A body cools from `50^(@)C` to `40^(@)C` in 5 min. The surroundings temperature is `20^(@)C`. In what further times (in minutes) will it cool to `30^(@)C` ?

Options

A. 5
B. `15/2`
C. `25/3`
D. 10

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 own temperature and the ambient temperature.

### Step-by-step Solution:

1. **Identify the given values:**
- Initial temperature \( T_1 = 50^\circ C \)
- Final temperature \( T_2 = 40^\circ C \)
- Surrounding temperature \( \theta_0 = 20^\circ C \)
- Time taken to cool from \( 50^\circ C \) to \( 40^\circ C \) is \( t = 5 \) minutes.

2. **Apply Newton's Law of Cooling for the first case:**
\[ \frac{T_1 - T_2}{t} = k \left( \frac{T_1 + T_2}{2} - \theta_0 \right) \]
Substituting the values:
\[ \frac{50 - 40}{5} = k \left( \frac{50 + 40}{2} - 20 \right) \]
Simplifying this gives:
\[ \frac{10}{5} = k \left( 45 - 20 \right) \]
\[ 2 = k \cdot 25 \]
Therefore, we find:
\[ k = \frac{2}{25} \]

3. **Now consider the second case: cooling from \( 40^\circ C \) to \( 30^\circ C \):**
- Here, \( T_1 = 40^\circ C \) and \( T_2 = 30^\circ C \).
- Let the time taken to cool from \( 40^\circ C \) to \( 30^\circ C \) be \( t' \).

4. **Apply Newton's Law of Cooling for the second case:**
\[ \frac{T_1 - T_2}{t'} = k \left( \frac{T_1 + T_2}{2} - \theta_0 \right) \]
Substituting the values:
\[ \frac{40 - 30}{t'} = k \left( \frac{40 + 30}{2} - 20 \right) \]
Simplifying gives:
\[ \frac{10}{t'} = k \left( 35 - 20 \right) \]
\[ \frac{10}{t'} = k \cdot 15 \]

5. **Substituting the value of \( k \) into the equation:**
\[ \frac{10}{t'} = \frac{2}{25} \cdot 15 \]
\[ \frac{10}{t'} = \frac{30}{25} \]
\[ \frac{10}{t'} = 1.2 \]

6. **Solving for \( t' \):**
\[ t' = \frac{10}{1.2} = \frac{100}{12} = \frac{25}{3} \text{ minutes} \]

### Final Answer:
The time taken to cool from \( 40^\circ C \) to \( 30^\circ C \) is \( \frac{25}{3} \) minutes, which is approximately \( 8.33 \) minutes.