NTA Abhyas JEE Main2020PhysicsMechanical Properties of FluidsPractice
A thin cylindrical rod PQ of length L and density d 1 is pivoted at its lowest point P , inside a stationary homogeneous and non-viscous liquid of density d 2 . The rod is always fully submerged inside the liquid and is free to rotate in a vertical plane about a horizontal axis passing through P . If d 1 < d 2 , then the time period of small angular oscillations of the rod about its vertical equilibrium position
Options
- AT = 2 π 2 L 3 g d 1 d 2 - d 1
- BT = 2 π L 3 g d 2 d 2 - d 1
- CT = 2 π 2 L g d 1 d 2 - d 1
- DT = 2 π 2 L 3 g d 2 - d 1 d 1
Correct answer
A. T = 2 π 2 L 3 g d 1 d 2 - d 1
Step-by-step solution
Consider the diagram in a displaced position. The weight and upthrust F B , both pass through the centre of gravity G W = π r 2 L   d 1 g F B = π r 2 L d 2 g The restoring torque about the pivot is τ = - π r 2 L g d 2 - d 1 L 2 θ α = - π r 2 L g d 2 - d 1 L 2 θ π r 2 L d 1   L 2 3 α = - 3 g 2 L d 2 - d 1 d 1 θ T = 2 π 2 L 3 g d 1 d 2 - d 1