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A thin rod hinged at the bottom of a tank of two immiscible liquids: the period of its small oscillations
JEE Advanced 2026
Paper 1 · Question 9
Physics
Small oscillations of a hinged rod in two immiscible liquids - buoyancy torques in layered liquids, moment of inertia and the period
Moderate
A thin rod of density rho, hinged at the bottom of a tank with liquid 6 rho up to L/2 and 2 rho above, stands upright and rocks when nudged. Buoyancy of each liquid and the weight give a restoring torque rho A g L^2 theta, I = rho A L^3/3, so T = (2 pi/sqrt 3) sqrt(L/g) - n = 1.73, timed by the run.
Open the interactive simulationSub-topics covered
- Buoyancy acts at the centre of each immersed part; in layered liquids each layer pushes on its own part
- Torque about a hinge from vertical forces: F d sin(theta)
- Moment of inertia of a rod about one end: M L^2/3
- Small oscillations: omega^2 = (restoring torque per radian)/I
What it teaches
- Lower half in 6 rho: buoyancy 3 rho A g L at L/4 gives (3/4) rho A g L^2 theta; upper half in 2 rho: rho A g L at 3L/4 gives (3/4) rho A g L^2 theta
- The weight rho A L g at L/2 gives (1/2) rho A g L^2 theta the other way; net restoring torque rho A g L^2 theta
- I = rho A L^3/3, so omega^2 = 3g/L and T = (2 pi/sqrt 3) sqrt(L/g): n = sqrt 3 = 1.73
- The tilt-dependent submerged length L/(2 cos theta) changes the torque only at order theta^3
Simulation ID
- ADV-2026-P1-PHY-Q09 · revision 1
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