An open circular cylinder 40 cm diameter and 100 cm long contains water up to a height of 80 cm. The maximum speed at which the cylinder is to be rotated about its vertical axis so that no water spills will be (Take g = 10 m/s²)
- A√200 rad/sCorrect
- B√180 rad/s
- C√160 rad/s
- D√140 rad/s
Solution
Concept
Maximum speed of rotation without spilling
Formula
Height of paraboloid h = ω²R²/(2g); rise at wall = ω²R²/(4g)
Step-by-step solution
1. Free board = 100 − 80 = 20 cm = 0.2 m. 2. For an open cylinder, the rise at the wall above the original level = ω²R²/(4g). 3. 0.2 = ω² × (0.2)² / (4 × 10), so ω² = 0.2 × 40 / 0.04 = 200. 4. ω = √200 rad/s.
Explanation
When a partly filled cylinder rotates, the liquid at the axis falls and at the wall rises by the same amount, ω²R²/(4g). The limit without spilling is that this rise equals the free board of 20 cm.
Shortcut
Rise at wall = fall at axis = ω²R²/4g = free board.
Common mistake
Using h = ω²R²/2g (total height of the paraboloid) as the free board.
Correct answer: A) √200 rad/s