A solid sphere and a hollow sphere of the same material and size are heated to the same temperature and allowed to cool in the same surroundings. If the temperature difference between the surroundings and each sphere is T, then :

Options

A. The hollow sphere will cool at a faster rate for all values of T
B. The solid sphere will cool at a faster rate for all values of T
C. Both spheres will cool at the same rate for all values of T
D. Both spheres will cool at the same rate only for small values of T

Explanation

To solve the problem, we need to analyze the cooling rates of a solid sphere and a hollow sphere made of the same material and size, both heated to the same temperature and allowed to cool in the same surroundings.

### Step-by-Step Solution:

1. **Understand the Setup**:
- We have two spheres: a solid sphere (let's call it Sphere A) and a hollow sphere (Sphere B).
- Both spheres are made of the same material, have the same size, and are heated to the same initial temperature.
- They are allowed to cool in the same surroundings, with a temperature difference (T) between the spheres and their surroundings.

2. **Cooling Law**:
- According to Newton's Law of Cooling, the rate of heat loss of a body is proportional to the temperature difference between the body and its surroundings.
- The formula for the rate of cooling can be expressed as:
\[ \frac{dT}{dt} = -k (T - T_0) \]
where \( T \) is the temperature of the sphere, \( T_0 \) is the temperature of the surroundings, and \( k \) is a constant that depends on the properties of the sphere.

3. **Factors Affecting Cooling Rate**:
- The cooling rate also depends on the mass and specific heat of the spheres.
- The mass of a solid sphere is greater than that of a hollow sphere of the same size because the hollow sphere has an empty interior.

4. **Mass Calculation**:
- For a solid sphere, the mass \( m_A \) can be calculated using the volume and density:
\[ m_A = \rho \cdot V_A = \rho \cdot \left(\frac{4}{3} \pi r^3\right) \]
- For a hollow sphere, the mass \( m_B \) is less because it has a shell structure:
\[ m_B = \rho \cdot V_B = \rho \cdot \left(\frac{4}{3} \pi (R^3 - r^3)\right) \]
where \( R \) is the outer radius and \( r \) is the inner radius.

5. **Rate of Cooling Comparison**:
- Since the hollow sphere has less mass, it will lose heat more quickly than the solid sphere when subjected to the same temperature difference.
- Therefore, the rate of cooling for the hollow sphere will be greater than that of the solid sphere.

6. **Conclusion**:
- The hollow sphere will cool at a faster rate than the solid sphere for all values of the temperature difference \( T \).

### Final Answer:
The hollow sphere will cool at a faster rate than the solid sphere for all values of \( T \).