Question

The terminal velocity of a liquid drop of radius $r$ falling through air is $v$. If two such drops are combined to form a bigger drop, the terminal velocity with which the bigger drop falls through air is (ignore any buoyant force due to air):
(A) \[\sqrt 2 v\]
(B) \[2\nu \]
(C) \[\sqrt[3]{4}\nu \]
(D) \[\sqrt[3]{2}\nu \]

Answer 1

Hint: The measure of resistance in the gradual deformation due to the shear or stress is known as viscosity. We can solve this problem by finding the volume of the sphere.

Formula used: To solve this type of problems we use the following formula.
Terminal velocity ${v_{ter\min al}} = \dfrac{{2\left( {\rho - \sigma } \right)g{r^2}}}{{9\mu }}$ ; where, $\rho $ is density of sphere, $\sigma $ is the density of water, $\mu $ is viscosity.
Volume of sphere $V = \dfrac{{4\pi {r^3}}}{3}$; where r is radius of sphere.

Complete step-by-step answer:
It is given in the question that two drops are falling with terminal velocity v. Now these two drops combine to form a bigger drop, we have to find the terminal velocity of bigger drops ignoring the buoyant force due to air.
We know the volume of a sphere is given by the formula $V = \dfrac{{4\pi {r^3}}}{3}$.
When the two drops combine the volume of bigger drops will be twice of $V = \dfrac{{4\pi {r^3}}}{3}$
$ \Rightarrow {V_{Bigger - drop}} = 2 \times \dfrac{{4\pi {r^3}}}{3}$
Let us rewrite the above expression.
$ \Rightarrow {V_{Bigger - drop}} = \dfrac{{4\pi {{\left( {{2^{\dfrac{1}{3}}}r} \right)}^3}}}{3}$ (1)
Therefore, from equation (1) we can conclude that the radius of the bigger sphere becomes ${2^{\dfrac{1}{3}}}$ times of the original spheres radius.
Now let use the formula of terminal velocity which is ${v_{ter\min al}} = \dfrac{{2\left( {\rho - \sigma } \right)g{r^2}}}{{9\mu }}$.
From this formula let us write the relation between terminal velocity and radius of sphere.
$ \Rightarrow {v_{ter\min al}} \propto {r^2}$
Hence, the terminal velocity of the bigger sphere is given below.
$ \Rightarrow {v_{ter\min al}} \Rightarrow {\left( {{2^{\dfrac{1}{3}}}} \right)^2}$
$ \Rightarrow {2^{\dfrac{1}{3}}}$
$\therefore \sqrt[3]{4}$

Hence, option (C) $\sqrt[3]{4}$ is the correct option.

Note: When a sphere falls its velocity increases until it reaches steady velocity known as terminal velocity.
At terminal velocity, the drag force is balanced by gravitational velocity.
The gathering of the information about a given material using the viscosity helps the manufactures to know the behavior of those materials in the real world.