Topic: Fluids
Concepts: Buoyant force, Statics, Torque from weight

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Solution:

We choose the hinge as our pivot point and balance torques on the rod. Only the weight of the rod and the buoyant force contribute to the torque. The weight $W$ acts downward at the CM of the rod, distance $l/2$ away from the pivot. The buoyant force $F_B$ acts upward at the CM of the underwater portion. If $f$ is the fraction of the rod above water, then $F_B$ acts at a distance

away from the pivot. Balancing torques,

The weight is given by

where $A$ is the cross-sectional area of the rod. By Archimedes’ principle,

Substituting into the earlier equation,

Since $\rho_{\text{rod}}=\frac{5}{9}\rho_{\text{water}}$,

Solving for $f$,

so the answer is D.