Consider a solid sphere placed in surrounding with small temperature difference between sphere's surface and surrounding. If `DeltaT` and `r` represent temperature difference (between sphere and surrounding) and radius of sphere respectively, rate of cooling of the sphere is directly proportional to:

Options

A. `(DeltaT)^(4)`
B. `DeltaT`
C. `r^(2)`
D. `1/r`

Explanation

To solve the problem, we need to analyze the rate of cooling of a solid sphere placed in a surrounding medium with a small temperature difference (ΔT) between the sphere's surface and the surrounding environment.

### Step-by-Step Solution:

1. **Understanding the Concept of Cooling**:
The rate of cooling of an object is governed by Newton's Law of Cooling, which states that the rate of change of temperature of an object is directly proportional to the temperature difference between the object and its surroundings.

2. **Setting Up the Equation**:
According to Newton's Law of Cooling:
\[ \frac{d\theta}{dt} \propto \Delta T \]
where \( \Delta T \) is the temperature difference between the sphere and the surrounding.

3. **Surface Area of the Sphere**:
The surface area \( A \) of a solid sphere is given by:
\[ A = 4\pi r^2 \]
where \( r \) is the radius of the sphere.

4. **Mass of the Sphere**:
The mass \( m \) of the sphere can be expressed as:
\[ m = \rho V = \rho \left(\frac{4}{3}\pi r^3\right) \]
where \( \rho \) is the density of the sphere.

5. **Combining the Equations**:
The rate of heat loss \( \frac{dQ}{dt} \) can be expressed as:
\[ \frac{dQ}{dt} = hA\Delta T \]
where \( h \) is the heat transfer coefficient.

6. **Substituting for Area and Mass**:
Substituting the expressions for area and mass into the equation gives:
\[ \frac{dQ}{dt} = h(4\pi r^2)\Delta T \]
and
\[ m = \rho\left(\frac{4}{3}\pi r^3\right) \]

7. **Rate of Cooling**:
The rate of cooling can be expressed as:
\[ \frac{d\theta}{dt} = \frac{dQ/dt}{m \cdot s} \]
where \( s \) is the specific heat capacity.

8. **Final Expression**:
Combining everything, we get:
\[ \frac{d\theta}{dt} \propto \frac{4\pi r^2 \Delta T}{\rho\left(\frac{4}{3}\pi r^3\right) s} \]
Simplifying this gives:
\[ \frac{d\theta}{dt} \propto \frac{\Delta T}{r} \]

9. **Conclusion**:
Therefore, the rate of cooling of the sphere is directly proportional to the temperature difference \( \Delta T \) and inversely proportional to the radius \( r \) of the sphere.

### Final Answer:
The rate of cooling of the sphere is directly proportional to \( \frac{\Delta T}{r} \).