JEE Advanced 2026 · Paper 1 · Physics Q9

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PhysicsNumerical+4Question 25 of 48

A tank contains two immiscible liquids of densities $6\rho$ and $2\rho$ . The higher density liquid is filled up to a height $L/2$ from the bottom. A thin rod of density $\rho$ 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 $\dfrac{2\pi}n\sqrt{\dfrac Lg}$ , where $g$ is the acceleration due to gravity. The value of $n$ is:

A tank: the lower half (height L/2) holds the denser liquid (dark), the upper part the lighter liquid (light grey). A thin vertical rod of length L stands hinged at the bottom centre, fully immersed; a small double arrow at its top shows it oscillating.

Your answer

Answer and worked solution

Official answer: 1.7

Take the rod's cross-section $A$ and tilt it by a small angle $\theta$ . Torques about the hinge (to first order in $\theta$ ):

Weight $\rho ALg$ at $\frac L2$ , tipping it over: $-\rho ALg\cdot\frac L2\theta$ .
Buoyancy, lower half $6\rho A\frac L2g$ at $\frac L4$ , restoring: $+\frac34\rho AL^2g\,\theta$ .
Buoyancy, upper half $2\rho A\frac L2g$ at $\frac{3L}4$ , restoring: $+\frac34\rho AL^2g\,\theta$ .

(The shift of the liquid interface along the tilted rod changes these only at order $\theta^2$ .)

Net restoring torque: $\left(\frac34+\frac34-\frac12\right)\rho AL^2g\,\theta=\rho AL^2g\,\theta$ .

With $I=\frac13\rho AL\cdot L^2$ :
$\omega^2=\frac{\rho AL^2g}{\rho AL^3/3}=\frac{3g}{L},\qquad T=\frac{2\pi}{\sqrt3}\sqrt{\frac Lg}.$
So $n=\sqrt3\approx1.73$ .

Topics:Simple and compound pendulum Buoyancy and Archimedes' principle Torque and rotational dynamics

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JEE Advanced 2026 · Paper 1, Physics question 9