a
$(a)$ Given, $V_{o}=\left\{\begin{array}{l}+10 V \text { if } V_{+} > V_{-} \\ -10 V \text { if } V_{+} < V_{-}\end{array}\right.$
Initially, when the capacitor is fully charged,
$V_{-}=V_{C}=V_{o}$
From the circuit,
$V_{+}=V_{o}\left(\frac{R}{R+R}\right)=\frac{V_{0}}{2}$
As, $V_{-}>V_{+}$, so $\quad V_{o}=-10 \,V$
For $V_{o}=-10 V , \quad V_{+}=\frac{-10}{2}=-5 \,V$
and $\quad V_{-}=-10 \,V$
i.e., $\quad V_{+}>V_{-}$
and $\quad V_{o}=+10 \,V$
As, the values are fixed, so correct graph is shown in option $(a)$
