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Q31P

Expert-verifiedFound in: Page 440

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 $500\text{\hspace{0.17em}kg}/{\text{m}}^{\text{3}}$.

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

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

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

${\mathit{B}}{\mathbf{=}}{\mathit{d}}{\mathit{g}}{\mathit{v}}$ ..…(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 **${\mathit{v}}$** and density **${\mathit{d}}$** is,**

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

Let the volume of the sphere be $V$ and its density be ${d}_{s}$. Volume of water displaced is $\raisebox{1ex}{$V$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$. 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

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

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

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

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

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

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

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