91Ó°ÊÓ

The weight of the fluid displaced by the immersed portion of the body equals the upward force experienced by an object immersed in a fluid. The buoyant or up thrust force is the name given to this force.

When the cylinder is in static equilibrium, the net force acting on it is zero.

$\begin{array}{r}\mathrm{Î\pounds }{F}_{\mathrm{y}}=0\\ F_{\mathrm{B}1}+F_{\mathrm{B}2}−F_{\mathrm{G}}=0\end{array}$

$F_{B1}$is a buoyant force caused by a less dense liquid ${\mathrm{ÒÏ}}_1$,

Rewrite the above equation as follows ;

${\mathrm{ÒÏ}}_{1}A{d}_{1}g+{\mathrm{ÒÏ}}_{2}A{d}_{2}g=\mathrm{ÒÏ}A\left(d_1+d_2\right)g$

${\mathrm{ÒÏ}}_1{d}_1+{\mathrm{ÒÏ}}_2{d}_2=\mathrm{ÒÏ}\left(d_1+d_2\right)$

$\frac{{\mathrm{ÒÏ}}_1}{{\mathrm{ÒÏ}}_2}{d}_1+d_2=\frac{\mathrm{ÒÏ}}{{\mathrm{ÒÏ}}_2}\left(d_1+d_2\right)$

Here, $d_1$is the cylinder's height immersed in a less dense liquid, $d_2$is the cylinder's height immersed in a denser liquid, and ${\mathrm{ÒÏ}}_{}$ is the density of the cylinder's substance.

The total height of the cylinder is

$\begin{array}{r}I=d_1+d_2\\ d_1=l−d_2\end{array}$

Substitute $d_{1=}l-d_{2}andl=d_1+d_2$in the equation$\frac{{\mathrm{ÒÏ}}_1}{{\mathrm{ÒÏ}}_2}{d}_1+d_2=\frac{\mathrm{ÒÏ}}{{\mathrm{ÒÏ}}_2}\left(d_1+d_2\right)$

$\begin{array}{r}\frac{{\mathrm{ÒÏ}}_1}{{\mathrm{ÒÏ}}_2}\left(l−d_2\right)+d_2=\frac{\mathrm{ÒÏ}}{{\mathrm{ÒÏ}}_2}l\\ d_2\left(1−\frac{{\mathrm{ÒÏ}}_1}{{\mathrm{ÒÏ}}_2}\right)+\frac{{\mathrm{ÒÏ}}_1}{{\mathrm{ÒÏ}}_2}=\frac{\mathrm{ÒÏ}}{{\mathrm{ÒÏ}}_2}l\end{array}$

$\begin{array}{c}{d}_2\left(1−\frac{{\mathrm{ÒÏ}}_1}{{\mathrm{ÒÏ}}_2}\right)=\frac{\mathrm{ÒÏ}−{\mathrm{ÒÏ}}_1}{{\mathrm{ÒÏ}}_2}l\\ d_2\left(\frac{{\mathrm{ÒÏ}}_2−{\mathrm{ÒÏ}}_1}{{\mathrm{ÒÏ}}_2}\right)=\frac{\mathrm{ÒÏ}−{\mathrm{ÒÏ}}_1}{{\mathrm{ÒÏ}}_2}l\end{array}$

$d_2\left({\mathrm{ÒÏ}}_2−{\mathrm{ÒÏ}}_1\right)=\left(\mathrm{ÒÏ}−{\mathrm{ÒÏ}}_1\right)l$

$d_2=\left(\frac{\mathrm{ÒÏ}-{\mathrm{ÒÏ}}_1}{{\mathrm{ÒÏ}}_2-{\mathrm{ÒÏ}}_1}\right)l$

The percentage of the cylinder that is submerged in the denser liquid is,

$f=\frac{d_2}{l}$

Substitute $d_2=\left(\frac{\mathrm{ÒÏ}−{\mathrm{ÒÏ}}_1}{{\mathrm{ÒÏ}}_2−{\mathrm{ÒÏ}}_1}\right)l$in the above equation

$f=\frac{\left(\frac{\mathrm{ÒÏ}−{\mathrm{ÒÏ}}_1}{{\mathrm{ÒÏ}}_2−{\mathrm{ÒÏ}}_1}\right)l}{l}$

$=\frac{\mathrm{ÒÏ}−{\mathrm{ÒÏ}}_1}{{\mathrm{ÒÏ}}_2−{\mathrm{ÒÏ}}_1}$