Two drops of the same radius are falling through air with a steady velocity of \[5\] cm per sec. If the two drops coalesce, the terminal velocity would be
A) \[10\] cm per sec
B) \[2.5\] cm per sec
C) \[5 \times {(4)^{\dfrac{1}{3}}}\] cm per sec
D) \[5\sqrt 3 \] cm per sec

Hint:When two drops coalesce to one single drop, the volume must be conserved. Find the volume of the drop formed and thus find the radius of the drop formed. The terminal velocity \[v\] of each individual drop is given as:

\[v = \dfrac{{\left( {\dfrac{2}{9}} \right){r^2}g\left( {\rho - \sigma } \right)}}{\eta }\]
Here, \[r\] is the radius of the drop formed
\[g\] is acceleration due to gravity.
\[\rho \] is the density of the sphere
\[\sigma \] is the density of fluid
\[\eta \] is the coefficient of viscosity.

Complete step by step solution:
In order to find the terminal velocity, we need to find the radius of the newly formed drop. As there is no loss in volume thus the initial and final volume must be equal.

Let \[{r_i}\] be the initial radius of each individual drop thus, the initial volume \[{V_i}\] of two drops will be:
\[{V_i} = 2 \times \dfrac{4}{3}\left( {\pi {r_i}^3} \right)\]
Let the final radius of the drop formed by \[R\] , its volume \[V\] will be given as:
\[V = \dfrac{4}{3}\left( {\pi {R^3}} \right)\]
As, the volumes must be equal thus we have:
\[{v_i} = V\]
\[ \Rightarrow 2 \times \dfrac{4}{3}\left( {\pi {r_i}^3} \right) = \dfrac{4}{3}\left( {\pi {R^3}} \right)\]
\[ \Rightarrow R = {2^{\dfrac{1}{3}}}{r_i}\]
--equation \[1\]

The terminal velocity of the initial drop is given as:
\[{v_i} = \dfrac{{\left( {\dfrac{2}{9}} \right){r_i}^2g\left( {\rho - \sigma } \right)}}{\eta } = 5\]

--equation $2$

The final terminal velocity \[{v_f}\] will be:
\[{v_f} = \dfrac{{\left( {\dfrac{2}{9}} \right){R^2}g\left( {\rho - \sigma } \right)}}{\eta }\]
Substituting \[R = {2^{\dfrac{1}{3}}}{r_i}\] , we get
\[{v_f} = \dfrac{{\left( {\dfrac{2}{9}} \right){{\left( {{2^{\dfrac{1}{3}}}{r_i}} \right)}^2}g\left( {\rho -
\sigma } \right)}}{\eta }\]

\[ \Rightarrow {v_f} = \dfrac{{\left( {\dfrac{2}{9}} \right){r_i}^2g\left( {\rho - \sigma } \right)}}{\eta }
\times {2^{\dfrac{2}{3}}}\]
\[ \Rightarrow {v_f} = \dfrac{{\left( {\dfrac{2}{9}} \right){r_i}^2g\left( {\rho - \sigma } \right)}}{\eta }
\times {4^{\dfrac{1}{3}}}\]
Using equation $2$ we can have:
\[{v_f} = 5 \times {4^{\dfrac{1}{3}}}\]
Therefore, the final terminal velocity will be \[5 \times {(4)^{\dfrac{1}{3}}}\] cm per sec.

Option C is the correct option.

Note:The volume remains constant as there is no loss in the volume. When drop coalesce there will be a difference in the total initial energy and the final energy. Also, when two drops coalesce into one drop, the radius of the final drop is not twice the radius of initial drop rather it is lesser than twice the radius.