A block of metal is heated to a temperature much higher than the room temperature and placed in an evacuated cavity. The curve which correctly represents the rate of cooling ( $T$ is temperature of the block and $t$ is the time.)

Key Idea Newton's law of cooling is given by expression, $$ -\frac{d T}{d t}=k^{\prime}\left(T-T_0\right) $$ where, $k^{\prime}=\frac{k}{m s}$ and negative sign shows the rate of heat loss. The given expression can be rearranged by integrating as, $$ \int \frac{d T}{T-T_0}=-k^{\prime} \int d t $$ $$ \begin{array}{rlrl} \log _e\left(T-T_0\right) & =-k^{\prime} t+\log _e A\left(\because \log _e A=\text { Constant }\right) \\ & -\frac{d T}{\left(T-T_0\right)} & =k^{\prime} d t \\ \Rightarrow & \ln \left(T-T_0\right) & =-k^{\prime} t \\ \Rightarrow & & T & =e^{-k^{\prime} t}+T_0 \text { at } t \rightarrow \infty \\ \Rightarrow & & T & =T_0 \text {, at } t=0 \Rightarrow T \rightarrow \infty \end{array} $$ Hence, the graph as shown below, shows temperature of a body (done) varies exponentially with time from $T$ to $T_0\left(T_0 < T\right)$.
Thus, the correct option is (b).