Chapter 9, Problem 104P

(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 104P

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.

  ${\displaystyle \sum F_y}=0$        (I)

Here, ${\displaystyle \sum F_y}$ is the net external force on the system in $y$ direction.

Write the expression for ${\displaystyle \sum F_y}$.

  ${\displaystyle \sum F_y}=F_B-m_{h}g$        (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 expression 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 $\rho$.

  $\rho =\left(\text{S}\cdot \text{G}\right){\rho }_{\omega }$        (IV)

Here, $\text{S}\cdot \text{G}$ is the specific gravity and ${\rho }_{\omega }$ is the density of water.

Write the equation for the volume of the liquid displaced

  $V=V_h-Ah$        (V)

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.

Substitute the expression (IV) and (V) in expression (III).

  $F_B=\left(S\cdot G\right){\rho }_{\omega }g\left(V_h-Ah\right)$        (VI)

Substitute the expression (VI) in expression (II) and finally in expression (I)

  $\begin{array}{l}{\displaystyle \sum F_y}=\left(S\cdot G\right){\rho }_{\omega }g\left(V_h-Ah\right)-m_{h}g\\ \left(S\cdot G\right){\rho }_{\omega }g\left(V_h-Ah\right)-m_{h}g=0\end{array}$        (VII)

Conclusion:

Substitute $1.00$ for $\text{S}\cdot \text{G}$ in expression (VII) and re-write for $h$

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

Substitute $\text{0}{\text{.400cm}}^{\text{2}}$  for $A$ , $8.80c{m}^3$ for $V_h$ , $4.80g$ for $m_h$ and $1.00g/c{m}^3$ for ${\rho }_{\omega }$ in equation (VIII)

  $\begin{array}{c}h=\frac{1}{\text{0}{\text{.400cm}}^{\text{2}}}\left(8.80c{m}^3-\frac{4.80g}{1.00\left(1.00g/c{m}^3\right)}\right)\\ =10.0\text{cm}\end{array}$

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 specify gravity of the alcohol

Answer to Problem 104P

The specific gravity of the alcohols is $0.814$ .

Explanation of Solution

Re-write the expression (VII) for $\left(S\cdot G\right)$

  $\begin{array}{c}\left(S\cdot G\right){\rho }_{\omega }g\left(V_h-Ah\right)=m_{h}g\\ S\cdot G=\frac{m_h}{g\left(V_h-Ah\right)}\end{array}$        (I)

Conclusion:

Substitute $\text{0}{\text{.400cm}}^{\text{2}}$  for $A$ , $1.00g/c{m}^3$ for ${\rho }_{\omega }$   and $8.80c{m}^3$ for $V_h$  in equation (VII)

  $\begin{array}{c}S\cdot G_{\min }=\frac{4.80g}{\left(1.00g/c{m}^3\right)\left(8.80c{m}^3\right)}\\ =0.545\end{array}$

Therefore, the specific gravity of the alcohol is $\mathrm{0.814.}$

(c)

To determine

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

Answer to Problem 104P

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

Explanation of Solution

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

    $V=V_h$

Here, $V$ is the volume of the liquid displaced and $V_h$ is the volume of the hydrometer.

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

Conclusion:

Substitute $\text{4}\text{.8g}$  for $m_h$ , $8.80c{m}^3$ for $V_h$ , $7.25\text{cm}$ for $h$,  $4.80g$ for $m_h$ and $1.00g/c{m}^3$ for ${\rho }_{\omega }$ in equation (VIII)

  $\begin{array}{c}S\cdot G=\frac{4.80g}{\left(1.00g/c{m}^3\right)\left(8.80c{m}^3-\left(\text{0}{\text{.400cm}}^{\text{2}}\right)\left(7.25\text{cm}\right)\right)}\\ =0.814\end{array}$

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