A body cools in a surrounding of constant temperature `30^@C` Its heat capacity is `2 J//^@C`. Initial temeprature of cooling is valid. The body cools to `38^@C` in 10 min
The temperature of the body In `.^@C` denoted by `theta`. The veriation of `theta` versus time t is best denoted as

Options

A.
B.
C.
D.

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 surrounding it.

### Step-by-Step Solution:

1. **Identify Given Data:**
- Surrounding temperature, \( T_s = 30^\circ C \)
- Heat capacity of the body, \( C = 2 \, J/^\circ C \)
- Initial temperature of the body, \( \theta_1 \) (unknown)
- Final temperature of the body after 10 minutes, \( \theta_2 = 38^\circ C \)
- Time, \( t = 10 \, minutes = 600 \, seconds \)

2. **Apply Newton's Law of Cooling:**
According to Newton's Law of Cooling, we can express the change in temperature over time as:
\[ \frac{d\theta}{dt} = -k(\theta - T_s) \]
where \( k \) is the cooling constant.

3. **Integrate the Equation:**
Rearranging gives:
\[ \frac{d\theta}{\theta - T_s} = -k \, dt \]
Integrating both sides, we get:
\[ \ln|\theta - T_s| = -kt + C \]
where \( C \) is the integration constant.

4. **Solve for the Constant:**
At \( t = 0 \), let \( \theta = \theta_1 \):
\[ \ln|\theta_1 - T_s| = C \]
So, we have:
\[ \ln|\theta - T_s| = -kt + \ln|\theta_1 - T_s| \]

5. **Find the Cooling Constant \( k \):**
At \( t = 600 \, seconds \), \( \theta = 38^\circ C \):
\[ \ln|38 - 30| = -600k + \ln|\theta_1 - 30| \]
Simplifying gives:
\[ \ln(8) = -600k + \ln|\theta_1 - 30| \]

6. **Relate Initial and Final Temperatures:**
Rearranging gives:
\[ \ln|\theta_1 - 30| = \ln(8) + 600k \]
Thus, we can express \( \theta_1 \) in terms of \( k \):
\[ \theta_1 = 30 + 8e^{600k} \]

7. **Determine the Nature of the Graph:**
The temperature \( \theta \) decreases over time and approaches the surrounding temperature \( T_s \). The graph of \( \theta \) versus \( t \) will be a curve that starts at \( \theta_1 \) and approaches \( T_s \) asymptotically.

8. **Conclusion:**
The variation of \( \theta \) versus \( t \) is best represented as a decreasing curve that flattens out as it approaches \( 30^\circ C \).

### Final Answer:
The variation of \( \theta \) versus time \( t \) is best denoted by a decreasing curve that approaches \( 30^\circ C \).