A wooded plank of length 1m and uniform cross-section is hinged at one end to the bottom of a tank as shown in figure tank is filled with water upto a height 0.5m 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 `costheta=0^(@)`)

Explanation

The force acting on the plane are shown in the ure. The height of water level is l=0.5m. The length of the plank is 1.0=2l. The weight of the lane acts through the centre B of the plank. We have OB=l. The buoyant fore F acts through the point A which is the middle point of the dipped part OC of the plank.
We have `OA=(OC)/2=l/(2costheta) `
Let the mass per unit length of the plane be rho`.
it weight `mg=2lrhog`.
The mass of the part OC of the plank `=(l/(costheta))rho`.
The mass of water displaced `=1/0.5 l/(costheta)rho=(2/p)/(costheta)`
The buoyant force F is therefore `F=(2lrhog)/(costheta)`
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)=((2lrho)/(costheta))(l/(2costheta))`
`or, cos^2theta= 1/2`
`or costheta=1/sqrt2, or theta=45^@`