A uniform rod of density ρ is placed in a wide tank containing a liquid of density ρ$$0($$ρ$$0$$ρ$$)$$. The depth of liquid in the tank is half the length of the rod. The rod is in equilibrium, with its lower end resting on the bottom of the tank. In this position the rod makes an angle θ with the horizontal

Question

A uniform rod of density ρ is placed in a wide tank containing a liquid of density  ρ$$0($$ρ$$0$$>ρ$$)$$. The depth of liquid in the tank is half the length of the rod. The rod is in equilibrium, with its lower end resting on the bottom of the tank. In this position the rod makes an angle  θ with the horizontal

Infinity Learn · Accepted Answer

Let L $$=$$ PQ $$=$$ length of rod∴$$SP=SQ=L2Weight$$ of rod, $$W=Al$$ρg , actingAt point SAnd force of buoyancy,$$FB=Al$$ρ$$0g$$ , [l $$=$$ PR]which acts at $$mid-point$$ of PR.For rotational equilibrium,Alρ$$0g$$×$$l2cos$$⁡θ$$=AL$$ρg×$$L2cos$$⁡θ⇒$$l2L2=$$ρρ$$0$$⇒$$lL=$$ρρ$$0From$$ figure, $$sin$$θ$$=hl=L2l=12$$ρ$$0$$ρ

Remember concepts with our Masterclasses.

Ready to Test Your Skills?

free study material