A hollow spherical body of inner and outer radii 6 cm & 8 cm respectively floats half submerged in water the density of material of the sphere is

Given that a hollow sphere is floating in water by half of it submerged.

Let inner radius r₁ = 6 cm

Outer radius r₂ = 8 cm

Volume of sphere $=\frac{4}{3}\pi r_2{}^3$

$\text{V}=\frac{4}{3}\times \frac{22}{7}\times 8\times 8\times 8=2,145.5\text{cc}$

Volume of the material of hollosphere

${\text{V}}^{|}=\frac{4}{3}\pi \left(r_2{}^3-r_1{}^3\right)$

$\Rightarrow \frac{4}{3}\times \frac{22}{7}\left(8^3-6^3\right)$

$\Rightarrow \frac{4}{3}\times \frac{22}{7}\times 8^3-\frac{4}{3}\times \frac{22}{7}\times 6^3$

$\Rightarrow 2145.5\text{cc}-905.2\text{cc}$

$\therefore {\text{V}}^{\text{|}}=1240.4\text{cc}$

It is given to be floating in water by half submerged

$\therefore$  Weight of sphere = Weight of liquid displaced

$\text{mg}={\text{V}}_{\text{immersed}}\times {\rho }_w\times g$

$\therefore \text{m}={\text{V}}_{\text{immersed}}\times {\rho }_w\times g$

Let mass of sphere = ${\text{V}}^{\text{|}}\times \text{density}\text{of}\text{material}\text{of}\text{sphere}$

                                                                                                                $={\text{V}}_{\text{d}}$

We can have ${\text{V}}_{\text{immersed}}=\frac{\text{V}}{\text{2}}$

$\Rightarrow {\text{V}}^{\text{|}}\text{d}=\frac{\text{V}}{\text{2}}\times {\rho }_w$                              $\left(\because {\rho }_w=1\text{g/cc}\right)$

$\Rightarrow 1240.4\times \text{d}=\frac{2145.5}{2}\times 1\text{g/cc}$

$\Rightarrow \text{d}=\frac{2145.5}{2\times 1240.4}\times 1$

$\text{d}=0.865\text{g/cc}$                                  $\left(\because 1\text{g/cc}={10}^3{\text{kg/m}}^{\text{3}}\right)$

$\Rightarrow \text{d}=0.865\times {10}^3{\text{kg/m}}^{\text{3}}$

$\Rightarrow \text{d}=865{\text{kg/m}}^{\text{3}}$