A wooden plank of length 1m and uniform cross-section is hinged at one end to the bottom of a tank as shown in fig. The tank is filled with water upto a hight 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 case `theta=theta^@`)

Answer

D

Explanation

`OA=(OB)/(2)=(0.5 sec theta)/(2)=0.25 sec(theta)`
About point `O`, clockwise moment of
`w=` anticlockwise moment of `F`.
`:. w(L/2 sin theta)=F(OA sin theta)=F(0.25 sin theta. sec theta)`
Given, `L=1m`
`:. cos theta =(F)/(2w)=((0.5 sec theta)(A)(1.0)g)/(2(1)(A)(0.5)g)`
or, `cos^(2)theta=1/2, cos theta=(1)/(sqrt(2))`
`:. theta=45^(@)`