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Q.

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 the figure. The tank is filled with water up to a height of 0.5 m. The specific gravity of the plank is 0.5. If the angle θ by the inclination of that the plank makes with the vertical in the equilibrium position (exclude the case θ = 0). Find the value of 1cos2⁡θ.

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answer is 2.

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Detailed Solution

Let y be the length of the plank inside water.y=0.5cos⁡θLet A be the cross-sectional area of the plank. Then the buoyant force on it is Fb=Vρωg=(Ay)ρwgSince the plank is in rotational equilibrium, so ∑τ→0=0 or mg×l2sin⁡θ−Fb×y2sin⁡θ=0 or mgl−Fb×y=0 or (Al×0.5)gl−(Ayρwyg=0 or 0.5l2=y2 or 0.5×(1)2=0.5cos⁡θ2 or cos2⁡θ=12
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