The excess temperature of a body falls from `12^(@)C` to `6^(@)C` in 5 minutes, then the time to fall the excess temperature from `6^(@)C` to `3^(@)C` is (assume the newton's cooling is valid)

Options

A. a. 10 minutes
B. b. 7.5 minutes
C. c. 5 minutes
D. d. 2.5 minutes

Answer

To solve the problem, we will apply 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 ambient temperature.

### Step-by-Step Solution:

1. **Understand the Problem Statement**:
- The excess temperature of the body falls from \(12^\circ C\) to \(6^\circ C\) in \(5\) minutes.
- We need to find the time taken for the excess temperature to fall from \(6^\circ C\) to \(3^\circ C\).

2. **Define Excess Temperature**:
- Let the ambient temperature (surrounding temperature) be \(T_s\).
- The initial excess temperature when the body is at \(12^\circ C\) is \(T_{body} - T_s\) (which equals \(12 - T_s\)).
- The excess temperature when the body is at \(6^\circ C\) is \(6 - T_s\).
- The excess temperature when the body is at \(3^\circ C\) is \(3 - T_s\).

3. **Apply Newton's Law of Cooling**:
- According to Newton's law, we can express the change in temperature as:
\[ \frac{dT}{dt} = -k(T - T_s) \]
- Rearranging gives:
\[ \frac{dT}{T - T_s} = -k \, dt \]

4. **Set Up the First Equation**:
- For the first interval (from \(12^\circ C\) to \(6^\circ C\)):
\[ \int_{12}^{6} \frac{dT}{T - T_s} = -k \int_{0}^{5} dt \]
- This simplifies to:
\[ \ln(6 - T_s) - \ln(12 - T_s) = -5k \]
- Using properties of logarithms:
\[ \ln\left(\frac{6 - T_s}{12 - T_s}\right) = -5k \]

5. **Set Up the Second Equation**:
- For the second interval (from \(6^\circ C\) to \(3^\circ C\)):
\[ \int_{6}^{3} \frac{dT}{T - T_s} = -k \int_{0}^{t} dt \]
- This simplifies to:
\[ \ln(3 - T_s) - \ln(6 - T_s) = -kt \]
- Using properties of logarithms:
\[ \ln\left(\frac{3 - T_s}{6 - T_s}\right) = -kt \]

6. **Relate the Two Equations**:
- From the first equation, we have:
\[ -5k = \ln\left(\frac{6 - T_s}{12 - T_s}\right) \]
- From the second equation:
\[ -kt = \ln\left(\frac{3 - T_s}{6 - T_s}\right) \]
- We can express \(kt\) in terms of \(k\) from the first equation:
\[ kt = \frac{1}{5} \ln\left(\frac{3 - T_s}{6 - T_s}\right) \cdot 5 \]

7. **Calculate the Time**:
- By comparing the two equations, we find that the time \(t\) for the temperature to fall from \(6^\circ C\) to \(3^\circ C\) is the same as the time taken to fall from \(12^\circ C\) to \(6^\circ C\) multiplied by the ratio of the logarithms:
\[ t = 5 \text{ minutes} \]

### Final Answer:
The time taken for the excess temperature to fall from \(6^\circ C\) to \(3^\circ C\) is **5 minutes**.