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Chapter 11: Problem 42

A hydrometer is a device used to measure the density of a liquid. It is a cylindrical tube weighted at one end, so that it floats with the heavier end downward. The tube is contained inside a large "medicine dropper," into which the liquid is drawn using the squeeze bulb (see the drawing). For use with your car, marks are put on the tube so that the level at which it floats indicates whether the liquid is battery acid (more dense) or antifreeze (less dense). The hydrometer has a weight of \(W=5.88 \times 10^{-2} \mathrm{N}\) and a crosssectional area of \(A=7.85 \times 10^{-5} \mathrm{m}^{2} .\) How far from the bottom of the tube should the mark be put that denotes (a) battery acid \(\left(\rho=1280 \mathrm{kg} / \mathrm{m}^{3}\right)\) and \((\mathrm{b})\) antifreeze \(\left(\rho=1073 \mathrm{kg} / \mathrm{m}^{3}\right) ?\)

Short Answer

Expert verified
(a) 0.0596 m for battery acid, (b) 0.0704 m for antifreeze.

Step by step solution

01

Understanding the Principle

The hydrometer works on the principle of buoyancy, where the weight of the liquid displaced by the submerged part of the hydrometer equals the weight of the hydrometer itself. This can be expressed with the equation: \[ W = \rho \cdot g \cdot V \]where \(V\) is the volume of the liquid displaced, \(\rho\) is the density of the liquid, and \(g\) is the acceleration due to gravity (approximately \(9.81 \, \text{m/s}^2\)).
02

Calculating Volume Displaced

To find the volume \(V\) of the liquid displaced, we rearrange the equation:\[ V = \frac{W}{\rho \cdot g} \]Substituting the weight of the hydrometer \(W = 5.88 \times 10^{-2} \, \mathrm{N}\):\( V = \frac{5.88 \times 10^{-2}}{\rho \cdot 9.81} \).
03

Finding Depth for Battery Acid

For battery acid with \(\rho = 1280 \, \text{kg/m}^3\), substitute \(\rho\) into the volume equation:\[ V = \frac{5.88 \times 10^{-2}}{1280 \times 9.81} \approx 4.68 \times 10^{-6} \, \text{m}^3 \].Since \(V = A \cdot h\), where \(A\) is the cross-sectional area and \(h\) is the depth, we can find \(h\) by:\[ h = \frac{V}{A} = \frac{4.68 \times 10^{-6}}{7.85 \times 10^{-5}} \approx 0.0596 \, \text{m} \].
04

Finding Depth for Antifreeze

For antifreeze with \(\rho = 1073 \, \text{kg/m}^3\), substitute \(\rho\) into the volume equation:\[ V = \frac{5.88 \times 10^{-2}}{1073 \times 9.81} \approx 5.53 \times 10^{-6} \, \text{m}^3 \].Using \(V = A \cdot h\), find \(h\) by:\[ h = \frac{V}{A} = \frac{5.53 \times 10^{-6}}{7.85 \times 10^{-5}} \approx 0.0704 \, \text{m} \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Buoyancy
Buoyancy is a fundamental principle in physics that describes the upward force exerted by a fluid, such as a liquid or gas, on an object submerged in it. This force enables objects to float or to sink, depending on their density relative to the fluid. In the context of a hydrometer, buoyancy is the principle that allows the device to float in a liquid to measure its density. When the hydrometer is placed in a liquid:
  • The weight of the displaced liquid equals the weight of the hydrometer.
  • This is why a hydrometer can float at different depths in liquids of varying densities.
The balance between the weight of the liquid displaced and the weight of the hydrometer is expressed in the equation: \[ W = \rho \cdot g \cdot V \] - Here, the weight \( W \) of the hydrometer is supported by the buoyant force provided by the volume \( V \) of the displaced liquid multiplied by the liquid's density \( \rho \) and gravity \( g \).
Understanding this equation helps in designing the scale on the hydrometer, so it correctly indicates the density of the liquid based on how deep it floats.
Density measurement
Measuring density with a hydrometer is a straightforward process, thanks to the principles of buoyancy and volume displacement. The hydrometer is calibrated to read density directly. Different fluids will cause the hydrometer to float at different levels:
  • If a fluid is denser, such as battery acid, the hydrometer will not sink as far.
  • Conversely, in less dense fluids like antifreeze, the hydrometer will float lower.
The scale on a hydrometer is typically marked with density values. By reading directly where the fluid's surface touches the scale, one can determine the density of the fluid: - In our exercise, specific densities are given for battery acid and antifreeze鈥�1280 kg/m鲁 and 1073 kg/m鲁, respectively.
This calibration is crucial for practical applications, like checking battery health or the concentration of antifreeze.
Hydrometers are not one-size-fits-all; specific applications might require custom calibration depending on the density range of the fluids measured.
Volume displacement
Volume displacement is a key concept when understanding how hydrometers operate. It refers to the amount of liquid pushed out of the way by an object's submerged portion. In our exercise:- The formula \[ V = \frac{W}{\rho \cdot g} \] allows calculation of the liquid volume displaced, which is essential to finding how deep the hydrometer will sink in a particular liquid.
  • This volume is critical as it is directly linked to the position where the hydrometer stabilizes in the liquid.
  • The volume \( V \) is also related to the cross-sectional area \( A \) of the hydrometer through the equation \( V = A \cdot h \).
  • Here, \( h \) is the depth or height from the bottom of the hydrometer submerged.
In practice, the volume displacement helps users understand and predict the floating behavior of the hydrometer across fluids of various densities.
This is why finding the displaced volume is a crucial step in determining how far a hydrometer will sink, enabling accurate density readings for different liquids.

