the radii of the bigger drop and one smaller drop.

Step 2: Express the relation between the terminal velocity and the radius of the drop.

The terminal velocity of each smaller drops is given to be $v = \dfrac{2}{9} \times {r^2}\left( {\dfrac{{\rho - \sigma }}{\eta }} \right)g$ where $\rho $ is the density of the water, $\sigma $ is the density of air, $\eta $ is its viscosity and $g$ is the acceleration due to gravity of the drop.

Let ${v_1} = \dfrac{2}{9} \times {R^2}\left( {\dfrac{{\rho - \sigma }}{\eta }} \right)g$ be the terminal velocity of the bigger drop which is to be determined.

From the above two relations, we have $v \propto {r^2}$ and ${v_1} \propto {R^2}$ , so we obtain the ratio of the two velocities as $\dfrac{{{v_1}}}{v} = \dfrac{{{R^2}}}{{{r^2}}}$ .

$ \Rightarrow {v_1} = {\left( {\dfrac{R}{r}} \right)^2}v$ -------- (4)

Substituting equation (3) in (4) we get, ${v_1} = {2^{{{\left( {\dfrac{1}{3}} \right)}^2}}}v = {2^{\left( {\dfrac{2}{3}} \right)}}v$

$\therefore $ the new velocity is obtained to be ${v_1} = {2^{\left( {\dfrac{2}{3}} \right)}}v$ .

**Note:**We assume the drops of water to have a spherical shape. So we express the volume of the bigger drop and the two smaller drops by the formula for the volume of a sphere. The terminal velocity of a drop refers to the steady speed attained by the drop as it falls freely in any medium (here it is air) and can no longer achieve further acceleration.

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