Question

A small sphere of radius $'r'$ falls from rest in a viscous liquid. As a result, heat is produced due to viscous force. The rate of production of heat when the sphere attains its terminal velocity, is proportional to

• A. $r^4$
• B. $r^3$
• C. $r^5$
• D. $r^2$

Answer 1

The Correct Option is C

The question is about the rate of production of heat when a small sphere reaches its terminal velocity in a viscous liquid. We need to find out how this rate of heat production is proportional to the radius of the sphere.

To solve this, let's consider the physics of a sphere falling in a viscous liquid:

1. Terminal Velocity: When the sphere falls in the viscous liquid, it eventually reaches a constant speed known as terminal velocity. This occurs when the gravitational force mg is balanced by the viscous drag and buoyant force. For a sphere of radius r, the viscous drag force F_d is given by Stokes' Law:
   F_d = 6 \pi \eta r v,
   where \eta is the viscosity of the liquid and v is the terminal velocity.
2. Gravitational Force: The gravitational force acting on the sphere is:
   F_g = \frac{4}{3} \pi r^3 \rho g,
   where \rho is the density of the sphere and g is the acceleration due to gravity.
3. Terminal Velocity Calculation: At terminal velocity, F_g = F_d. Solving this, we get:
   \frac{4}{3} \pi r^3 \rho g = 6 \pi \eta r v
   Simplifying, v = \frac{2}{9} \frac{r^2 \rho g}{\eta}
4. Rate of Production of Heat: The rate of heat production (power) P due to viscous force is the work done per unit time, which is the product of the viscous force and terminal velocity:
   P = F_d v = (6 \pi \eta r v) v = 6 \pi \eta r v^2
5. Substitute the expression for v:
   P = 6 \pi \eta r \left(\frac{2}{9} \frac{r^2 \rho g}{\eta}\right)^2
   Simplifying, P \propto r^5

Therefore, the rate of production of heat is proportional to r^5 as per the options given.

Hence, the correct answer is r^5.