91Ó°ÊÓ

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

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

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

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

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

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, $\mathrm{ÒÏ}=\frac{m}{V}$ (ii)

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

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

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

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

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

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

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

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

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

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

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