B

Step 1: The potential energy (PE) at height h when the sphere is displaced is given by PE = mgh.
Step 2: When the sphere is released, it converts this potential energy back to kinetic energy (KE) near the lowest point: KE = \frac{1}{2}mv^2. At the lowest point, all PE becomes KE. Thus, mgh = \frac{1}{2}mv^2.
Step 3: The speed v at the lowest point is then v = \sqrt{2gh}.
Step 4: When the sphere reaches the bar A, it momentarily comes to rest, and all kinetic energy will convert back into potential energy as it rises again. Let the height it reaches after hitting the bar be h'.
Step 5: Since we are assuming a perfectly elastic collision, the energy relation holds: \frac{1}{2}mv^2 = mg h' => mgh = mg h' => h' = h.
Step 6: Therefore, the height to which it will rise after being interrupted by the bar will also be h.
Thus, the answer is B.