Question

When a hydrometer having a stem diameter of 0.30 in. is placed in water, the stem protrudes 3.15 in. above the water surface. If the water is replaced with a liquid having a specific gravity of 1.19, how much of the stem would protrude above the liquid surface? The hydrometer weighs 0.042 lb

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Answer 1

Concepts and reason

The concept required to solve this question is Archimedes’ principle.

Archimedes’ Principle: It states that when a body is partially or fully submerged in a fluid, then a force acts on it in an upward direction. This force is called the buoyant force. The magnitude of this buoyant force is equal to the weight of the fluid displaced by the body. In other words, the volume of the displaced fluid is equal to the volume of the part of the body that is submerged.

First of all, calculate the volume of water displaced by the fluid. and use the volume of the water displaced to calculate the volume of the fluid displaced. Use the expression for the change in volume of the hydrometer above the free surface when it placed in the fluid from water. This is done in order to calculate the height of hydrometer above the free surface of fluid.

Fundamentals

The expression of volume of water displaced by an object is written as,

${V_w} = \frac{m}{{{\rho _w}}}$

Here, ${V_w}$ is the volume of water displaced by the object, $m$ is the mass of the object ad ${\rho _w}$ is the density of water.

Specific gravity of the fluid is expressed as,

$S = \frac{{{V_w}}}{{{V_f}}}$

Here, ${V_f}$ is volume of the fluid displaced and ${V_w}$ is volume of water displaced by the object.

Perform the unit conversion for the density of water.

$\begin{array}{c}\\{\rho _w} = 62.42{\rm{ lb/f}}{{\rm{t}}^{\rm{3}}}\\\\ = 62.42{\rm{ lb/f}}{{\rm{t}}^{\rm{3}}}{\left( {\frac{{1{\rm{ ft}}}}{{12{\rm{ in}}}}} \right)^3}\\\\ = 0.036123{\rm{ lb/i}}{{\rm{n}}^{\rm{3}}}\\\end{array}$

Write the expression for the volume of water displaced.

${V_w} = \frac{m}{{{\rho _w}}}$

Here, ${V_w}$ is the volume of the water displaced and m is mass of the hydrometer.

Substitute $0.042{\rm{ lb}}$ for $m$ and $0.036123{\rm{ lb/i}}{{\rm{n}}^{\rm{3}}}$ for ${\rho _w}$ .

$\begin{array}{c}\\{V_w} = \frac{{0.042{\rm{ lb}}}}{{0.036123{\rm{ lb/i}}{{\rm{n}}^{\rm{3}}}}}\\\\ = 1.1627{\rm{ i}}{{\rm{n}}^{\rm{3}}}\\\end{array}$

Write the expression for the volume of the fluid displaced by hydrometer.

${V_l} = \frac{{{V_w}}}{S}$

Here, ${V_l}$ is the volume of the fluid displaced and m is mass of the hydrometer.

Substitute $1.1627{\rm{ i}}{{\rm{n}}^{\rm{3}}}$ for ${V_w}$ and $1.19$ for $S$ .

$\begin{array}{c}\\{V_l} = \frac{{1.1627{\rm{ i}}{{\rm{n}}^{\rm{3}}}}}{{1.19}}\\\\ = 0.977{\rm{ i}}{{\rm{n}}^{\rm{3}}}\\\end{array}$

Write the expression for change in volume of the hydrometer above the free surface when it placed in the fluid from the water.

$\frac{{\pi {d^2}}}{4}\left( {{h_l} - {h_w}} \right) = {V_w} - {V_l}$

Here, ${h_l}$ is height of hydrometer above the free surface in the fluid, ${h_w}$ is height above the free surface in water.

Substitute $0.977{\rm{ i}}{{\rm{n}}^{\rm{3}}}$ for ${V_l}$ , $1.1627{\rm{ i}}{{\rm{n}}^{\rm{3}}}$ for ${V_w}$ , $0.30{\rm{ in}}$ for $d$ and $3.15{\rm{ in}}$ for ${h_i}$ . $\begin{array}{l}\\\frac{{\pi {{\left( {0.30{\rm{ in}}} \right)}^2}}}{4}\left( {{h_l} - 3.15{\rm{ in}}} \right) = 1.1627{\rm{ i}}{{\rm{n}}^{\rm{3}}} - 0.977{\rm{ i}}{{\rm{n}}^{\rm{3}}}\\\\{h_l} - 3.15 = 2.626\\\\{h_l} = 5.776{\rm{ in}}\\\end{array}$

Ans:

The height of stem above the liquid surface is $5.776{\rm{ in}}$ .