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Question
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.
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Solution
Given:
Inner radius of the hollow spherical body, r1 = 6 cm
Outer radius of the hollow spherical body, r2 = 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:
\[\text{Weight of the liquid displaced }= \left( \frac{\text{V}}{2} \right)( \rho_\text{w} ) \times \text{g}\]
\[\text{ We know }: \]
\[\text{ Upward thrust = Weight of the liquid displaced }\]
\[ \therefore \left( \frac{4}{3}\pi r_3^2 - \frac{4}{3}\pi r_1^3 \right)\rho = \left( \frac{1}{2} \right)\frac{4}{3}\pi r_2^3 \times \rho_\text{w} \]
\[ \Rightarrow \left( r_2^3 - r_1^3 \right) \times \rho = \left( \frac{1}{2} \right) r_2^3 \times 1\]
\[ \Rightarrow (8 )^3 - (6 )^3 \times \rho = \left( \frac{1}{2} \right)(8 )^3 \times 1\]
\[ \Rightarrow \rho = \frac{512}{2 \times (512 - 216)}\]
\[ = \frac{512}{2 \times 296}=0.865 \text{ gm/cc =865kg/m}^3\]
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