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 (in degree) that the plank makes with the vertical in the equilibrium position (Exclude the case $θ = 0$ ).

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answer is 45.

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Detailed Solution

Let y is the length of the plank inside water

$∴ y = \frac{0.5}{\cos θ}$

Let A be the cross-sectional area of the plank, then buoyant force on it
$F_{b} = V ρ_{w} g = (A y) ρ_{w} g$
Since plank is in rotational equilibrium, so

$\sum {\vec{τ}}_{o} = 0$

$\begin{array}{l} \Rightarrow m g \times \frac{ℓ}{2} \sin θ - F_{b} \times \frac{y}{2} \sin θ = 0 \\ \Rightarrow m g ℓ - F_{b} \times y = 0 \\ \Rightarrow (A ℓ \times 0.5) g ℓ - (A y) (1) y = 0 \\ \Rightarrow 0.5 ℓ^{2} = y^{2} \\ \Rightarrow 0.5 \times {(1)}^{2} = {(\frac{0.5}{\cos θ})}^{2} \\ \Rightarrow \cos^{2} θ = \frac{1}{2} \\ \Rightarrow c o s θ = \frac{1}{\sqrt{2}} \\ \Rightarrow θ = 45^{\circ} \end{array}$