Due to its higher density, cold water stays close to the bottom of a rectangular vessel which is filled up to the height of h = 30 cm. We assume that the density of water in the vessels grows linearly with increasing depth. At the water level, the density is equal to ${\mathrm{\rho }}_1=996{\mathrm{kgm}}^{-3}$at top, while the density ${\mathrm{\rho }}_{\mathrm{b}}$ at the bottom of the vessel is unknown. Determine this density using the fact, that a homogeneous rod with density ${\mathrm{\rho }}_{\mathrm{r}}=997{\mathrm{kgm}}^{-3}$ and length h immersed in the water and fixed by one of its ends at the water level makes an angle of $\mathrm{\varphi }={60}^{\circ }$ with the vertical. If ${\text{ρ}}_{\mathrm{b}}=996+\mathrm{x}$ , then the value of $\mathrm{x}$ is

At depth x, the density is determined by $\mathrm{\rho }\left(\mathrm{x}\right)={\mathrm{\rho }}_1+\frac{\mathrm{x}}{\mathrm{h}}\left({\mathrm{\rho }}_{\mathrm{b}}-{\mathrm{\rho }}_1\right)$

 For the rod to be at equilibrium, it is required that the total torque acting on the rod is zero, i.e.,

$\int \mathrm{dM}=0$

We can express the elementary torque acting on an infinitely small section of the rod as $\mathrm{dM}=\mathrm{x}\left({\mathrm{dF}}_{\mathrm{vz}}-{\mathrm{dF}}_{\mathrm{g}}\right)$ where the elementary forces ${\mathrm{dF}}_{\mathrm{vx}}$ are given by

          ${\mathrm{dF}}_{\mathrm{vz}}=\frac{\mathrm{mg}\rho \left(\mathrm{x}\right)}{{\rho }_{\mathrm{r}}\mathrm{h}}\mathrm{dx},{\mathrm{dF}}_{\mathrm{g}}=\frac{\mathrm{mg}}{\mathrm{h}}\mathrm{dx}$, 

Where m is the mass of the rod. Integrating from 0 to $\mathrm{hcos}\mathrm{\varphi }$ (where our x–axis is directed vertically downwards and $\mathrm{hcos}\mathrm{\varphi }$ is the x–coordinate of the lower end of the rod), we have

        $\begin{array}{l}{\int }_0^{\mathrm{hcos}\mathrm{\varphi }} \mathrm{x}\left(\frac{\mathrm{mg}}{{\mathrm{\rho }}_{\mathrm{r}}\mathrm{\ell }}\left({\mathrm{\rho }}_1+\frac{\mathrm{x}}{\mathrm{h}}\left({\mathrm{\rho }}_{\mathrm{b}}-{\mathrm{\rho }}_1\right)\right)-\frac{\mathrm{mg}}{\mathrm{\ell }}\right)\mathrm{dx}=0\\ \frac{{\mathrm{\rho }}_1}{2{\mathrm{\rho }}_{\mathrm{r}}\mathrm{\ell }}+\frac{\mathrm{hcos}\mathrm{\varphi }}{3{\mathrm{p}}_{\mathrm{r}}\mathrm{h\ell }}\left({\mathrm{\rho }}_{\mathrm{b}}-{\mathrm{\rho }}_1\right)-\frac{1}{2\mathrm{\ell }}=0\end{array}$

          It remains to express ${\mathrm{\rho }}_{\mathrm{b}}$ and substitute the numerical values. Eventually, we obtain

          ${\mathrm{\rho }}_{\mathrm{b}}=\frac{3}{2\cos \mathrm{\varphi }}\left({\mathrm{\rho }}_{\mathrm{r}}-{\mathrm{\rho }}_1\right)+{\mathrm{\rho }}_1=999\mathrm{kg}\cdot {\mathrm{m}}^{-3}$