Question

$n$ drops of water, each of radius $2mm$, fall through air at a terminal velocity of $8cm{{s}^{-1}}$. If they coalesce to form a single drop, then the terminal velocity of the combined drop is $32cm{{s}^{-1}}$. The value of $n$ will then be,

Answer 1

Hint: When $n$ drops of water are merged into form a single water droplet, then the volume of that single droplet will be equivalent to the sum of the volume of each of the tiny droplets. Here we can get a relation between the both the radius of single as well as tiny droplets. The terminal velocity is directly proportional to the square of the radius of the droplet. This will be used to arrive at the value of $n$. These all may help you to solve this question.

Complete step by step answer:
First of all we can see that, when $n$ drops of water are merged into form a single water droplet, then the volume of that single droplet will be equivalent to the sum of the volume of each of the tiny droplets.
That is we can write that,
$\dfrac{4}{3}\pi {{R}^{3}}=n\times \dfrac{4}{3}\pi {{r}^{3}}$
Where $R$be the radius of the single droplet which is formed by the merging of tiny droplets and $r$ be the radius of each tiny droplet of water.
That is we can simplify this as,
${{R}^{3}}=n\times {{r}^{3}}$
This can be written as,
$R={{n}^{\dfrac{1}{3}}}\times r$
As the radius of tiny droplets are given as,
$r=2mm$
Therefore,
$R=2{{n}^{\dfrac{1}{3}}}mm$
As we all know, the terminal velocity is directly proportional to the square of the radius of the body. Therefore we can write that,
${{v}_{0}}\propto {{r}^{2}}$
Where ${{v}_{0}}$ be the terminal velocity of the tiny droplets.
And also
${{v}_{0}}^{\prime }\propto {{R}^{2}}$
Where ${{v}_{0}}^{\prime }$ be the terminal velocity of the single droplet.
Comparing both of these equation will give,
$\dfrac{{{v}_{0}}^{\prime }}{{{v}_{0}}}=\dfrac{{{R}^{2}}}{{{r}^{2}}}$
Substituting the values in it,
$\dfrac{{{v}_{0}}^{\prime }}{{{v}_{0}}}=\dfrac{4{{n}^{\dfrac{2}{3}}}}{4}$
As the terminal velocity of the tiny droplets are,
${{v}_{0}}=8cm{{s}^{-1}}$
And the terminal velocity of the single droplet is,
${{v}_{0}}^{\prime }=32cm{{s}^{-1}}$
Therefore we can substitute these values in the equation,
$\begin{align}
  & \dfrac{32}{8}={{n}^{\dfrac{2}{3}}} \\
 & n={{\left( 4 \right)}^{\dfrac{3}{2}}}=\sqrt{64}=8 \\
 & n=8 \\
\end{align}$
Therefore the correct answer is obtained.

Note:
Terminal velocity is defined as the maximum speed of the body which is moving inside the liquid medium. It happens when the sum of the drag force and the buoyancy is equivalent to the downward force of gravity experienced on the body.
As the resultant force on the body is zero, the body will be having zero acceleration.