A body cools from a temperature `3 T` to `2 T` in `10` minutes. The room temperature is `T`. Assume that Newton's law of cooling is applicable. The temperature of the body at the end of next `10` minutes will be

Options

A. `(4)/(3) T`
B. `T`
C. `(7)/(4) T`
D. `(3)/(2) T`

Explanation

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 (room temperature).

### Step-by-Step Solution:

1. **Identify Initial and Final Temperatures**:
- Initial temperature of the body, \( T_1 = 3T \)
- Final temperature of the body after 10 minutes, \( T_2 = 2T \)
- Room temperature, \( T_r = T \)

2. **Calculate the Change in Temperature**:
- The change in temperature over the first 10 minutes is:
\[ \Delta T = T_1 - T_2 = 3T - 2T = T \]

3. **Apply Newton's Law of Cooling**:
- According to Newton's Law of Cooling:
\[ \frac{dT}{dt} = -k(T - T_r) \]
- For the first 10 minutes, we can write:
\[ \frac{T_1 - T_r}{T_2 - T_r} = e^{-kt} \]
- Substituting the values:
\[ \frac{3T - T}{2T - T} = e^{-k \cdot 10} \]
\[ \frac{2T}{T} = e^{-10k} \]
\[ 2 = e^{-10k} \]

4. **Take the Natural Logarithm**:
- Taking the natural logarithm of both sides:
\[ \ln(2) = -10k \]
- Therefore:
\[ k = -\frac{\ln(2)}{10} \]

5. **Find the Temperature After the Next 10 Minutes**:
- Let the temperature of the body after the next 10 minutes be \( T' \).
- Using Newton's Law of Cooling again:
\[ \frac{T_2 - T_r}{T' - T_r} = e^{-kt} \]
- Substituting the values:
\[ \frac{2T - T}{T' - T} = e^{-10k} \]
\[ \frac{T}{T' - T} = 2 \]
- Rearranging gives:
\[ T = 2(T' - T) \]
\[ T = 2T' - 2T \]
\[ 3T = 2T' \]
- Therefore:
\[ T' = \frac{3T}{2} \]

### Final Answer:
The temperature of the body at the end of the next 10 minutes will be \( \frac{3T}{2} \).

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