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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 upto a height 0.5 m. The specific gravity of the plank is 0.5. Find the angle θ that the plank makes with the vertical in the equilibrium position. (Exclude the case θ=0∘)

A
30
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B
45
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C
60
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D
90
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Solution

## The correct option is B 45∘For equilibrium Fnet=0 and τnet=0 Let FB be the buoyant force acting on the plank and m be the mass of plank. Taking moment about O, mg×l2sinθ=FB(l−x2)sinθ ⇒mg×l=FB(l−x) .......(1) Also FB= wt. of fluid displaced ⇒FB=[(l−x)A]×ρwg And, m=(lA)0.5ρw where A is the area of cross section of the rod. Putting the value of FB and m in Eq.(1), (lA)0.5ρwg×l=[(l−x)A]ρwg×(l−x) Here, l=1 m ∴(1−x)2=0.5⇒x=0.293 m From the diagram, cosθ=0.51−x=0.50.707⇒θ≈45∘ Hence, option (b) is the correct answer.

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