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.

Options

A. `r^(5)`
B. `r^(2)`
C. `r^(3)`
D. `r^(4)`

Answer

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

### Step 1: Understand the Forces Acting on the Sphere
When a small sphere falls through a viscous liquid, it experiences three main forces:
- The gravitational force acting downward (weight of the sphere).
- The buoyant force acting upward (due to the liquid).
- The viscous drag force acting upward (due to the resistance of the liquid).

At terminal velocity, the net force acting on the sphere is zero, meaning the downward gravitational force is balanced by the upward buoyant force and the viscous drag force.

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

### Step 3: Determine Terminal Velocity
The terminal velocity \( v_t \) of a sphere falling through a viscous liquid is given by:
\[ v_t = \frac{2}{9} \frac{(r^2)(\rho_s - \rho_l)g}{\eta} \]
where:
- \( \rho_s \) is the density of the sphere,
- \( \rho_l \) is the density of the liquid,
- \( g \) is the acceleration due to gravity.

### Step 4: Relate Viscous Force to Terminal Velocity
At terminal velocity, the viscous force equals the net downward force (weight of the sphere minus buoyant force). Therefore, we can express:
\[ F_v = 6 \pi \eta r v_t \]
At terminal velocity, this force is constant.

### Step 5: Calculate the Rate of Production of Heat
The rate of production of heat (power \( P \)) due to viscous drag can be expressed as:
\[ P = F_v \cdot v_t \]
Substituting the expression for viscous force:
\[ P = (6 \pi \eta r v_t) \cdot v_t = 6 \pi \eta r v_t^2 \]

### Step 6: Substitute for Terminal Velocity
Now substituting the expression for \( v_t \):
\[ P = 6 \pi \eta r \left(\frac{2}{9} \frac{(r^2)(\rho_s - \rho_l)g}{\eta}\right)^2 \]
Simplifying this expression:
\[ P = 6 \pi \eta r \cdot \frac{4}{81} \frac{(r^4)(\rho_s - \rho_l)^2 g^2}{\eta^2} \]
\[ P = \frac{24 \pi (\rho_s - \rho_l)^2 g^2}{81 \eta} r^5 \]

### Conclusion
Thus, the relation between the rate of production of heat \( P \) and the radius \( r \) of the sphere at terminal velocity is:
\[ P \propto r^5 \]