Question

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.

Short answer

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

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

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 $\mathit{d}$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)

(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 $\raisebox{1ex}{$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

$$\begin{gathered} d_{s}Vg=1000\text{\hspace{0.17em}kg}/{\text{m}}^{\text{3}}\times \frac{V}{2}\times g \\ {d}_s=500\text{\hspace{0.17em}kg}/{\text{m}}^{\text{3}} \end{gathered}$$

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 $1250\text{\hspace{0.17em}kg}/{\text{m}}^{\text{3}}$.

(b) Determining the density of sphere:

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