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. r³
• B. r⁵
• C. r²
• D. r⁴

Answer 1

The Correct Option is B

Let's analyze the problem of a small sphere falling through a viscous liquid. The sphere eventually reaches a terminal velocity due to the balance of forces. The heat produced as the sphere falls at this velocity is a result of the work done against the viscous force.

1. When the sphere moves through a viscous liquid, it experiences a viscous force given by Stokes' law, which is: F = 6 \pi \eta r v_t, where:
   • \eta is the viscosity of the liquid,
   • r is the radius of the sphere,
   • v_t is the terminal velocity of the sphere.
2. The power or rate of heat production P due to viscous force is given by the work done per unit time: P = F \times v_t = 6 \pi \eta r v_t^2.
3. At terminal velocity, the viscous force balances the gravitational force minus the buoyant force. Expressing terminal velocity v_t in terms of the radius can be crucial to determining how the power is related to r.
4. From the balance of forces at terminal velocity: F_{\text{gravity}} - F_{\text{buoyant}} = F_{\text{viscous}} This simplifies to: \left( \frac{4}{3} \pi r^3 \rho g - \frac{4}{3} \pi r^3 \rho_l g \right) = 6 \pi \eta r v_t, where \rho is the density of the sphere, and \rho_l is the density of the liquid.
5. Solving for v_t gives: v_t = \frac{2}{9} \frac{r^2 (\rho - \rho_l) g}{\eta}
6. Substituting v_t into the expression for P gives: P = 6 \pi \eta r \left( \frac{2}{9} \frac{r^2 (\rho - \rho_l) g}{\eta} \right)^2
7. Simplifying this expression, we arrive at: P \propto r^5

Therefore, the rate of production of heat when the sphere attains its terminal velocity is proportional to r^5.