Consider two concentric thin wall steel tubes, the larger one closed at the bott
ID: 1849930 • Letter: C
Question
Consider two concentric thin wall steel tubes, the larger one closed at the bottom and the inner one open. The tubes are filled with a fluid. A well fitting cylindrical plug that slides easily is dropped in the inner tube and a well fitting annular ring (doughnut) which slides easily is dropped between the two tubes. The weight of the doughnut plug (F2) is 3X the weight of the cylindrical plug (F1). After oscillating a little the plugs settle down concentrically with one another as shown. What must be the relationship between the radii of the tubes? (Ignore the wall thickness of the tubes)Explanation / Answer
In this question the for the given concentric equilibrium, the ratio of their weights should be equal to ratio of there cross sectional area. Let inner radius = Ra and outer radius = Rb. then, 3 = [pi*(Rb^2-Ra^2)]/pi*Ra^2 On simplifying we get Rb^2/Ra^2 = 4 and so Rb/Ra = 2.