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^@C` is

Options

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

Explanation

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

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Initial temperature (T1) = 62 °C
- Final temperature (T2) = 61 °C
- Surrounding temperature (θ) = 30 °C
- Time taken to cool from 62 °C to 61 °C = T minutes

2. **Apply Newton's Law of Cooling for the First Case:**
According to Newton's Law of Cooling:
\[ \frac{T1 - T2}{T} \propto \frac{T1 + T2}{2} - θ \]
Substituting the values:
\[ \frac{62 - 61}{T} \propto \frac{62 + 61}{2} - 30 \]
Simplifying:
\[ \frac{1}{T} \propto 61.5 - 30 = 31.5 \]
Thus, we can write:
\[ \frac{1}{T} = k \cdot 31.5 \quad \text{(where k is a proportionality constant)} \]

3. **Identify the Second Case:**
Now, we need to find the time taken (t) for the body to cool from 46 °C to 45 °C.
- Initial temperature (T1) = 46 °C
- Final temperature (T2) = 45 °C

4. **Apply Newton's Law of Cooling for the Second Case:**
Using the same formula:
\[ \frac{T1 - T2}{t} \propto \frac{T1 + T2}{2} - θ \]
Substituting the values:
\[ \frac{46 - 45}{t} \propto \frac{46 + 45}{2} - 30 \]
Simplifying:
\[ \frac{1}{t} \propto 45.5 - 30 = 15.5 \]
Thus, we can write:
\[ \frac{1}{t} = k \cdot 15.5 \]

5. **Relate the Two Cases:**
From the first case, we have:
\[ \frac{1}{T} = k \cdot 31.5 \]
From the second case, we have:
\[ \frac{1}{t} = k \cdot 15.5 \]
Dividing the two equations:
\[ \frac{1/T}{1/t} = \frac{31.5}{15.5} \]
This simplifies to:
\[ \frac{t}{T} = \frac{15.5}{31.5} \]

6. **Solve for t:**
Rearranging gives:
\[ t = T \cdot \frac{15.5}{31.5} \]
Since we know that T is the time taken to cool from 62 °C to 61 °C, we can conclude that:
\[ t = T \]

### Final Answer:
Therefore, the time taken by the body to cool from 46 °C to 45 °C is equal to T minutes.