If a metallic sphere gets cooled from `62^(@)C` to `50^(@)C` in minutes 10 and in the next 10 minutes gets cooled to `42^(@)C` , then the temperature of the surroundings is

Options

A. `30^(@)C`
B. `36^(@)C`
C. `26^(@)C`
D. `20^(@)C`

Answer

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 (temperature of the surroundings).

### Step-by-Step Solution:

1. **Identify the Initial and Final Temperatures**:
- In the first 10 minutes, the sphere cools from \( \theta_1 = 62^\circ C \) to \( \theta_2 = 50^\circ C \).
- In the next 10 minutes, it cools from \( \theta_1 = 50^\circ C \) to \( \theta_2 = 42^\circ C \).

2. **Set Up the First Equation Using Newton's Law of Cooling**:
- According to Newton's Law of Cooling:
\[ \frac{\theta_1 - \theta_2}{t} = k \left( \frac{\theta_1 + \theta_2}{2} - \theta_n \right) \]
- For the first cooling period (10 minutes):
\[ \frac{62 - 50}{10} = k \left( \frac{62 + 50}{2} - \theta_n \right) \]
- Simplifying the left side:
\[ \frac{12}{10} = k \left( \frac{112}{2} - \theta_n \right) \]
\[ \frac{6}{5} = k (56 - \theta_n) \]
- This is our **Equation 1**:
\[ \frac{6}{5} = k (56 - \theta_n) \tag{1} \]

3. **Set Up the Second Equation for the Next Cooling Period**:
- For the second cooling period (also 10 minutes):
\[ \frac{50 - 42}{10} = k \left( \frac{50 + 42}{2} - \theta_n \right) \]
- Simplifying the left side:
\[ \frac{8}{10} = k \left( \frac{92}{2} - \theta_n \right) \]
\[ \frac{4}{5} = k (46 - \theta_n) \]
- This is our **Equation 2**:
\[ \frac{4}{5} = k (46 - \theta_n) \tag{2} \]

4. **Divide Equation 1 by Equation 2**:
- Dividing both equations:
\[ \frac{\frac{6}{5}}{\frac{4}{5}} = \frac{k(56 - \theta_n)}{k(46 - \theta_n)} \]
- The \( k \) cancels out:
\[ \frac{6}{4} = \frac{56 - \theta_n}{46 - \theta_n} \]
- Simplifying:
\[ \frac{3}{2} = \frac{56 - \theta_n}{46 - \theta_n} \]

5. **Cross Multiply and Simplify**:
- Cross multiplying gives:
\[ 3(46 - \theta_n) = 2(56 - \theta_n) \]
\[ 138 - 3\theta_n = 112 - 2\theta_n \]
- Rearranging:
\[ 138 - 112 = 3\theta_n - 2\theta_n \]
\[ 26 = \theta_n \]

6. **Conclusion**:
- The temperature of the surroundings \( \theta_n \) is \( 26^\circ C \).

### Final Answer:
The temperature of the surroundings is \( 26^\circ C \).