IIT JEE Physics (1978 to 2018: 41 Years) Topic-wise Complete Solutions

# 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}