A body cools from `50^@C` to `49^@C` in 5 s. How long will it take to cool from `40^@C` to `39.5^@C`? Assume the temperature of surroundings to be `30^@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 temperature and the surrounding temperature.

### Step-by-Step Solution:

**Step 1: Understand the cooling process in the first case.**
- The body cools from \(50^\circ C\) to \(49^\circ C\) in \(5\) seconds.
- The average temperature during this cooling is:
\[ \text{Average Temperature} = \frac{50 + 49}{2} = 49.5^\circ C \]
- The temperature difference between the average temperature and the surrounding temperature (\(30^\circ C\)) is:
\[ \Delta T_1 = 49.5 - 30 = 19.5^\circ C \]

**Step 2: Set up the equation for the first case using Newton's Law of Cooling.**
- The rate of cooling can be expressed as:
\[ \frac{50 - 49}{5} \propto 19.5 \]
- This simplifies to:
\[ \frac{1}{5} \propto 19.5 \]

**Step 3: Understand the cooling process in the second case.**
- The body cools from \(40^\circ C\) to \(39.5^\circ C\).
- The average temperature during this cooling is:
\[ \text{Average Temperature} = \frac{40 + 39.5}{2} = 39.75^\circ C \]
- The temperature difference between the average temperature and the surrounding temperature is:
\[ \Delta T_2 = 39.75 - 30 = 9.75^\circ C \]

**Step 4: Set up the equation for the second case.**
- The rate of cooling can be expressed as:
\[ \frac{40 - 39.5}{T} \propto 9.75 \]
- This simplifies to:
\[ \frac{0.5}{T} \propto 9.75 \]

**Step 5: Relate the two cases.**
- From the first case, we have:
\[ \frac{1}{5} = k \cdot 19.5 \quad \text{(where \(k\) is the proportionality constant)} \]
- From the second case:
\[ \frac{0.5}{T} = k \cdot 9.75 \]

**Step 6: Set up the ratio of the two cases.**
- Dividing the second equation by the first gives:
\[ \frac{0.5/T}{1/5} = \frac{9.75}{19.5} \]
- Simplifying this:
\[ \frac{0.5 \cdot 5}{T} = \frac{9.75}{19.5} \]

**Step 7: Solve for \(T\).**
- Rearranging gives:
\[ T = \frac{0.5 \cdot 5 \cdot 19.5}{9.75} \]
- Simplifying further:
\[ T = \frac{2.5 \cdot 19.5}{9.75} = \frac{48.75}{9.75} = 5 \text{ seconds} \]

Thus, the time it will take to cool from \(40^\circ C\) to \(39.5^\circ C\) is **5 seconds**.