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

#### Solution

Given:

Inner radius of the hollow spherical body, *r*_{1} = 6 cm

Outer radius of the hollow spherical body, *r*_{2} = 8 cm

Let the density of the material of the sphere be ρ and the volume of the water displaced by the hollow sphere be *V*.

If `rho _w` is the density of water, then:

\[\text{Weight of the liquid displaced }= \left( \frac{\text{V}}{2} \right)( \rho_\text{w} ) \times \text{g}\]

\[\text{ We know }: \]

\[\text{ Upward thrust = Weight of the liquid displaced }\]

\[ \therefore \left( \frac{4}{3}\pi r_3^2 - \frac{4}{3}\pi r_1^3 \right)\rho = \left( \frac{1}{2} \right)\frac{4}{3}\pi r_2^3 \times \rho_\text{w} \]

\[ \Rightarrow \left( r_2^3 - r_1^3 \right) \times \rho = \left( \frac{1}{2} \right) r_2^3 \times 1\]

\[ \Rightarrow (8 )^3 - (6 )^3 \times \rho = \left( \frac{1}{2} \right)(8 )^3 \times 1\]

\[ \Rightarrow \rho = \frac{512}{2 \times (512 - 216)}\]

\[ = \frac{512}{2 \times 296}=0.865 \text{ gm/cc =865kg/m}^3\]