A body cools from `50.0^(@)C` to` 49.9^(@)C` in 5s. How long will it take to cool from `40.0^(@)C` to `39.9^(@)C` ? Assume the temperature of the surroundings to be `30.0^(@)C` and Newton's law of cooling to be valid:

Options

A. 2.5 s
B. 10 s
C. 20 s
D. 5 s

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 own temperature and the ambient temperature.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Initial temperature (T1) = 50.0°C
- Final temperature (T2) = 49.9°C
- Time taken (t1) = 5 seconds
- Surrounding temperature (T0) = 30.0°C

2. **Apply Newton's Law of Cooling for the First Case:**
According to Newton's Law of Cooling:
\[ \frac{T1 - T2}{t1} = k \left(\frac{T1 + T2}{2} - T0\right) \]
Substituting the values:
\[ \frac{50.0 - 49.9}{5} = k \left(\frac{50.0 + 49.9}{2} - 30.0\right) \]
Simplifying:
\[ \frac{0.1}{5} = k \left(\frac{99.9}{2} - 30.0\right) \]
\[ \frac{0.1}{5} = k \left(49.95 - 30.0\right) \]
\[ \frac{0.1}{5} = k \cdot 19.95 \]
\[ 0.02 = k \cdot 19.95 \]
Thus, we can express k:
\[ k = \frac{0.02}{19.95} \]

3. **Identify the Second Case:**
- Initial temperature (T3) = 40.0°C
- Final temperature (T4) = 39.9°C
- Time taken (t2) = ? (to be determined)

4. **Apply Newton's Law of Cooling for the Second Case:**
Using the same formula:
\[ \frac{T3 - T4}{t2} = k \left(\frac{T3 + T4}{2} - T0\right) \]
Substituting the values:
\[ \frac{40.0 - 39.9}{t2} = k \left(\frac{40.0 + 39.9}{2} - 30.0\right) \]
Simplifying:
\[ \frac{0.1}{t2} = k \left(\frac{79.9}{2} - 30.0\right) \]
\[ \frac{0.1}{t2} = k \left(39.95 - 30.0\right) \]
\[ \frac{0.1}{t2} = k \cdot 9.95 \]

5. **Set Up the Ratio of the Two Cases:**
From the first case, we have:
\[ \frac{0.1}{5} = k \cdot 19.95 \]
From the second case:
\[ \frac{0.1}{t2} = k \cdot 9.95 \]

Dividing the first equation by the second:
\[ \frac{\frac{0.1}{5}}{\frac{0.1}{t2}} = \frac{k \cdot 19.95}{k \cdot 9.95} \]
The k and 0.1 cancel out:
\[ \frac{t2}{5} = \frac{19.95}{9.95} \]
Thus:
\[ t2 = 5 \cdot \frac{19.95}{9.95} \]

6. **Calculate t2:**
\[ t2 = 5 \cdot 2.005 = 10.025 \text{ seconds} \]

### Final Answer:
It will take approximately **10.03 seconds** for the body to cool from 40.0°C to 39.9°C.