A liquid in a beaker has temperature `theta(t)` at time t and `theta_0` is temperature of surroundings, then according to Newton's law of cooling the correct graph between `log_e( theta-theta_0)` and t is :

Options

A.
B.
C.
D.

Answer

To solve the problem, we will use Newton's Law of Cooling, which states that the rate of heat loss of a body is directly proportional to the difference in temperature between the body and its surroundings. Let's go through the steps to derive the relationship and determine the correct graph.

### Step-by-Step Solution:

1. **Understanding Newton's Law of Cooling**:
According to Newton's Law of Cooling, the rate of change of temperature of the liquid is given by:
\[ \frac{d\theta}{dt} = -k(\theta - \theta_0) \]
where \( \theta \) is the temperature of the liquid at time \( t \), \( \theta_0 \) is the temperature of the surroundings, and \( k \) is a positive constant.

2. **Rearranging the Equation**:
We can rearrange the equation to separate the variables:
\[ \frac{d\theta}{\theta - \theta_0} = -k \, dt \]

3. **Integrating Both Sides**:
Now we will integrate both sides. The left side integrates with respect to \( \theta \) and the right side with respect to \( t \):
\[ \int \frac{d\theta}{\theta - \theta_0} = -k \int dt \]
This gives us:
\[ \log(\theta - \theta_0) = -kt + C \]
where \( C \) is the constant of integration.

4. **Exponentiating Both Sides**:
To eliminate the logarithm, we exponentiate both sides:
\[ \theta - \theta_0 = e^{-kt + C} = e^C e^{-kt} \]
Let \( C' = e^C \), then:
\[ \theta - \theta_0 = C' e^{-kt} \]

5. **Taking the Logarithm**:
We can express \( \log(\theta - \theta_0) \) in terms of \( t \):
\[ \log(\theta - \theta_0) = \log(C') - kt \]

6. **Identifying the Form**:
This equation can be rearranged to:
\[ \log(\theta - \theta_0) = -kt + \log(C') \]
This is in the form of a linear equation \( y = mx + b \) where:
- \( y = \log(\theta - \theta_0) \)
- \( x = t \)
- \( m = -k \) (the slope)
- \( b = \log(C') \) (the y-intercept)

7. **Graph Characteristics**:
The graph of \( \log(\theta - \theta_0) \) versus \( t \) will be a straight line with a negative slope (since \( k > 0 \)).

8. **Conclusion**:
Therefore, the correct graph will be a straight line that slopes downwards. Among the given options, the first graph represents this relationship correctly.

### Final Answer:
The correct graph between \( \log(\theta - \theta_0) \) and \( t \) is the first option.