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

Newton’s law of cooling states that the rate of fall of temperature of a body is directly proportional to the temperature excess above the surroundings:

$$\frac{d\theta}{dt} \;=\; -k\bigl(\,\theta(t)-\theta_0\,\bigr) \qquad -(1)$$
where $$k \gt 0$$ is the cooling constant.

Rewrite equation (1) by separating variables:

$$\frac{d\theta}{\theta-\theta_0} \;=\; -k\,dt \qquad -(2)$$

Integrate both sides from the initial temperature $$\theta_i$$ at $$t=0$$ to temperature $$\theta$$ at time $$t$$:

$$\int_{\theta_i}^{\theta}\frac{d\theta'}{\theta'-\theta_0} \;=\; -k\int_{0}^{t} dt'$$

This gives:

$$\ln\!\bigl(\theta-\theta_0\bigr) - \ln\!\bigl(\theta_i-\theta_0\bigr) \;=\; -kt$$

Re-arrange to the form that will be plotted:

$$\ln\!\bigl(\theta-\theta_0\bigr) \;=\; -kt + \ln\!\bigl(\theta_i-\theta_0\bigr) \qquad -(3)$$

Equation (3) is of the form $$y = mx + c$$, a straight line where:

• Vertical axis $$\;y = \ln(\theta-\theta_0)$$
• Horizontal axis $$\;x = t$$
• Slope $$\;m = -k$$ (negative)
• Intercept $$\;c = \ln(\theta_i-\theta_0)$$ (positive for $$\theta_i \gt \theta_0$$)

Hence, the required graph is a straight line with a negative slope cutting the vertical axis at $$\ln(\theta_i-\theta_0)$$.

Therefore, the correct choice is:
Option A which is: a straight line with negative slope.