91Ó°ÊÓ

1. Density of liquid,${\mathrm{ÒÏ}}_{\mathrm{w}}=800\mathrm{kg}/{\mathrm{m}}^3$
2. Inner radius of hollow sphere,$r_i=8.0\mathrm{cm}\mathrm{or}0.08\mathrm{m}$
3. Outer radius of hollow sphere,$r_o=9.0\mathrm{cm}\mathrm{or}0.09\mathrm{m}$

We can use the concept of buoyancy, which states that the downward force or weight of the body is balanced by the upward buoyancy force for the object to float. The weight of the displaced fluid is equal to the weight of the object in this condition. By using submerged volume, we can find the mass of the sphere as well as the density of the material.

Formula:

Volume of a hollow sphere,$V=\frac{4}{3}\mathrm{Ï€}\left({\mathrm{r}}_{out}^3-r_{in}^3\right)$ (i) Downward force on a body,$F_{buoyancy}=\mathrm{ÒÏgV}$ (ii)

Buoyant force exerted by fluid on body, (iii)

Volume of sphere, $V=\frac{4}{3}\mathrm{Ï€}{r}^3$ (iv)

Let’s assume V is the volume of the sphere. Since half of the sphere is submerged, volume of the water displaced is V/2. We can find the weight of this displaced water as

$W={\mathrm{ÒÏ}}_{w}g\frac{V}{2}$ (v)

This would provide the force of buoyancy due to which the sphere floats.

So, the weight of the sphere is balanced by the force of buoyancy, hence can be written using equation (ii) & (v) as:

$\begin{array}{l}{W}_s=W\\ m_{s}g={\mathrm{ÒÏ}}_{w}g\frac{W}{2}\end{array}$

$m_S={\mathrm{ÒÏ}}_w\frac{V}{2}$ (vi)

Volume can be written in terms of the outer radius of the sphere using equation (iv), we get

$V=\frac{4}{3}\mathrm{Ï€}{r}_o^3$

Substituting the value in the above equation (vi) and using the given values, we have

$$\begin{gathered} m_s={\mathrm{ÒÏ}}_w\frac{1}{2}\left(\frac{4}{3}\mathrm{Ï€}{r}_o^3\right) \\ =800kg/m^3×\frac{2}{3}×\mathrm{Ï€}×{\left(0.09m\right)}^3 \\ =1.22kg \end{gathered}$$

Hence, the mass of the sphere is 1.22 kg

We know that

$\mathrm{Density}=\frac{\mathrm{Mass}}{\mathrm{Volume}}$ (vii)

Using equation (i), we get the volume of the hollow sphere as:

$$\begin{gathered} Volume=\frac{4}{3}×\left(3.14\right)×{\left(0.09m\right)}^3-{\left(0.80m\right)}^3 \\ =9.09×{10}^{-4}{m}^3 \end{gathered}$$

Now, using the above value and the given values in equation (vii), we get

$$\begin{gathered} \mathrm{Density}=\frac{1.22\mathrm{kg}}{9.09×{10}^{-4}{\mathrm{m}}^3} \\ =1.3×{10}^3\mathrm{kg}/{\mathrm{m}}^3 \end{gathered}$$

Hence, the density of the material is $1.3×{10}^{3}kg/m^3$