A body takes T minutes to cool from `62^@C` to `61^@C` when the surrounding temperature is `30^@C`. The time taken by the body to cool from `46^@C` to `45.5^@C` is

Options

A. Greater than T minutes
B. Equal to T minutes
C. Less than T minutes
D. None of these

Answer

To solve the problem, we will use Newton's Law of Cooling, which states that the rate of cooling of a body is proportional to the difference between its temperature and the surrounding temperature.

### Step-by-step Solution:

1. **Identify the given data:**
- Initial temperature \( T_1 = 62^\circ C \)
- Final temperature \( T_2 = 61^\circ C \)
- Surrounding temperature \( T_s = 30^\circ C \)
- Time taken to cool from \( T_1 \) to \( T_2 \) is \( t \) minutes.

2. **Calculate the change in temperature for the first case:**
\[ \Delta T_1 = T_1 - T_2 = 62 - 61 = 1^\circ C \]

3. **Calculate the average temperature for the first case:**
\[ T_{avg1} = \frac{T_1 + T_2}{2} = \frac{62 + 61}{2} = 61.5^\circ C \]

4. **Apply Newton's Law of Cooling for the first case:**
According to Newton's Law:
\[ \frac{\Delta T_1}{t} = k (T_{avg1} - T_s) \]
Substituting the values:
\[ \frac{1}{t} = k (61.5 - 30) \]
\[ \frac{1}{t} = k \cdot 31.5 \quad \text{(Equation 1)} \]

5. **Now consider the second case:**
- Initial temperature \( T_3 = 46^\circ C \)
- Final temperature \( T_4 = 45.5^\circ C \)

6. **Calculate the change in temperature for the second case:**
\[ \Delta T_2 = T_3 - T_4 = 46 - 45.5 = 0.5^\circ C \]

7. **Calculate the average temperature for the second case:**
\[ T_{avg2} = \frac{T_3 + T_4}{2} = \frac{46 + 45.5}{2} = 45.75^\circ C \]

8. **Apply Newton's Law of Cooling for the second case:**
\[ \frac{\Delta T_2}{t_0} = k (T_{avg2} - T_s) \]
Substituting the values:
\[ \frac{0.5}{t_0} = k (45.75 - 30) \]
\[ \frac{0.5}{t_0} = k \cdot 15.75 \quad \text{(Equation 2)} \]

9. **Divide Equation 1 by Equation 2:**
\[ \frac{\frac{1}{t}}{\frac{0.5}{t_0}} = \frac{k \cdot 31.5}{k \cdot 15.75} \]
The \( k \) cancels out:
\[ \frac{1}{t} \cdot \frac{t_0}{0.5} = \frac{31.5}{15.75} \]

10. **Simplify the right side:**
\[ \frac{31.5}{15.75} = 2 \]
Therefore:
\[ \frac{t_0}{t} = 2 \cdot 0.5 \]
\[ t_0 = t \]

### Conclusion:
The time taken by the body to cool from \( 46^\circ C \) to \( 45.5^\circ C \) is equal to \( t \) minutes.