Question

A less-dense liquid of density ${\rho }_1$floats on top of a more-dense liquid of density ${\rho }_2$. A uniform cylinder of length $L$and density ${\rho }_{}$, with ${\rho }_1<\rho <{\rho }_2$, floats at the interface with its long axis vertical. What fraction of the length is in the more-dense liquid?

Short answer

Therefore, the fraction of the length of the cylinder in the more dense liquid is$\frac{\rho -{\rho }_1}{{\rho }_2-{\rho }_1}$

Step-by-step solution

Step :1 Introduction  

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.

Step :2 Explanation 

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

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

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

$F_{B2}$is The buoyant force is caused by the density of the liquid being more dense.${\rho }_2$, and $F_G$is weight of the cylinder.

Rewrite the above equation as follows ;

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

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

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

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 ${\rho }_{}$ is the density of the cylinder's substance.

Step :3 Height of the cylinder 

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

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

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

$d_2\left({\rho }_2-{\rho }_1\right)=\left(\rho -{\rho }_1\right)l$

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

Step :4 Fraction 

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

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

Substitute $d_2=\left(\frac{\rho -{\rho }_1}{{\rho }_2-{\rho }_1}\right)l$in the above equation

$f=\frac{\left(\frac{\rho -{\rho }_1}{{\rho }_2-{\rho }_1}\right)l}{l}$

$=\frac{\rho -{\rho }_1}{{\rho }_2-{\rho }_1}$