Question

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?

Short answer

The radius of the iron ball is$1.45\times {10}^{-3}\mathrm{m}$

Step-by-step solution

The given data

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,${\mathrm{\rho }}_{\mathrm{Fe}}=7.9\frac{\mathrm{g}}{{\mathrm{cm}}^3}\mathrm{or}7900\frac{\mathrm{kg}}{{\mathrm{m}}^3}$

Understanding the concept of Archimedes Principle

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)

Calculation of the radius of iron ball

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$1.45\times {10}^{-3}\mathrm{m}$