A body is heated to a temperature `40^(@)C` and kept in a chamber maintained at `20^@C`. If the temperature of the body decreases to `36^@C` in 2 minutues, then the time after which the temperature will further decrease by `4^@C`, is

Options

A. 2 minutes
B. 2 minutes 33 seconds
C. 2 minutes 55 seconds
D. 3 minutues

Answer

To solve the problem, we will use Newton's Law of Cooling, which states that the rate of cooling of a body is proportional to the difference in temperature between the body and its surroundings.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Initial temperature of the body, \( \theta_1 = 40^\circ C \)
- Surrounding temperature, \( \theta_0 = 20^\circ C \)
- Temperature after 2 minutes, \( \theta_2 = 36^\circ C \)
- Time taken to cool from \( 40^\circ C \) to \( 36^\circ C \), \( t_1 = 2 \) minutes.

2. **Calculate the Average Temperature:**
- The average temperature during the first cooling period can be calculated as:
\[ \theta_{avg1} = \frac{\theta_1 + \theta_2}{2} = \frac{40 + 36}{2} = 38^\circ C \]

3. **Apply Newton's Law of Cooling:**
- According to Newton's Law of Cooling:
\[ \frac{d\theta}{dt} = -k(\theta - \theta_0) \]
- For the first cooling period:
\[ \frac{d\theta}{dt} = -k(\theta_{avg1} - \theta_0) = -k(38 - 20) = -k(18) \]
- The rate of cooling can be expressed as:
\[ \frac{d\theta}{dt} = \frac{\Delta \theta}{\Delta t} = \frac{36 - 40}{2} = -2 \text{ degrees per minute} \]
- Thus, we have:
\[ -2 = -k(18) \implies k = \frac{2}{18} = \frac{1}{9} \text{ per minute} \]

4. **Calculate Time for Further Cooling:**
- Now, we need to find the time taken for the temperature to decrease from \( 36^\circ C \) to \( 32^\circ C \).
- The average temperature during this cooling period is:
\[ \theta_{avg2} = \frac{36 + 32}{2} = 34^\circ C \]
- Applying Newton's Law again:
\[ \frac{d\theta}{dt} = -k(\theta_{avg2} - \theta_0) = -k(34 - 20) = -k(14) \]
- The rate of cooling is:
\[ \frac{d\theta}{dt} = \frac{32 - 36}{t_2} = \frac{-4}{t_2} \]
- Setting the two expressions for the rate of cooling equal:
\[ \frac{-4}{t_2} = -k(14) \implies \frac{4}{t_2} = k(14) \]
- Substituting \( k = \frac{1}{9} \):
\[ \frac{4}{t_2} = \frac{1}{9}(14) \implies t_2 = \frac{4 \times 9}{14} = \frac{36}{14} = \frac{18}{7} \text{ minutes} \]

5. **Convert Time to Minutes and Seconds:**
- \( t_2 = \frac{18}{7} \) minutes can be converted to minutes and seconds:
\[ 2 \text{ minutes } + \frac{4}{7} \text{ minutes} \]
- Converting \( \frac{4}{7} \) minutes to seconds:
\[ \frac{4}{7} \times 60 \approx 34.29 \text{ seconds} \]
- Therefore, the total time is approximately:
\[ 2 \text{ minutes } 34 \text{ seconds} \]

### Final Answer:
The time after which the temperature will further decrease by \( 4^\circ C \) is approximately **2 minutes 34 seconds**.