A uniform rod of density ρ is placed in a wide tank containing a liquid of density ρ0(ρ0>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
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a
sinθ=12ρ0/ρ
b
sinθ=12 . ρ0ρ
c
sinθ=ρ/ρ0
d
sinθ=ρ0/ρ
answer is A.
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Detailed Solution
Let L = PQ = length of rod∴SP=SQ=L2Weight of rod, W=Alρg , actingAt point SAnd force of buoyancy,FB=Alρ0g , [l = PR]which acts at mid-point of PR.For rotational equilibrium,Alρ0g×l2cosθ=ALρg×L2cosθ⇒l2L2=ρρ0⇒lL=ρρ0From figure, sinθ=hl=L2l=12ρ0ρ