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Most popular questions from this chapter

(a) The volume flow rate in an artery supplying the brain is \(3.6 \times 10^{-6} \mathrm{m}^{3} / \mathrm{s} .\) If the radius of the artery is \(5.2\) \(\mathrm{mm},\) determine the average blood speed. (b) Find the average blood speed at a constriction in the artery if the constriction reduces the radius by a factor of \(3 .\) Assume that the volume flow rate is the same as that in part (a).

If a scuba diver descends too quickly into the sea, the internal pressure on each eardrum remains at atmospheric pressure, while the external pressure increases due to the increased water depth. At sufficient depths, the difference between the external and internal pressures can rupture an eardrum. Eardrums can rupture when the pressure difference is as little as \(35\) \(\mathrm{kPa}\). What is the depth at which this pressure difference could occur? The density of seawater is \(1025\) \(\mathrm{kg} /\mathrm{m}^{3}\).

A room has a volume of \(120 \mathrm{m}^{3} .\) An air-conditioning system is to replace the air in this room every twenty minutes, using ducts that have a square cross section. Assuming that air can be treated as an incompressible fluid, find the length of a side of the square if the air speed within the ducts is (a) \(3.0 \mathrm{m} / \mathrm{s}\) and (b) \(5.0 \mathrm{m} / \mathrm{s}\).

Neutron stars consist only of neutrons and have unbelievably high densities. A typical mass and radius for a neutron star might be \(2.7 \times 10^{28}\) \(\mathrm{kg}\) and \(1.2 \times 10^{3}\) \(\mathrm{m} .\) (a) Find the density of such a star. (b) If a dime \(\left(V=2.0 \times 10^{-7} \mathrm{m}^{3}\right)\) were made from this material, how much would it weigh (in pounds)?

The drawing shows a hydraulic system used with disc brakes. The force \(\overrightarrow{\mathbf{F}}\) is applied perpendicularly to the brake pedal. The pedal rotates about the axis shown in the drawing and causes a force to be applied perpendicularly to the input piston (radius \(\left.=9.50 \times 10^{-3} \mathrm{m}\right)\) in the master cylinder. The resulting pressure is transmitted by the brake fluid to the output plungers (radii \(\left.=1.90 \times 10^{-2} \mathrm{m}\right)\), which are covered with the brake linings. The linings are pressed against both sides of a disc attached to the rotating wheel. Suppose that the magnitude of \(\overrightarrow{\mathbf{F}}\) is \(9.00\) \(\mathrm{N}\). Assume that the input piston and the output plungers are at the same vertical level, and find the force applied to each side of the rotating disc.
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