A rod of length 6 m has specific gravity $\mathrm{\rho }(=25/36)$. One end of the rod is tied to a 5 m long rope, which in turn is tied to the floor of a pool 10 m deep, as shown. Find the length (in m) of the part of rod which is out of water.

Assume x is the portion of rod outside water

$\theta$ is the angle made by rod with horizontal
Torque acting on the rod about the point connecting rope and rod is zero

$$\begin{gathered} {\mathrm{F}}_{\mathrm{B}}.{\mathrm{l}}_{\perp }=\mathrm{mg} {\mathrm{r}}_{\perp } \\ (6-\mathrm{x}){\mathrm{A\rho }}_{\mathrm{\omega }}\mathrm{g}\cdot \frac{6-\mathrm{x}}{2}\mathrm{cos\theta }=6{\mathrm{A\rho }}_{\mathrm{r}}\mathrm{g}.3\mathrm{cos\theta } \\ \frac{{\left(6-\mathrm{x}\right)}^2}{2}=18\frac{{\mathrm{\rho }}_{\mathrm{r}}}{\mathrm{\rho }}=18\times \frac{25}{36} \\ \therefore 6-\mathrm{x}=5 \\ \mathrm{x}=1 \mathrm{m} \end{gathered}$$