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 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 Data:**
- Initial temperature of the body, \( T_i = 40^\circ C \)
- Ambient temperature, \( T_a = 20^\circ C \)
- Temperature after 2 minutes, \( T_f = 36^\circ C \)
- Time taken for the first temperature drop, \( t_1 = 2 \) minutes.

2. **Apply Newton's Law of Cooling:**
According to Newton's Law of Cooling:
\[ \frac{dT}{dt} = -k(T - T_a) \]
where \( k \) is the cooling constant.

3. **Integrate the Equation:**
Rearranging and integrating gives:
\[ \ln(T - T_a) = -kt + C \]
where \( C \) is the integration constant.

4. **Set Up the First Case:**
For the first case, from \( T_i = 40^\circ C \) to \( T_f = 36^\circ C \):
\[ \ln(36 - 20) - \ln(40 - 20) = -k(2) \]
Simplifying this:
\[ \ln(16) - \ln(20) = -2k \]
\[ \ln\left(\frac{16}{20}\right) = -2k \]
\[ \ln\left(\frac{4}{5}\right) = -2k \]

5. **Set Up the Second Case:**
For the second case, from \( T_f = 36^\circ C \) to \( T_{f2} = 32^\circ C \):
\[ \ln(32 - 20) - \ln(36 - 20) = -kt_2 \]
Simplifying this:
\[ \ln(12) - \ln(16) = -kt_2 \]
\[ \ln\left(\frac{12}{16}\right) = -kt_2 \]
\[ \ln\left(\frac{3}{4}\right) = -kt_2 \]

6. **Relate the Two Cases:**
Now, we have two equations:
- From the first case: \( -2k = \ln\left(\frac{4}{5}\right) \)
- From the second case: \( -kt_2 = \ln\left(\frac{3}{4}\right) \)

Dividing the second equation by the first:
\[ \frac{-kt_2}{-2k} = \frac{\ln\left(\frac{3}{4}\right)}{\ln\left(\frac{4}{5}\right)} \]
This simplifies to:
\[ \frac{t_2}{2} = \frac{\ln\left(\frac{3}{4}\right)}{\ln\left(\frac{4}{5}\right)} \]
Therefore:
\[ t_2 = 2 \cdot \frac{\ln\left(\frac{3}{4}\right)}{\ln\left(\frac{4}{5}\right)} \]

7. **Calculate the Values:**
Using logarithm values:
- \( \ln\left(\frac{3}{4}\right) \approx -0.2877 \)
- \( \ln\left(\frac{4}{5}\right) \approx -0.2231 \)

Thus:
\[ t_2 = 2 \cdot \frac{-0.2877}{-0.2231} \approx 2 \cdot 1.292 \approx 2.584 \text{ minutes} \]

8. **Convert to Minutes and Seconds:**
\( 2.584 \) minutes is approximately \( 2 \) minutes and \( 35 \) seconds.

### Final Answer:
The time after which the temperature will further decrease by \( 4^\circ C \) is approximately **2 minutes and 35 seconds**.