# Stoke's Law

## Problems from IIT JEE

**Problem (IIT JEE 2004): **
A small sphere falls from rest in a viscous liquid. Due to friction, heat is produced. Find the relation between the rate of production of heat and the radius of the sphere at terminal velocity.

**Solution: **
The forces acting on the sphere are its weight $\frac{4}{3}\pi r^3 \rho g$ downwards, buoyancy force $\frac{4}{3}\pi r^3 \sigma g$ upwards, and viscous force $6\pi\eta r v$ upwards. The sphere attains the terminal velocity $v_t$ when the resultant force on it is zero i.e.,
\begin{align}
\label{ilb:eqn:1}
\tfrac{4}{3}\pi r^3 \rho g=\tfrac{4}{3}\pi r^3 \sigma g+6\pi\eta r v_t.
\end{align}
Solve above equation to get the terminal velocity,
\begin{align}
v_t=\frac{2r^2(\rho-\sigma)g}{9\eta}.
\end{align}
The rate of heat generation is equal to the rate of work done by the viscous force which, in turn, is equal to its power. Thus,
\begin{align}
\frac{\mathrm{d}Q}{\mathrm{d}t}=(6\pi\eta rv_t) v_t=\frac{8\pi(\rho-\sigma)^2 g^2 r^5}{27\eta}.\nonumber
\end{align}