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Found in: Page 440

### Physics For Scientists & Engineers

Book edition 9th Edition
Author(s) Raymond A. Serway, John W. Jewett
Pages 1624 pages
ISBN 9781133947271

# A plastic sphere floats in water with ${\mathbf{50}}{\mathbf{%}}$ of its volume submerged. This same sphere floats in glycerin with ${\mathbf{40}}{\mathbf{%}}$ of its volume submerged. Determine the densities of (a) the glycerin and (b) the sphere.

(a) The density of glycerin is $1250\text{\hspace{0.17em}kg}/{\text{m}}^{\text{3}}$.

(b) The density of the sphere is $500\text{\hspace{0.17em}kg}/{\text{m}}^{\text{3}}$.

See the step by step solution

## Given data:

Percentage of volume of sphere submerged in water is $50%$.

Percentage of volume of sphere submerged in glycerin is $40%$.

## Buoyancy and gravitational force:

The buoyant force from a liquid of density ${\mathbit{d}}$ when volume of the liquid displaced ${\mathbit{v}}$ is,

${\mathbit{B}}{\mathbf{=}}{\mathbit{d}}{\mathbit{g}}{\mathbit{v}}$ ..…(I)

Here, $g$ is the acceleration due to gravity having value,

$g=9.8\text{\hspace{0.17em}m}/{\text{s}}^{\text{2}}$

The gravitational force on a body of volume ${\mathbit{v}}$ and density ${\mathbit{d}}$ is,

role="math" localid="1663748724511" ${\mathbit{F}}{\mathbf{=}}{\mathbit{d}}{\mathbit{v}}{\mathbit{g}}$ ..... (II)

## (a) Determining the density of glycerin:

Let the volume of the sphere be $V$ and its density be ${d}_{s}$. Volume of water displaced is $V}{2}$. Density of water is $1000\text{\hspace{0.17em}kg}/{\text{m}}^{\text{3}}$. Since the sphere is in equilibrium, equate equations (I) and (II) to get

Volume of displaced glycerin is .

Let the density of glycerin be .

Since the sphere is in equilibrium, equate equations (I) and (II) to get

${d}_{s}Vg=1000\text{\hspace{0.17em}kg}/{\text{m}}^{\text{3}}×\frac{V}{2}×g\phantom{\rule{0ex}{0ex}}{d}_{s}=500\text{\hspace{0.17em}kg}/{\text{m}}^{\text{3}}$

Thus, the required density is,

$\frac{4}{10}V=0.4V$

Let the density of glycerin be ${d}_{g}$.

Since the sphere is in equilibrium, equate equations (I) and (II) to get

$\begin{array}{rcl}{d}_{s}Vg& =& {d}_{g}×0.4V×g\end{array}$

$\begin{array}{rcl}{d}_{g}& =& \frac{{d}_{s}}{0.4}\\ & =& \frac{500}{0.4}\text{\hspace{0.17em}kg}/{\text{m}}^{\text{3}}\\ & =& 1250\text{\hspace{0.17em}kg}/{\text{m}}^{\text{3}}\end{array}$

Thus, the required density is $1250\text{\hspace{0.17em}kg}/{\text{m}}^{\text{3}}$.

## (b) Determining the density of sphere:

This has already been calculated in the previous step. The required density is $500\text{\hspace{0.17em}kg}/{\text{m}}^{\text{3}}$.