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 force acting at the pivoted end of the rod in terms of mass `m` of the rod.

Options

A. `(sqrt(2)-1)mg`
B. `(sqrt(2)+1)mg`
C. `(sqrt(3)+1)mg`
D. `(sqrt(2)+3)mg`

Answer

A

Explanation

Weight of the rod `w=LArhog`
(`A=`area of cross section `rho =`density of material)
Submerged length of the given by
`OT=L/(2sintheta)`
Buoyant force `B=(L/(2sintheta))Arho_(w)g`
From Archimedes' principle (`rhow=` density of water).

This force acts at `P`, where `OP=(OT)/2=l/(4sintheta)`
Balancing torque about `O` is
`WOQcostheta=BOPcostheta`
`(LArhog)L/2=(L/(sintheta))Arho_(w)g(L/(4sintheta))`
`sin^(2)theta =(rho_(w))/(4rho)impliessin^(2)theta=1/2impliestheta=45^@`
Also balancing forces `R+W=B`
`R=B-W=(rhoArho_(w)g)/(2sintheta)-LArhog`
`=LArhog[1/(sintheta)-1]` (`:'rho_(w)=2rho)=(sqrt(2)-1)mg`