A body takes 10 min to cool douwn from `62^@C` to `50^@C`. If the temperature of surrounding is `26^@C` then in the next 10 minutes temperature of the body will be

Options

A. `38^(@)C`
B. `40^(@)C`
C. `42^(@)C`
D. `44^(@)C`

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 ambient temperature.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Initial temperature of the body, \( \theta_1 = 62^\circ C \)
- Final temperature of the body after 10 minutes, \( \theta_2 = 50^\circ C \)
- Ambient temperature, \( \theta_0 = 26^\circ C \)
- Time interval, \( t = 10 \) minutes

2. **Apply Newton's Law of Cooling:**
The formula can be expressed as:
\[ \frac{d\theta}{dt} = -k(\theta - \theta_0) \]
Rearranging gives:
\[ \frac{\theta_2 - \theta_1}{t} = -k\left(\frac{\theta_1 + \theta_2}{2} - \theta_0\right) \]

3. **Substitute the Known Values:**
\[ \frac{50 - 62}{10} = -k\left(\frac{62 + 50}{2} - 26\right) \]
Simplifying the left side:
\[ \frac{-12}{10} = -k\left(\frac{112}{2} - 26\right) \]
\[ -1.2 = -k(56 - 26) \]
\[ -1.2 = -k(30) \]

4. **Solve for \( k \):**
\[ k = \frac{1.2}{30} = \frac{12}{300} = \frac{1}{25} \]

5. **Calculate the Temperature After the Next 10 Minutes:**
Let the temperature after the next 10 minutes be \( \theta \). We can apply the same formula again:
\[ \frac{\theta - \theta_2}{10} = -k\left(\frac{\theta_2 + \theta}{2} - \theta_0\right) \]
Substituting the known values:
\[ \frac{\theta - 50}{10} = -\frac{1}{25}\left(\frac{50 + \theta}{2} - 26\right) \]

6. **Simplify and Solve for \( \theta \):**
Multiply both sides by 10:
\[ \theta - 50 = -\frac{10}{25}\left(\frac{50 + \theta}{2} - 26\right) \]
\[ \theta - 50 = -\frac{2}{5}\left(\frac{50 + \theta}{2} - 26\right) \]
\[ \theta - 50 = -\frac{2}{5}\left(\frac{50 + \theta - 52}{2}\right) \]
\[ \theta - 50 = -\frac{2}{5}\left(\frac{\theta - 2}{2}\right) \]
Cross-multiplying gives:
\[ 5(\theta - 50) = -\theta + 2 \]
\[ 5\theta - 250 = -\theta + 2 \]
\[ 6\theta = 252 \]
\[ \theta = \frac{252}{6} = 42^\circ C \]

### Final Answer:
The temperature of the body after the next 10 minutes will be \( 42^\circ C \).