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 $\ln (\theta -{\theta }_0)$ and $t$ is:

According to Newton's Law of Cooling:$\frac{d\theta }{dt}=-k(\theta -{\theta }_0)$$\frac{d\theta }{\theta -{\theta }_0}=-kdt$ Integrating both sides:$\ln (\theta -{\theta }_0)=-kt+C$ This is an equation of the form $y=mx+c$, which represents a straight line.Here, $y=\ln (\theta -{\theta }_0)$, $x=t$, and the slope $m=-k$ (which is negative).Thus, the graph is a straight line with a negative slope.