A solid copper cube of edges 1 cm is suspended in an evacuated enclosure. Its temperature is found to fall from `100^@C` to `99^@C` in 100 s. Another solid copper cube of edges 2 cm, with similar surface nature, is suspended in a similar manner. The time required for this cube to cool from `100^@C` to `99^@C` will be approximately

Options

A. 25 s
B. 50 s
C. 200 s
D. 400 s

Explanation

To solve the problem, we need to determine the time required for a larger copper cube (with edges of 2 cm) to cool from 100°C to 99°C, given that a smaller cube (with edges of 1 cm) takes 100 seconds to cool through the same temperature difference.

### Step-by-Step Solution:

1. **Identify the properties of the cubes**:
- The first cube has a side length \( L_1 = 1 \, \text{cm} \).
- The second cube has a side length \( L_2 = 2 \, \text{cm} \).
- Both cubes are made of copper and have the same specific heat capacity \( C \) and density \( \rho \).

2. **Calculate the volume of each cube**:
- Volume of the first cube:
\[ V_1 = L_1^3 = 1^3 = 1 \, \text{cm}^3 \]
- Volume of the second cube:
\[ V_2 = L_2^3 = 2^3 = 8 \, \text{cm}^3 \]

3. **Determine the mass of each cube**:
- Mass of the first cube:
\[ m_1 = V_1 \cdot \rho = 1 \cdot \rho = \rho \, \text{g} \]
- Mass of the second cube:
\[ m_2 = V_2 \cdot \rho = 8 \cdot \rho = 8\rho \, \text{g} \]

4. **Calculate the heat lost by each cube**:
- Heat lost by the first cube when cooling from 100°C to 99°C:
\[ \Delta Q_1 = m_1 \cdot C \cdot \Delta T = \rho \cdot C \cdot 1 \]
- Heat lost by the second cube:
\[ \Delta Q_2 = m_2 \cdot C \cdot \Delta T = 8\rho \cdot C \cdot 1 \]

5. **Use the Stefan-Boltzmann law**:
- The rate of heat loss is proportional to the surface area and the fourth power of the temperature.
- Surface area of the first cube:
\[ A_1 = 6L_1^2 = 6 \cdot 1^2 = 6 \, \text{cm}^2 \]
- Surface area of the second cube:
\[ A_2 = 6L_2^2 = 6 \cdot 2^2 = 24 \, \text{cm}^2 \]

6. **Set up the time equations**:
- Time for the first cube to cool:
\[ t_1 = \frac{\Delta Q_1}{A_1 \cdot \sigma \cdot T^4} = \frac{\rho C \cdot 1}{6 \cdot \sigma \cdot T^4} \]
- Time for the second cube to cool:
\[ t_2 = \frac{\Delta Q_2}{A_2 \cdot \sigma \cdot T^4} = \frac{8\rho C \cdot 1}{24 \cdot \sigma \cdot T^4} \]

7. **Relate the two times**:
- From the equations, we can find the ratio of the two times:
\[ \frac{t_1}{t_2} = \frac{\Delta Q_1 / (A_1 \cdot \sigma \cdot T^4)}{\Delta Q_2 / (A_2 \cdot \sigma \cdot T^4)} = \frac{\Delta Q_1 \cdot A_2}{\Delta Q_2 \cdot A_1} \]
- Substituting the values:
\[ \frac{100}{t_2} = \frac{\rho C \cdot 6}{8\rho C \cdot 1} \cdot \frac{24}{6} \]
- Simplifying gives:
\[ \frac{100}{t_2} = \frac{6 \cdot 24}{8 \cdot 1} = \frac{144}{8} = 18 \]
- Therefore, \( t_2 = \frac{100}{18} \) which simplifies to:
\[ t_2 = 200 \, \text{seconds} \]

### Final Answer:
The time required for the second cube to cool from 100°C to 99°C is approximately **200 seconds**.