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

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. **Identify Initial Conditions**:
- Initial temperature of the body, \( \theta_1 = 3T \)
- Final temperature after 10 minutes, \( \theta_2 = 2T \)
- Room temperature (ambient temperature), \( \theta_s = T \)
- Time interval, \( t = 10 \) minutes

2. **Apply Newton's Law of Cooling**:
According to Newton's law of cooling:
\[ \frac{d\theta}{dt} = -k(\theta - \theta_s) \]
where \( k \) is a constant.

3. **Calculate the Average Temperature**:
The average temperature of the body during the cooling process from \( 3T \) to \( 2T \) is:
\[ \theta_{avg} = \frac{\theta_1 + \theta_2}{2} = \frac{3T + 2T}{2} = \frac{5T}{2} \]

4. **Set Up the Cooling Equation**:
Using the temperatures and the time interval:
\[ \frac{2T - 3T}{10} = -k\left(\frac{5T}{2} - T\right) \]
Simplifying the left side:
\[ \frac{-T}{10} = -k\left(\frac{5T}{2} - T\right) \]
The right side simplifies to:
\[ \frac{5T}{2} - T = \frac{5T}{2} - \frac{2T}{2} = \frac{3T}{2} \]
Thus, we have:
\[ \frac{-T}{10} = -k\left(\frac{3T}{2}\right) \]

5. **Solve for \( k \)**:
Rearranging gives:
\[ k = \frac{T/10}{3T/2} = \frac{1}{10} \cdot \frac{2}{3} = \frac{1}{15} \]

6. **Determine the Temperature After the Next 10 Minutes**:
Now, we need to find the temperature after another 10 minutes, starting from \( 2T \):
- New initial temperature \( \theta_2 = 2T \)
- We will use the same ambient temperature \( T \) and the same \( k \).

The average temperature during this cooling period is:
\[ \theta_{avg} = \frac{2T + \theta_f}{2} \]
where \( \theta_f \) is the final temperature we want to find.

The cooling equation becomes:
\[ \frac{\theta_f - 2T}{10} = -k\left(\frac{2T + \theta_f}{2} - T\right) \]
Simplifying the right side:
\[ \frac{\theta_f - 2T}{10} = -\frac{1}{15}\left(\frac{2T + \theta_f}{2} - T\right) \]
\[ = -\frac{1}{15}\left(\frac{2T + \theta_f - 2T}{2}\right) = -\frac{1}{15}\left(\frac{\theta_f}{2}\right) \]
Thus, we have:
\[ \frac{\theta_f - 2T}{10} = -\frac{\theta_f}{30} \]

7. **Cross-Multiply and Solve for \( \theta_f \)**:
Cross-multiplying gives:
\[ 3(\theta_f - 2T) = -\theta_f \]
\[ 3\theta_f - 6T = -\theta_f \]
\[ 4\theta_f = 6T \]
\[ \theta_f = \frac{6T}{4} = \frac{3T}{2} \]

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