A uniform rod of length `L` pivoted at the bottom of a pond of depth `L//2` stays in stable equilibrium as shown in the figure.

Find the angle `theta` if the density of the material of the rod is half of the density of water.

Explanation

(a) Weight of the rod W = LA`rho` g
(A = area of cross section, `rho` = density of material)
Submerged length of the rod
`OT = (L)/(2 sin theta)`
Buoyant force, `B = ((L)/(2 sin theta)) A rho_(w)g`
(from Archimedies principle) (`rho_(w)` = density of water)
This acts at P, where `OP = (OT)/(2) = (l)/(4 sin theta)`
Balancing torque about O
`W. OQ cos theta = B. OP cos theta`
`(LA rho g) L//2 = ((L)/(2 sin theta))A rho_(w) g((L)/(4 sin theta))`
`sin^(2) theta = (rho_(w))/(4 rho)`
`rArr" "sin^(2)theta = 1//2 rArr theta = 45^(@)`

(b) Also, balancing forces
R + W = B
`R = B - W = (rho A rho_(w)g)/(2 sin theta) - LA rho g`
`= LA rho g [(1)/(sin theta) - 1]" "(because rho_(w) = 2rho)`
`= ( sqrt(2) - 1) mg`