Question 3.13: A cylinder of diameter d with specific gravity s floats on w......

From Figure 3.31,

OG = \frac{l}{2}

Let y be the depth of immersion. Then

Weight of cylinder = Weight of water displaced

(\frac{\pi }{4} )d^{2}lws=(\frac{\pi }{4} )d^{2}wy

Therefore,                                        y=sl      or       OB = \frac{sl}{2}

Now,

OM = OB + BM = \frac{sl}{2}+\frac{({\pi d^{4} }/{64})}{({\pi d^{2} }/{4})sl{}}=\frac{sl}{2}+\frac{d^{2} }{16sl}

For stable equilibrium, we have

OM > OG

\frac{sl}{2}+\frac{d^{2} }{16sl}\gt\frac{l}{2}

Solving for 1, we get

l\lt \frac{d}{\sqrt{8s(1-s)} }