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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 $\frac{1}{\cos^{2} θ}$

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answer is 2.

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Detailed Solution

Let y be the length of the plank inside water, $y = \frac{0.5}{\cos θ}$

Let A be the cross-sectional area of the plank. Then the buoyant force on it is

$F_{b} = {Vρ}_{ω} g = (Ay) ρ_{ω} g$

Since the plank is in rotational equilibrium, so $\sum \vec{τ_{0}} = 0$ or

$mg \times \frac{1}{2} \sin θ - F_{b} \times \frac{y}{2} \sin θ = 0$

$\Rightarrow (A l \times 0.5 ρ_{w} g) \times \frac{l sinθ}{2} = ({Ayρ}_{w} g) \times \frac{ysinθ}{2}$

$\Rightarrow 1 \times 0.5 \times \frac{1}{2} = {(\frac{0.5}{\cos θ})}^{2} \times \frac{1}{2} (∵ l = 1)$

$\Rightarrow \cos^{2} θ = \frac{1}{2} \Rightarrow θ = 45^{o}$

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