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//4`
C. `25//3`
D. 10

Explanation

To solve the problem of how long it will take for the body to cool from 40°C to 30°C, we can use Newton's Law of Cooling. The law states that the rate of heat loss of a body is directly proportional to the difference in temperature between the body and its surroundings, provided this difference is small.

### Step-by-Step Solution:

1. **Identify the known values**:
- Initial temperature (T1) = 50°C
- Final temperature (T2) = 40°C
- Surrounding temperature (Ts) = 20°C
- Time taken to cool from 50°C to 40°C (t1) = 5 minutes

2. **Calculate the temperature difference**:
- For the first interval (from 50°C to 40°C):
- ΔT1 = T1 - Ts = 50°C - 20°C = 30°C
- ΔT2 = T2 - Ts = 40°C - 20°C = 20°C

3. **Use Newton's Law of Cooling**:
The formula according to Newton's Law of Cooling can be expressed as:
\[ \frac{T1 - Ts}{T2 - Ts} = \frac{t1}{t2} \]
where \( t2 \) is the time taken to cool from 40°C to 30°C.

4. **Set up the equation**:
\[ \frac{30°C}{20°C} = \frac{5 \text{ min}}{t2} \]

5. **Cross-multiply to solve for \( t2 \)**:
\[ 30 \cdot t2 = 20 \cdot 5 \]
\[ 30t2 = 100 \]
\[ t2 = \frac{100}{30} = \frac{10}{3} \text{ minutes} \]

6. **Convert to a more understandable form**:
\[ t2 = 3 \frac{1}{3} \text{ minutes} \text{ or } 3.33 \text{ minutes} \]

### Final Answer:
The time taken for the body to cool from 40°C to 30°C is \( \frac{10}{3} \) minutes or approximately 3.33 minutes.