A $30cm$-tall, $4.0cm$-diameter plastic tube has a sealed bottom. $250g$of lead pellets are poured into the bottom of the tube, whose mass is $30g$, then the tube is lowered into a liquid. The tube floats with$5.0cm$ extending above the surface. What is the density of the liquid?

Archimedes principle states that the upward force acts on the object when an object partly or fully immersed in a fluid. This force is called buoyant force. Mathematically, the buoyant force can be expressed as,

$F_{\mathrm{B}}=\rho V/\mathrm{g}$

Here, $\rho$is the density of fluid, $V$is volume of displaced fluid, and$g$ is acceleration due to gravity.

Let $r$be the radius of the hemisphere. The diameter of the hemisphere is $0.8cm$. Thus, the radius of the hemisphere is,

$\begin{array}{r}r=\frac{8.0\mathrm{cm}}{2}\end{array}$

$=4.0\mathrm{cm}\left(\frac{1\mathrm{m}}{100\mathrm{cm}}\right)$

$=0.04\mathrm{m}$

The volume of the hemisphere is half of the sphere. Thus, the volume of hemisphere is,

$V=\frac{2}{3}\pi r^3$

Substitute $0.04\mathrm{m}\text{for}r$in above equation

$V=\frac{2}{3}\pi (0.04\mathrm{m}{)}^3$

$=1.34\times {10}^{-4}{\mathrm{m}}^3$

The weight of the rock and the hemisphere bowl

$\begin{array}{r}{F}_g=mg+Mg\\ F_s=g(m+M)\end{array}$

$\begin{array}{r}{F}_{\mathrm{z}}=mg+Mg\\ F_{\mathrm{s}}=g(m+M)\end{array}$

Here, $m$is the mass of the rock, $M$is the mass of the hemisphere, and $g$is the acceleration due to gravity.

After the rock is placed in the bottom of a plastic boat with maximum weight, the hemisphere gets totally dipped in the water without sinking it, and the water level is just in level with the hemisphere. So, the total buoyant force exerted on the hemisphere is equal to the total weight of the system of rock and the hemisphere bowl. Thus.

$F_{\mathrm{B}}=F_{\mathrm{g}}$

Substitute $\rho Vg\text{for}{F}_{\mathrm{B}},g(m+M)\text{for}{F}_{\mathrm{g}}$in above equation

$\begin{array}{r}\rho Vg=(m+M)g\\ \end{array}$

$\rho V=m+M$

Rewrite the equation in ${\rho }_{\mathrm{w}}V=m+M\text{for}m$

$m={\rho }_{\mathrm{w}}V-M$

Substitute $1000\mathrm{kg}/{\mathrm{m}}^3\text{for}\rho ,1.34\times {10}^{-4}{\mathrm{m}}^3\text{for}V\text{, and}21\mathrm{g}\text{for}M$in above equation

$M=\left(1000\mathrm{kg}/{\mathrm{m}}^3\right)\left(1.34\times {10}^{-4}{\mathrm{m}}^3\right)-\left(21\mathrm{g}\right)\left(\frac{{10}^{-3}\mathrm{kg}}{1\mathrm{g}}\right)$

$=0.113kg$

$\begin{array}{r}\\ =0.113\mathrm{kg}\left(\frac{1000\mathrm{g}}{1\mathrm{kg}}\right)\\ \end{array}$

$=113g$