Chapter 9, Problem 105P

(a)

To determine

The distance from the top of the cylinder at which a mark is to be placed to indicate a specific gravity of $1.00$.

Answer to Problem 105P

The distance from the top of the cylinder at which a mark is to be placed to indicate a specific gravity of $1.00$ is $10.0\text{ cm}$.

Explanation of Solution

Write the expression for the Newton’s second law.

$\Sigma F_y=0$ (I)

Here, $\Sigma F_y$ is the net external force on the system in $y$ direction

Write the expression for $\Sigma F_y$.

$\Sigma F_y=F_B-m_{h}g$ (II)

Here, $F_B$ is the buoyant force, $m_h$ is the mass of the hydrometer and $g$ is the acceleration due to gravity

Write the equation for $F_B$.

$F_B=\rho gV$ (III)

Here, $\rho$ is the density of the liquid and $V$ is the volume of the liquid displaced

Write the expression for $\rho$.

$\rho =\left(\text{S}\text{.G}\right){\rho }_w$

Here, $\left(\text{S}\text{.G}\right)$ is the specific gravity and ${\rho }_w$ is the density of water

Write the equation for the volume of the liquid displaced.

$V=V_h-Ah$

Here, $V_h$ is the volume of the hydrometer, $A$ is the cross-sectional area of the stem and $h$ is the height above the liquid

Put the above two equations in equation (III).

$F_B=\left(\text{S}\text{.G}\right){\rho }_{w}g\left(V_h-Ah\right)$ (II)

Put the above equation in equation (II).

$\Sigma F_y=\left(\text{S}\text{.G}\right){\rho }_{w}g\left(V_h-Ah\right)-m_{h}g$

Put the above equation in equation (I).

$\left(\text{S}\text{.G}\right){\rho }_{w}g\left(V_h-Ah\right)-m_{h}g=0$ (IV)

Given that the specific gravity to be indicated is $1.00$. Substitute $1.00$ for $\left(\text{S}\text{.G}\right)$ in equation (IV) and rewrite it for $h$.

$\begin{array}{c}1.00{\rho }_{w}g\left(V_h-Ah\right)-m_{h}g=0\\ 1.00{\rho }_{w}gAh=1.00{\rho }_{w}g{V}_h-m_{h}g\\ h=\frac{1.00{\rho }_{w}gV}{1.00{\rho }_{w}gA}-\frac{m_{h}g}{1.00{\rho }_{w}gA}\\ =\frac{1}{A}\left(V_h-\frac{m_h}{1.00{\rho }_w}\right)\end{array}$ (V)

Conclusion:

Given that the cross-sectional area of the stem is $0.400{\text{ cm}}^2$ , volume of the hydrometer is $8.80{\text{ cm}}^3$ and the mass of the hydrometer is $4.80\text{ g}$. The value of density of water is $1.00{\text{ g/cm}}^3$.

Substitute $0.400{\text{ cm}}^2$ for $A$ , $8.80{\text{ cm}}^3$ for $V_h$ , $4.80\text{ g}$ for $m_h$ and $1.00{\text{ g/cm}}^3$ for ${\rho }_w$ in equation (V) to find $h$.

$\begin{array}{c}h=\frac{1}{0.400{\text{ cm}}^2}\left(8.80{\text{ cm}}^3-\frac{4.80\text{ g}}{1.00\left(1.00{\text{ g/cm}}^3\right)}\right)\\ =10.0\text{ cm}\end{array}$

Therefore, the distance from the top of the cylinder at which a mark is to be placed to indicate a specific gravity of $1.00$ is $10.0\text{ cm}$.

(b)

To determine

The specific gravity of the alcohol.

Answer to Problem 105P

The specific gravity of the alcohol is $0.814$.

Explanation of Solution

Rewrite equation (IV) for $\left(\text{S}\text{.G}\right)$.

$\begin{array}{c}\left(\text{S}\text{.G}\right){\rho }_{w}g\left(V_h-Ah\right)=m_{h}g\\ \text{S}\text{.G}=\frac{m_{h}g}{{\rho }_{w}g\left(V_h-Ah\right)}\\ =\frac{m_h}{{\rho }_w\left(V_h-Ah\right)}\end{array}$ (VI)

Conclusion:

Given that the hydrometer floats with $7.25\text{ cm}$ of stem above the surface when placed in alcohol.

Substitute $4.80\text{ g}$ for $m_h$ , $1.00{\text{ g/cm}}^3$ for ${\rho }_w$ , $8.80{\text{ cm}}^3$ for $V_h$ ,$0.400{\text{ cm}}^2$ for $A$ and $7.25\text{ cm}$ for $h$ in equation (VI) to find $\text{S}\text{.G}$.

$\begin{array}{c}\text{S}\text{.G}=\frac{4.80\text{ g}}{\left(1.00{\text{ g/cm}}^3\right)\left(8.80{\text{ cm}}^3-\left(0.400{\text{ cm}}^2\right)\left(7.25\text{ cm}\right)\right)}\\ =0.814\end{array}$

Therefore, the specific gravity of the alcohol is $0.814$.

(c)

To determine

The lowest specific gravity that can be measured with the given hydrometer.

Answer to Problem 105P

The lowest specific gravity that can be measured with the given hydrometer is $0.545$.

Explanation of Solution

The volume of the displaced liquid will be equal to the volume of the hydrometer for lowest specific gravity that can be measured.

$V=V_h$

This implies the term $Ah$ should be removed from the denominator of equation (VI) to find the expression for the lowest specific gravity that can be measured.

$\text{S}{\text{.G}}_{\text{min}}=\frac{m_h}{{\rho }_w{V}_h}$ (VII)

Here, $\text{S}{\text{.G}}_{\text{min}}$ is the lowest specific gravity that can be measured

Conclusion:

Substitute $4.80\text{ g}$ for $m_h$ , $1.00{\text{ g/cm}}^3$ for ${\rho }_w$ and $8.80{\text{ cm}}^3$ for $V_h$ in equation (VII) to find $\text{S}{\text{.G}}_{\text{min}}$.

$\begin{array}{c}\text{S}{\text{.G}}_{\text{min}}=\frac{4.80\text{ g}}{\left(1.00{\text{ g/cm}}^3\right)\left(8.80{\text{ cm}}^3\right)}\\ =0.545\end{array}$

Therefore, the lowest specific gravity that can be measured with the given hydrometer is $0.545$.