Question 2.13: A cube of side a floats with one of its axes vertical in a l...

Let the cube float with h as the submerged depth as shown in Fig. 2.36. For equilibrium of the cube,

Weight = Buoyant force

or, $h=a\left(S_{c} / S_{L}\right)=a / x$

where $S_{L} / S_{c}=x$

The distance between the centre of buoyancy B and centre of gravity G becomes

Let M be the metacentre, then

The metacentric height $M G=B M-B G=\frac{a x}{12}-\frac{a}{2}\left(1-\frac{1}{x}\right)$

According to the given condition

or, $x^{2}-6 x+6=0$

which gives $x=\frac{6 \pm \sqrt{12}}{2}=4.732,1.268$

Hence $S_{L} / S_{c}=4.732 \quad \text { or } \quad 1.268$