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`)
,

Options

A. ` 15^@ `
B. `30 ^@ `
C. `45^@ `
D. `90^@`

Answer

C

Explanation

Three forces will act on the plank.
(a) Weight which will act at centre of plank.
(b) Upthrust which will act at centre of submerged portion.
`(c)` Force from the hinge at O.
Taking moments of all three forces about point O.
Moment of hinge force will be zero.
`therefore` Moment of W(clockwise) = Moment of F (anti-clockwise)
`therefore" "(Algrho)(1)/(2)sintheta=A(0.5sectheta)(rho_w)(g)((0.5sectheta)/(2))sintheta`, I=length of the plank
`therefore" "cos^(2)theta=((0.5)^(2)(1))/((1)^(2)(rho))=((0.5)^(2))/((0.5))=(1)/(2)" "(asI=1m)" "therefore" "costheta=(1)/(sqrt2)" "or" "theta=45^(@)`