Vertical Circular Motion

Based on NCERT Class 11, Chapter 5 (Laws of Motion), in vertical circular motion the critical condition occurs at the highest point, where both gravity and any contact force point downward toward the centre. At the minimum speed, the bucket provides no push on the water, so gravity alone supplies the entire centripetal force: mg = m v² / r. The mass cancels, giving v_min = √(g r). Substituting, v_min = √(10 × 0.9) = √9 = 3 m/s. At this speed the water stays in the bucket because the required inward force exactly equals its weight. Option 9 m/s mistakenly uses g r without the square root. Option 1.5 m/s halves the correct value. Option 6 m/s doubles it, perhaps using 2g r. A subtle but important idea is that the string tension or normal force can never be negative, because a string can only pull and a surface can only push; this physical constraint is exactly what sets the critical minimum speed at the top. Below this speed the object would need an outward push that no real contact can provide, so it leaves the circular path. Plausibility check: the minimum top speed must be modest for a small radius, and √(g r) yielding 3 m/s is dimensionally correct in m/s and matches the well-known critical-speed condition, confirming the answer.