A thin vertical uniform wooden rod is pivoted at the top and immersed in water as shown. The container is slowly raised. At a certain moment, the equilibrium becomes unstable. If density of water is $9/5$ times the density of wood, then ratio of total length of rod to the submerged length of rod, at that moment is
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where $F_{B}=\frac{9 p}{5}(x A) g$
Balancing torque about hinge
$\left(\rho(\mathrm{A}) \mathrm{g} \times\left(\frac{\ell}{2} \sin \theta\right)\right.$
$=\left[\frac{9 \rho}{5} \times(\mathrm{x} \mathrm{A})\right] \mathrm{g} \times\left[\left(\ell-\frac{\mathrm{x}}{2}\right) \sin \theta\right]$
[where $\theta$ is very small]
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