Chapter 9, Problem 37P

(a)

To determine

The specific gravity of the disk.

Answer to Problem 37P

The specific gravity of the disk is $0.910$.

Explanation of Solution

The specific gravity of a substance is the ratio of its density to the density of water.

Write the equation for specific gravity of the disk.

$\text{S}\text{.G}=\frac{\rho }{{\rho }_{\text{w}}}$ (I)

Here, $\rho$ is the density of the object and ${\rho }_{\text{w}}$ is the density of water.

Write the equation for density.

$\rho =\frac{m}{V}$ (II)

Here, $m$ is the mass of the disk and $V$ is the volume of the cylindrical disk.

Conclusion:

Substitute equation (II) in equation (I).

$\text{S}\text{.G}=\frac{m}{{\rho }_{\text{w}}V}$. (III)

Substitute $8.16\text{kg}$ for $m$, $1.00\times {10}^3{\text{kg/m}}^3$ for ${\rho }_{\text{w}}$, and $8.97\times {10}^{-3}{\text{m}}^3$ for $V$ in equation (III).

$\begin{array}{c}\text{S}\text{.G}=\frac{8.16\text{kg}}{\left(1.00\times {10}^3{\text{kg/m}}^3\right)\left(8.97\times {10}^{-3}{\text{m}}^3\right)}\\ =\frac{8.16\text{kg}}{8.97\text{kg}}\\ =0.910\end{array}$

Therefore, the specific gravity of the disk is $0.910$.

(b)

To determine

The distance of the water level at its bottom surface.

Answer to Problem 37P

The distance of the water level at its bottom surface is $1.28\text{cm}$.

Explanation of Solution

Write the equation for buoyant force is equal to the weight of the fluid displaced.

$\begin{array}{c}{\displaystyle \sum F}=0\\ {\rho }_{\text{f}}g{V}_{\text{i}}-{\rho }_{\text{o}}g{V}_{\text{o}}=0\end{array}$ (IV)

Here, ${\rho }_{\text{f}}$ is the density of the water at $0°\text{C}$, $g$ is the gravitational acceleration, $V_{\text{i}}$ is the volume of the fluid displaced after immersed, ${\rho }_{\text{o}}$ is the density of the object, and $V_{\text{o}}$ is the volume of the object.

Since the relationship between the fraction of a floating object’s volume that is submerged and the ratio of the object’s density to that of the fluid in which it floats is rewritten from the equation (IV).

$\begin{array}{c}{\rho }_{\text{f}}g{V}_{\text{f}}={\rho }_{\text{o}}g{V}_{\text{o}}\\ {\rho }_{\text{f}}{V}_{\text{f}}={\rho }_{\text{o}}{V}_{\text{o}}\\ \frac{V_{\text{f}}}{V_{\text{o}}}=\frac{{\rho }_{\text{o}}}{{\rho }_{\text{f}}}\end{array}$ (V)

The ratio of the density of the disk to that of water is equal to the fraction of the volume of the disk is submerged.

$\frac{V_{\text{water}}}{V_{\text{disk}}}=\frac{{\rho }_{\text{disk}}}{{\rho }_{\text{water}}}$ (VI)

Here, ${\rho }_{\text{disk}}$ is the density of the disk, $V_{\text{disk}}$ is the volume of the disk, ${\rho }_{\text{water}}$ is the density of the water, and $V_{\text{water}}$ is the volume of the water.

Compare equation (VI) and (I).

$\text{S}\text{.G}=\frac{V_{\text{water}}}{V_{\text{disk}}}$

Replace $dA$ for $V_{\text{water}}$ and $V$ for  $V_{\text{disk}}$ in the above equation.

$\text{S}\text{.G}=\frac{dA}{V}$ (VII)

Here, $d$ is the distance of the water level at its bottom surface.

Solve the equation (VII) for $d$.

$\begin{array}{c}\text{S}\text{.G}=\frac{dA}{V}\\ d=\left(\text{S}\text{.G}\right)\frac{V}{A}\end{array}$ (VIII)

Conclusion:

Substitute $0.640{\text{m}}^2$ for $A$, $0.910$ for $\text{S}\text{.G}$, and $8.97\times {10}^{-3}{\text{m}}^3$ for $V$ in equation (VIII).

$\begin{array}{c}d=\left(0.910\right)\frac{\left(8.97\times {10}^{-3}{\text{m}}^3\right)}{\left(0.640{\text{m}}^2\right)}\\ =\left(0.910\right)\left(14.02\times {10}^{-3}\text{m}\right)\\ =12.76\times {10}^{-3}\text{m}\end{array}$

Convert the above value into centimeter.

$\begin{array}{c}d=\text{1}\text{.28}\times {10}^{-2}\text{m}\\ \text{=1}\text{.28}\times {10}^{-2}\text{m}\left(\frac{{10}^2\text{cm}}{1\text{m}}\right)\\ =1.28\text{cm}\end{array}$

Therefore, the distance of the water level at its bottom surface is $1.28\text{cm}$.

(c)

To determine

The distance of the water level at its top surface.

Answer to Problem 37P

The distance of the water level at its top surface is $0.13\text{cm}$.

Explanation of Solution

From the equation (VII), find the distance of the top surface of the disk above the water level.

$\text{S}\text{.G}=\frac{dA}{V}$

Replace $hA$ for $V$ in the above equation.

$\begin{array}{c}\text{S}\text{.G}=\frac{dA}{hA}\\ =\frac{d}{h}\end{array}$ (IX)

Here, $h$ is the thickness of the disk

Rewrite equation (IX) by using equation (VIII).

$\begin{array}{c}h-d=h-\left(\text{S}\text{.G}\right)h\\ h=\frac{V}{A}\left(1-\text{S}\text{.G}\right)\end{array}$ (X)

Conclusion:

Substitute $0.640{\text{m}}^2$ for $A$, $0.910$ for $\text{S}\text{.G}$, and $8.97\times {10}^{-3}{\text{m}}^3$ for $V$ in equation (X).

$\begin{array}{c}h=\frac{8.97\times {10}^{-3}{\text{m}}^3}{0.640{\text{m}}^2}\left(1-0.910\right)\\ =14.02\times {10}^{-3}\text{m}\left(0.09\right)\\ =1.26\times {10}^{-3}\text{m}\end{array}$

Convert the above value into centimeter.

$\begin{array}{c}h=0.\text{126}\times {10}^{-2}\text{m}\\ \text{=0}\text{.13}\times {10}^{-2}\text{m}\left(\frac{{10}^2\text{cm}}{1\text{m}}\right)\\ =0.13\text{cm}\end{array}$

Therefore, the distance of the water level at its top surface is $0.13\text{cm}$.