Figure 14-48 shows an iron ball suspended by thread of negligible mass from an upright cylinder that floats partially submerged in water. The cylinder has a height of6.00 cm, a face area of$\mathbf{12}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{cm}\mathbf{2}$on the top and bottom, and a density of $\mathbf{0}\mathbf{.}\mathbf{3}\mathbf{}\mathbf{g}\mathbf{/}{\mathbf{cm}}^{\mathbf{3}}$, and 2.00 cmof its height is above the water surface. What is the radius of the iron ball?

i) The height of the cylinder, $\mathrm{h}=6.00\mathrm{cm}\mathrm{or}6.00\times {10}^{-2}\mathrm{m}$

ii) The face area of top and bottom of the cylinder,$A=12.0{\mathrm{cm}}^2\mathrm{or}12.0\times {10}^{-4}{\mathrm{m}}^2$

iii) The density of the cylinder,${\mathrm{\rho }}_{\mathrm{c}}=0.30\frac{\mathrm{g}}{{\mathrm{cm}}^3}\mathrm{or}300\frac{\mathrm{kg}}{{\mathrm{m}}^3}$

iv) The height of the cylinder above the water surface,$h_1=2.00cmor2.00\times {10}^{-2}m$

v) The density of the iron ball,${\mathrm{\rho }}_{\mathrm{Fe}}=7.9\frac{\mathrm{g}}{{\mathrm{cm}}^3}\mathrm{or}7900\frac{\mathrm{kg}}{{\mathrm{m}}^3}$

We can use the concept of Archimedes’ principle and expression of density. When the cylinder with the iron ball is fully submerged in water, then a buoyant force from the surrounding fluid acts on it. This force has a magnitude equal to the weight of the water displaced by the cylinder with the ball.

Formulae:

Force applied on body (or weight),$F_b=m_{f}g$ (i)

Density of a substance, $\rho =\frac{m}{V}$ (ii)

The volume of the cylinder is given as follows using given values:

$$\begin{gathered} V_c=A\times h \\ =12.0\times {10}^{-4}{\mathrm{m}}^2\times 6.00\times {10}^{-2}\mathrm{m} \\ =72.0\times {10}^{-6}{\mathrm{m}}^3 \end{gathered}$$

The height of cylinder below the water is given as follows using given values:

$$\begin{gathered} h_w=h-h_1 \\ =\left(6.00\times {10}^{-2}\mathrm{m}\right)-\left(2.00\times {10}^{-2}\mathrm{m}\right) \\ =4.00\times {10}^{-2}\mathrm{m} \end{gathered}$$

Then the volume of the cylinder below the water is given as:

$$\begin{gathered} V_c^{\text{'}}=A\times h_w \\ =12.0\times {10}^{-4}{\mathrm{m}}^2\times 4.00\times {10}^{-2}\mathrm{m} \\ =48.0\times {10}^{-6}{\mathrm{m}}^3 \end{gathered}$$

According to Newton’s second law, we can write the net force as:

$$\begin{gathered} F_{net}=ma \\ T+F_b=\left(m_c+m_b\right)g.....................................................\left(iii\right) \end{gathered}$$

The volume of the cylinder below the water is$V_c^{\text{'}}$. When the cylinder with ball is floating in water, then the buoyant force acts on them.Using equation (i) & (ii) in equation (iii), we get

$$\begin{gathered} {\rho }_{c}g{V}_c+{\rho }_{Fe}g{V}_b={\rho }_{w}g{V^{\text{'}}}_c+{\rho }_{w}g{V}_b \\ {\rho }_c{V}_c+{\rho }_{Fe}{V}_b={\rho }_w{V^{\text{'}}}_c+{\rho }_w{V}_b \\ {\rho }_{Fe}{V}_b-{\rho }_w{V}_b={\rho }_w{V^{\text{'}}}_c-{\rho }_w{V}_b \\ \left({\rho }_{Fe}-{\rho }_w\right)V_b={\rho }_w{V^{\text{'}}}_c-{\rho }_c{V}_c \\ {V}_b=\frac{{\rho }_w{V^{\text{'}}}_c-{\rho }_c{V}_c}{\left({\rho }_{Fe}-{\rho }_w\right)} \\ =\frac{\left(1000{\displaystyle \frac{kg}{m^3}}\times 72.0\times {10}^{-6}{m}^3\right)-\left(300{\displaystyle \frac{kg}{m^3}}\times 72.0\times {10}^{-6}{m}^3\right)}{\left(7900\frac{kg}{m^3}-1000{\displaystyle \frac{kg}{m^3}}\right)} \\ =7.304\times {10}^{-6}{\mathrm{m}}^3 \end{gathered}$$

The expression of the volume of the sphere is given as:

$$\begin{gathered} V_b=\frac{4}{3}{\mathrm{\pi r}}^3 \\ \mathrm{r}=\sqrt[3]{\frac{3{\mathrm{V}}_{\mathrm{b}}}{4\mathrm{\pi }}} \\ =\sqrt[3]{\frac{3\times 7.304\times {10}^{-6}{\mathrm{m}}^3}{4\times 3.142}} \\ =1.45\times {10}^{-3}\mathrm{m} \end{gathered}$$

Hence, the radius of iron ball is found to be$1.45\times {10}^{-3}\mathrm{m}$