B

Step 1: Analyze the system. The plate is suspended, and its weight acts downwards at its center of mass, which is at the midpoint.
Step 2: Calculate the torque (\tau) about point A when point B is removed. The weight of the plate (W = mg) can be calculated as:
W = 20 kg × 10 m/s² = 200 N.
Step 3: The distance from A to the center of mass is half the width of the plate (let's say width = L). Thus, distance (d) = L/2. The torque is given by \( \tau = W \cdot d = 200 N \cdot \frac{L}{2} = 100L \).
Step 4: Moment of inertia (I) of the plate about point A is calculated as: \( I = \frac{1}{3} m L^2 = \frac{1}{3} (20 kg) L^2 = \frac{20}{3} L^2 \).
Step 5: Using Newton's second law for rotation, the angular acceleration (\alpha) can be expressed as: \( \alpha = \frac{\tau}{I} = \frac{100L}{\frac{20}{3} L^2} = \frac{100 \cdot 3}{20L} = \frac{15}{L} rad/s².
Hence, substituting L with the requisite width will lead you to the specific angular acceleration. Therefore, the answer is option B.