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^@`)

Explanation

Let `A` : area of cross-section of rod
`L` : length of rod
`sigma` : density of material of rod
Weight of plank `W = sigma AL g` (vertically upward). Acting at distance `(1)/(2)((h)/(cos theta)) f rom O`.
Taking torque about `O` (net torque `= 0`)
`W xx (L)/(2) sin theta =F_B xx (1)/(2) (h)/(cos theta) sin theta`
`sigma AL g((L)/(2)) sin theta = rho_omega A (h)/(cos theta) g. (h)/(2 cos theta) sin theta`
`sigma = 0.5 gm//c c, L = 1m, rho_omega = 1 gm//c c, h = 0.5 m`
`0.5 = ((0.5)^2)/(cos^2 theta) rArr cos^2 theta = 0.5 = (1)/(2)`
`cos theta = (1)/(sqrt(2)) rArr theta = 4.5^@`.
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