A sphere, a cube and a thin circular plate are heated to the same temperature. All are made of the same material and have the equal masses. If `t_(1),t_(2) and t_(3)` are the respective time taken by the sphere, cube and the circular plate in cooling down to common temperature, then

Options

A. `t_(1) gt t_(2) gt t_(3)`
B. `t_(1) lt t_(2) lt t_(3)`
C. `t_(2) gt t_(1) gt t_(3)`
D. `t_(1) = t_(2) = t_(3)`

Explanation

To solve the problem, we need to analyze the cooling rates of the sphere, cube, and thin circular plate, all made of the same material and having equal masses. We will use the principles of heat transfer and the Stefan-Boltzmann law.

### Step-by-step Solution:

1. **Understanding the Cooling Process**:
- According to Newton's Law of Cooling, the rate of heat loss of a body is proportional to the surface area of the body and the temperature difference between the body and its surroundings.

2. **Identify the Surface Areas**:
- For a sphere with radius \( r \), the surface area \( A_1 \) is given by:
\[ A_1 = 4\pi r^2 \]
- For a cube with side length \( a \), the surface area \( A_2 \) is:
\[ A_2 = 6a^2 \]
- For a thin circular plate with radius \( R \) and negligible thickness, the surface area \( A_3 \) is:
\[ A_3 = \pi R^2 \]

3. **Relating Mass and Volume**:
- Since all three objects have the same mass \( m \) and are made of the same material, their volumes must also be equal.
- The volume of the sphere is \( V_1 = \frac{4}{3}\pi r^3 \).
- The volume of the cube is \( V_2 = a^3 \).
- The volume of the thin circular plate is \( V_3 = \pi R^2 t \) (where \( t \) is the thickness, which we can consider negligible).

4. **Cooling Rate**:
- The rate of cooling is proportional to the surface area. Thus, the time taken to cool down to a common temperature is inversely proportional to the surface area:
\[ \text{Rate of cooling} \propto A \]
- Therefore, the time taken to cool down can be expressed as:
\[ t \propto \frac{1}{A} \]

5. **Comparing Surface Areas**:
- We need to compare \( A_1 \), \( A_2 \), and \( A_3 \):
- The sphere has the smallest surface area for a given volume, followed by the cube, and the thin circular plate has the largest surface area for the same mass.
- Thus, we can conclude:
\[ A_1 < A_2 < A_3 \]

6. **Conclusion**:
- Since the time taken to cool is inversely proportional to the surface area, we have:
\[ t_1 > t_2 > t_3 \]
- Therefore, the time taken for the sphere \( t_1 \) is the longest, followed by the cube \( t_2 \), and the thin circular plate \( t_3 \) takes the least time to cool down.

### Final Result:
\[ t_1 > t_2 > t_3 \]