The population $p$ of the city at time $t$ is given by $\frac{d p}{d t}=\frac{p}{2}-100$. If initial population is 100 then $\mathrm{p}=$

The differential equation $\frac{dp}{dt} = \frac{p}{2} - 100$ with initial condition $p(0) = 100$ is separable.
Rewriting gives $\frac{dp}{dt} = \frac{p - 200}{2}$, leading to $\frac{dp}{p - 200} = \frac{1}{2} dt$.
Integrating both sides yields $\ln|p - 200| = \frac{1}{2}t + C$.

Exponentiating produces $|p - 200| = e^{C}e^{t/2}$.
The general solution is $p(t) = 200 + Ae^{t/2}$, where $A$ is a constant.

Applying the initial condition $p(0) = 100$ gives $100 = 200 + A$, so $A = -100$.
Thus, the particular solution is $p(t) = 200 - 100e^{t/2}$.

This corresponds to option B.

\boxed{B}