A small spherical drop falls from rest in viscous liquid. Due to the friction, heat is produced. The correct relation between the rate of production and the radius of the spherical drop at terminal velocity will be
\[
  (A)\dfrac{{dH}}{{dt}} \propto \dfrac{1}{{{r^5}}} \\
  (B)\dfrac{{dH}}{{dt}} \propto {r^4} \\
  (C)\dfrac{{dH}}{{dt}} \propto \dfrac{1}{{{r^4}}} \\
  (D)\dfrac{{dH}}{{dt}} \propto {r^5} \\
 \]

Hints:-
Heat (\[H\]) is the form of energy and rate of production of heat (\[\dfrac{{dH}}{{dt}}\]) is equal to the power in that process \[power(P) = force(f) \times velocity(v)\]. Terminal velocity is depends on the force, \[f = 6\pi \eta v\]

Complete step by step solution:-
Here we need to find the relation between rate of production of heat (\[\dfrac{{dH}}{{dt}}\]) and radius of sphere \[(r)\].
Here, we use a useful relation, Rate of production of heat\[ = \] power produced by the forces in this process.
Ie, \[\dfrac{{dH}}{{dt}} = P\]
But, \[power(P) = force(f) \times velocity(v) = {F_v} \times v\]
Where \[{F_v}\] is the viscous force. And \[v\]is the terminal velocity .
Viscous force is given by \[{F_v} = 6\pi \eta v\]
\[\eta \]is the coefficient of viscosity. It is constant for a particular liquid.
So, \[{F_v} \propto v\]
\[\dfrac{{dH}}{{dt}} = P = {F_v}v\]
\[
  \dfrac{{dH}}{{dt}} \propto v \times v \\
  \dfrac{{dH}}{{dt}} \propto {v^2} \\
 \]
Now, the terminal velocity is given by \[v = \dfrac{2}{9}{r^2}\dfrac{{({D_s} - {D_l})}}{\eta }g\]
\[g\]:- The acceleration due to gravity.
\[{D_s}\]:- The density of material of drop.
\[{D_l}\]:- The density of liquid.
These three terms are constants in this case so,
\[v \propto {r^2}\]
But,
\[
  \dfrac{{dH}}{{dt}} \propto {v^2} \\
  \dfrac{{dH}}{{dt}} \propto {({r^2})^2} \\
 \]
\[\dfrac{{dH}}{{dt}} \propto {r^4}\]
So the answer is (B) \[\dfrac{{dH}}{{dt}} \propto {r^4}\]

Note:- The heat is measured by the unit Joule \[\left( J \right)\] and power in Watt \[\left( W \right).\] Speed of rain drop is constant after some time because of this viscous force.