Two ideal diodes D_1 and D_2 are connected in two parallel branches...

The potential difference between nodes A and B is given by $V(t) = V_A - V_B = 200 \sin(50\pi t)$.

At time $t = 10 \text{ ms} = 10 \times 10^{-3} \text{ s} = 0.01 \text{ s}$, the voltage is:

$V(0.01) = 200 \sin(50\pi \times 0.01)$

$V(0.01) = 200 \sin(0.5\pi) = 200 \sin\left(\dfrac{\pi}{2}\right) = 200 \text{ V}$

Since $V(0.01) > 0$, we have $V_A - V_B > 0$, which implies $V_A > V_B$. Thus, node A is at a higher potential than node B.

From the circuit diagram, the anode (p-side) of diode $D_1$ is connected to node A and its cathode (n-side) is connected towards node B. Since $V_A > V_B$, $D_1$ is forward biased.

The cathode (n-side) of diode $D_2$ is connected to node A and its anode (p-side) is connected towards node B. Since $V_A > V_B$, $D_2$ is reverse biased.

Therefore, $D_1$ is forward biased and $D_2$ is reverse biased.

Answer: $D_1$ is forward biased, $D_2$ is reverse biased