A wooden plank OP of length 1m and uniform cross-section s is hinged at one end to the bottom of a tank. The tank is filled with water up to a height h m. The specific gravity $ρ_{s}$ of the plank is 0.5. In the equilibrium position, the plank makes angle $θ$ with the vertical as shown in figure. Then, if $θ = 45^{0}$ ,

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the value of h is 0.5 m.

horizontal component of reaction at O is zero.

Centre of buoyancy would get shifter if $ρ_{s}$ is altered

location of center of buoyancy is proportional to $(\frac{ρ_{s}}{ρ_{l}})$

answer is A, B, C.

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Detailed Solution

The situation is shown in the figure. The upthrust $F_{B}$ acts through G. G is the centre of gravity. The reaction R acts through O. Let the length of the rod inside water be $l^{1}$ Taking moments about O,

$F_{B} \times a^{1} = m g \times a a = (\frac{l}{2}) \sin θ ∴ l^{1} = l \sqrt{\frac{ρ_{s}}{ρ_{l}}} = 1 \times \sqrt{0.5}$

$A l s o ∴ h = l^{1} \cos θ = l^{1} \cos 45^{0} = 1 \times \sqrt{0.5} \times \frac{1}{\sqrt{2}} = \frac{1}{2} m$

$∴ = 0.5 m$

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