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Question

# A wooden plank of length 1 m and uniform cross section is hinged at one end to the bottom of a tank as shown in figure. The tank is filled with water up to a height of 0.5 m. The specific gravity of the plank is 0.5. Find the angle θ (in degrees) that the plank makes with the vertical in the equilibrium position (exclude the case θ=0).

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Solution

## weight of the plank W=Aldg, where d is the density of the plank, A is the uniform cross section area of plank. l is the total length of the plank. Force of buoyancy B=(A0.5secθ)ρg where 0.5secθ is the submerged length. Since the plank is at rest i.e. it is in rotational equilibrium, we have W×l2sinθ=B×0.5secθsinθ2⇒Aldg×l2sinθ=(A0.5secθ)ρg×0.5secθsinθ2⇒Al(0.5ρ)g×l2sinθ=(A0.5secθ)ρg×0.5secθsinθ2∵d=0.5ρ⇒sec2θ=2⇒cosθ=1√2⇒θ=45∘

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