01The question
Q.9 A tank contains two immiscible liquids of densities 6ρ and 2ρ. The higher density liquid is filled up to a height L/2 from the bottom. A thin rod of density ρ and length L is fully immersed and hinged at the bottom so that it can oscillate freely, as shown in the figure. If the rod is slightly disturbed from its equilibrium, the time period of small oscillations is (2π/n)√(L/g), where g is the acceleration due to gravity. The value of n is:
02The experiment
Answer—
Press START: the rod is nudged, it rocks, and the run times its oscillations.
Ready.
Nudge
Forces
Speed
03Detailed solution
Answer: n = — (= √3; the accepted range is 1.70 to 1.75)
- Why it stands up. The rod (density ρ) is lighter than both liquids, so buoyancy exceeds its weight. Tilted by θ about the hinge, it is pushed back upright. Take the cross-section as A.
- Torques about the hinge for a small tilt θ. Each vertical force F acting a distance d along the rod has lever arm d sin θ. The lower L/2 of the rod sits in the 6ρ liquid and the upper L/2 in the 2ρ liquid.
buoyancy (lower): 6ρ · A(L/2) · g · (L/4) = (3/4)ρAgL² buoyancy (upper): 2ρ · A(L/2) · g · (3L/4) = (3/4)ρAgL²weight: ρAL · g · (L/2) = (1/2)ρAgL² ⇒ restoring torque τ = (3/4 + 3/4 − 1/2) ρAgL² sin θ = ρAgL² θ.
- Moment of inertia about the hinge: I = (1/3)(ρAL)L² = ρAL³/3.
- Period. ω² = τ/(Iθ) = ρAgL²/(ρAL³/3) = 3g/L, so
T = 2π√(L/3g) = (2π/√3)√(L/g) ⇒ n = √3 ≈ 1.73.
- A detail. As the rod tilts, the interface stays at height L/2, so the length of rod in the dense liquid becomes L/(2 cos θ). That changes the torque only at order θ³. The small-oscillation period is unchanged, and the experiment's 12° nudge shows the tiny amplitude effect.
Understand the concept
For a body pivoted in a fluid, weight and buoyancy both give torques about the pivot, and here buoyancy wins. The rod behaves like an upside-down pendulum held up by the liquid. Each liquid pushes on the part of the rod it surrounds, at that part's centre, so the denser lower liquid acts on a short lever arm and the lighter upper one on a long arm. Here the two buoyancy torques happen to be equal. Their sum minus the weight's torque sets the "spring constant" κ = ρAgL², and the period is 2π√(I/κ). As in the usual textbook treatment, the liquid's drag and the fluid dragged along with the rod (added mass) are neglected.
Key takeaways
- Buoyancy acts at the centre of the displaced volume, part by part in layered liquids.
- Torque about a hinge from a vertical force F at distance d along the rod: F d sin θ.
- Small oscillations: ω² = (net restoring torque per radian)/I. Here ω² = 3g/L and n = √3.
- I of a rod about one end is ML²/3.
Verification: an independent derivation, a sympy route, a numerical integration of the full equation of motion (including the L/(2 cos θ) submerged length) with the period timed from zero crossings, and the page's own engine, run by tests/verify_p1phyq09.py. The official key (1.70 to 1.75) and the MathonGo solution (1.73) agree. Status: script-verified by the PrayogX simulation factory; no review by a person has been recorded. The model uses L = 1 m and g = 9.8 m/s², and n does not depend on either.
How to use this simulation
- Press ▶ START EXPERIMENT. The upright rod is nudged by the chosen angle and released.
- Watch the arrows: the two buoyancy forces (in each liquid) push the rod back, and its weight pulls it over.
- The trace on the right records the angle against time. The run times several swings and computes n = 2π/(T√(g/L)).
- Try Nudge 3°, 6° and 12°: small swings all give n ≈ 1.73.
- Replay runs it again. Reset clears it. Classroom fills the screen for a projector.