A uniform wooden plank (cuboid) is floating on water is tied to the bottom of the pool such that in static equilibrium the diagonal plane of symmetry of its vertical cross section coincides with the level surface of water, as shown in figure. What is  specific gravity of the wood?

 

Centre of buoyancy of the submerged portion can be found as $\left(\frac{o+c+a}{3},\frac{o+o+h}{3}\right)$ 

Centre of gravity is $\left(\frac{a}{2},0\right)$  when we take torque about (c, h) = 0

We get $mg\left(\frac{a}{2}-c\right)=F_B\left(\frac{c+a}{3}-c\right)$

$\Rightarrow F_B=\frac{3mg}{2}\Rightarrow \mathrm{Re}lativedensity=\frac{1}{3}$

Alternate Method :

It is easy to see that the block would have submerged less than half the volume if it had not been             tied. So relative density $<\frac{1}{2}$ . Only one of the options match