A small sphere falls from rest in a viscous liquid. Due to friction, heat is produced. Find the relation between the rate of produced heat and the radius of the sphere at terminal velocity.

Explanation

To find the relation between the rate of produced heat and the radius of the sphere at terminal velocity, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Forces**:
When a small sphere falls through a viscous liquid, it experiences a gravitational force downward and a viscous drag force upward. At terminal velocity (vt), these forces balance each other.

2. **Expression for Drag Force**:
The drag force (F) acting on a sphere moving through a viscous fluid is given by Stokes' law:
\[ F = 6 \pi \eta r v_t \]
where:
- \( \eta \) = dynamic viscosity of the liquid,
- \( r \) = radius of the sphere,
- \( v_t \) = terminal velocity.

3. **Finding Terminal Velocity**:
At terminal velocity, the gravitational force acting on the sphere is equal to the drag force:
\[ mg = F \]
For a sphere of radius \( r \):
\[ mg = \frac{4}{3} \pi r^3 \rho g \]
where \( \rho \) is the density of the sphere. The drag force can be expressed as:
\[ F = 6 \pi \eta r v_t \]
Setting these equal gives:
\[ \frac{4}{3} \pi r^3 \rho g = 6 \pi \eta r v_t \]

4. **Solving for Terminal Velocity**:
Rearranging the equation for \( v_t \):
\[ v_t = \frac{2}{9} \frac{r^2 (\rho - \sigma) g}{\eta} \]
where \( \sigma \) is the density of the liquid.

5. **Rate of Heat Production**:
The rate of heat produced due to viscous friction as the sphere moves through the liquid is given by:
\[ \frac{dq}{dt} = F v_t \]
Substituting the drag force:
\[ \frac{dq}{dt} = (6 \pi \eta r v_t) v_t = 6 \pi \eta r v_t^2 \]

6. **Substituting Terminal Velocity**:
Substitute the expression for \( v_t \):
\[ \frac{dq}{dt} = 6 \pi \eta r \left(\frac{2}{9} \frac{r^2 (\rho - \sigma) g}{\eta}\right)^2 \]
Simplifying this:
\[ \frac{dq}{dt} = 6 \pi \eta r \cdot \frac{4}{81} \frac{r^4 (\rho - \sigma)^2 g^2}{\eta^2} \]
\[ = \frac{24 \pi (\rho - \sigma)^2 g^2}{81 \eta} r^5 \]

7. **Final Relation**:
Thus, we find that the rate of heat production is proportional to \( r^5 \):
\[ \frac{dq}{dt} \propto r^5 \]

### Conclusion:
The relation between the rate of produced heat and the radius of the sphere at terminal velocity is:
\[ \frac{dq}{dt} \propto r^5 \]