Question

a sphere has a hollow portion which is one third of its total volume. it floats in water with four fifth of its volume immersed. the specific gravity of its material is

Answer 1

Given:

A sphere has a hollow portion which is one third of its total volume. It floats in water with four fifth of its volume immersed.

To find:

Specific gravity of the material.

Calculation:

Let external radius be R and internal radius be r ;

$\therefore \: \dfrac{4}{3} \pi {r}^{3} = \dfrac{1}{3} \times ( \dfrac{4}{3} \pi {R}^{3} )$

$= > {r}^{3} = \dfrac{1}{3} {R}^{3} \: \: \: \: ......(1)$

Now , since body is floating ;

$\therefore$Up-thrust = Weight of body

$= > \dfrac{4}{5} \times ( \dfrac{4}{3} \pi {R}^{3} ) \sigma g= \dfrac{4}{3} \pi( {R}^{3 } - {r}^{3} ) \rho g$

$= > \dfrac{4}{5} {R}^{3} \sigma = ( {R}^{3 } - {r}^{3} ) \rho$

$= > \dfrac{4}{5} {R}^{3} \sigma = ( {R}^{3 } - \dfrac{1}{3} {R}^{3} ) \rho$

$= > \dfrac{4}{5} {R}^{3} \sigma = \dfrac{2}{3} {R}^{3} \rho$

$= > \dfrac{4}{5} {R}^{3} (1) = \dfrac{2}{3} {R}^{3} \rho$

$= > \dfrac{4}{5} {R}^{3} = \dfrac{2}{3} {R}^{3} \rho$

$= > \dfrac{4}{5} = \dfrac{2}{3} \rho$

$= > \: \rho = \dfrac{6}{5}$

$= > \: \rho = 1.2 \: g {cc}^{ - 1}$

So, final answer is :

$\boxed{ \: \rho = 1.2 \: g {cc}^{ - 1} }$