The 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 `theta` that the plank makes with the vertical in the equilibrium position. (Exclude the case `theta=0`)
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Explanation

The force acting on the plank are shown in the figure. The height of water level is `l`. The length of the plank is `2l`. The weight of the plank acts through the centre `B` of the plank.
We have `OB = l`. The buoyant force `F` acts though the point `A` which is the middle point of the dippeed part `OC` of the plank. We have `OA = (OC)/(2) = (l)/(2 cos theta)`
Let the mass per unit length of the plank be `rho`.
Its weight `mg = 2lrhog`.
The mass of the part `OC` of the plank `= ((l)/(costheta))rho`.
The mass of water displaced `= (1)/(0.5) (l)/(cos theta)rho = (2lrho)/(cos theta)`
The buoyant force `F` is, therefore, `F = (2lrhog)/(cos theta)`.
Now, for equilibrium, the torque of `mg` about `O` should balance the torque of `F` about `O`.
So, `mg (OB) sintheta = F(OA) sintheta`
or, `(2lrho)l = ((2lrho)/(cos theta)) ((l)/(2 cos theta))` or, `cos^(2)theta = (1)/(2)` or, `costheta = (1)/sqrt(2)`
or, `theta = 45^(@)`.