Question

64 spherical rain drops of equal size are falling vertically through air with a terminal velocity 1.5m$s^{-1}$. If these drops collapse to form a big spherical drop, then terminal velocity of big drop is:
A) 8 m$s^{-1}$
B) 16 m$s^{-1}$
C) 24 m$s^{-1}$
D) 32 m$s^{-1}$

Answer 1

Hint
Terminal velocity is directly proportional to the square of the radius of the spherical body.
Since mass is conserved and density remains the same, upon coalescing, the volumes will add up. Thus, the volume of the bigger drop = 64 times the volume of each small drop. Since volume is directly proportional to cube of radius, this implies that the radius of the big drop is 4 times the radius of each small drop.
Therefore, terminal velocity of the big drop is 4 square = 16 times the terminal velocity of each small drop = $1.5 \times 16 = 24m{s^{ - 1}}$

Complete step by step answer
As, it is given that
Volume of big drop = 64× volume of small drop ………………………. (1)
As the drop is of spherical in shape and the volume of sphere is = $\dfrac{4}{3}\pi {r^3}$
Let the small drop is of radius r and big drop has radius R
From equation (1), we get
$ \Rightarrow \dfrac{4}{3}\pi {R^3} = 64 \times \dfrac{4}{3}\pi {r^3}$
$ \Rightarrow R = 4r$ ……………………. (2)
As, we know that the terminal velocity of the drop is
$v = \dfrac{{2{r^2}\left( {\rho - \sigma } \right)}}{{9\eta }}$
From this equation, we observe that velocity is proportional to the square of the radius of the raindrop i.e. $v \propto {r^2}$……………………. (3)
If $v_1$ and $v_2$ are the terminal velocity of the small and big rain drop respectively.
Now, using equation (3), we get
$ \Rightarrow \dfrac{{{v_1}}}{{{v_2}}} = {\left( {\dfrac{r}{R}} \right)^2}$
Substitute the value from equation (2), we get
$ \Rightarrow \dfrac{{{v_1}}}{{{v_2}}} = {\left( {\dfrac{1}{4}} \right)^2} = \dfrac{1}{{16}}$
$ \Rightarrow {v_2} = 16{v_1}$
As the terminal velocity of the small rain drop is ${v_1} = 1.5m{s^{ - 1}}$, therefore the terminal velocity of the big rain drop is
$ \Rightarrow {v_2} = 16 \times 1.5 = 24m{s^{ - 1}}$
Hence, the terminal velocity of a big raindrop is 24m/s.
Therefore, option (C) is correct.

Note
Rain drops fall with terminal velocity due to the viscosity of air. Terminal velocity is speed of the object when net force on the object is zero. In case of rain drops due to mass of drop weight acts in downward direction and viscous force of air acts in upward direction.
Rain drop maximum speed is dependent on multiple factors which include air temperature, water drop temperature, air density and atmospheric pressure. These factors affect the falling rain drop’s Reynolds number in air along with the water surface tension and viscosity which determines shape and breakup velocity of a drop.