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Chapter 2: Problem 98

A hydrometer with a stem diameter of \(9 \mathrm{~mm}\) is placed in distilled water, and the volume of the hydrometer below the water surface is estimated to be \(20 \mathrm{~cm}^{3}\). If the hydrometer is placed in a liquid with a specific gravity of 1.2 , how far above the liquid surface will the distilled water mark be located?

Short Answer

Expert verified
The distilled water mark will be 5.24 cm above the new liquid surface.

Step by step solution

01

Understand the Problem

First, identify that we need to calculate how much of the hydrometer will be above the surface when it is placed in a liquid with a specific gravity of 1.2, compared to when it is in water.
02

Calculate Displaced Volume in New Liquid

The volume of liquid displaced by the hydrometer must equal the hydrometer's weight. Since the hydrometer displaces 20 cm鲁 of water, it weighs as much as 20 cm鲁 of water. In the liquid with specific gravity 1.2, the same hydrometer displaces a smaller volume because it is denser. The specific gravity of 1.2 implies that the liquid is 1.2 times denser than water, so the displaced volume is calculated as \[ V_{new} = \frac{V_{water}}{1.2} = \frac{20}{1.2} \approx 16.67 \text{ cm}^3 \]
03

Identify the Volume of the Stem Above the New Liquid Surface

Since the hydrometer originally displaced 20 cm鲁 in water but now only displaces approximately 16.67 cm鲁, the remaining difference is the volume that corresponds to the portion of the stem above the liquid. Therefore, the volume of the stem above the liquid is \[ V_{above} = 20 - 16.67 = 3.33 \text{ cm}^3 \]
04

Calculate the Height Above the Liquid Surface

Given the stem diameter of 9 mm, convert this to radius in centimeters (since all our volumes are in cm鲁): \[ r = \frac{9}{2 \times 10} = 0.45 \text{ cm} \] The cross-sectional area of the stem is \[ A = \pi r^2 = \pi (0.45)^2 \approx 0.636 \text{ cm}^2 \] Now, calculate the height of 3.33 cm鲁 of stem above the surface:\[ h = \frac{V_{above}}{A} = \frac{3.33}{0.636} \approx 5.24 \text{ cm}\]

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Key Concepts

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

Hydrometer
A hydrometer is a simple device used to measure the density or specific gravity of liquids. It typically consists of a glass tube with a weighted bottom and a stem with a scale. When a hydrometer is placed in a liquid, it floats, and the depth to which it submerges is a direct measure of the liquid's density. The denser the liquid, the less the hydrometer will sink. This floating principle is based on Archimedes' principle. Understanding this assists in applications ranging from brewing beer to determining the concentration of solutions.
Specific Gravity
Specific gravity is an important concept in fluid mechanics, representing the ratio of the density of a liquid to the density of water at 4掳C (where \(\rho_{water} = 1 \, \text{g/cm}^3\)). It is a dimensionless quantity because it is a ratio, meaning it has no units. Specific gravity helps in quickly identifying how much denser a liquid is compared to water. If a substance has a specific gravity greater than 1, it is denser than water and a hydrometer would float higher in that liquid. A specific gravity less than 1 indicates a less dense liquid, causing the hydrometer to sink further.
Displaced Volume
When an object is immersed in a liquid, it displaces a volume of liquid equal to the volume of its submerged part. This displaced volume equates to the object's weight when floating, according to Archimedes' Principle. For instance, the hydrometer in the exercise displaces 20 cm鲁 of water, meaning the hydrometer weighs the same as 20 cm鲁 of water. In a denser liquid, such as one with a specific gravity of 1.2, the hydrometer displaces less volume because the liquid can support the hydrometer's weight with a smaller volume. This reduced displacement can be calculated to determine how far above the liquid surface the hydrometer stem extends.
Density
Density is defined as mass per unit volume, usually expressed in g/cm鲁 or kg/m鲁. It determines how much mass is contained in a given volume. The concept is crucial when working on problems involving floating objects like hydrometers. In the given exercise, density plays a critical role. Since the liquid with specific gravity 1.2 is denser than water, the same mass of the hydrometer requires a smaller volume to be displaced to float, leading to a different immersion depth compared to water. The key takeaway here is that different densities result in different displacements and consequently different readings on the hydrometer.

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Most popular questions from this chapter

A rectangular tank \(3 \mathrm{~m}\) long, \(0.8 \mathrm{~m}\) wide, and \(1.6 \mathrm{~m}\) deep is filled with kerosene to a depth of \(1.2 \mathrm{~m}\) and placed on a truck. Consider two orientations of the tank on the truck: (a) the long side aligned with the direction of truck motion and (b) the short side aligned with the direction of truck motion. Which orientation would allow the greatest truck acceleration without spillage? What is that acceleration?

(a) At what depth below the surface of a water body will the (gauge) pressure be equal to \(200 \mathrm{kPa}\) ? (b) If a 1.55-m-tall person orients himself vertically underwater in a pool, what pressure difference does he feel between his head and his toes? Assume water at \(20^{\circ} \mathrm{C}\).

A water truck is mounted with a cylindrical tank that has a diameter of \(2 \mathrm{~m}\) and a length of \(10 \mathrm{~m}\). The long axis of the tank is oriented with the direction of truck motion. If the tank is filled with water at \(20^{\circ} \mathrm{C}\) and the truck accelerates at a rate of \(2 \mathrm{~m} / \mathrm{s}^{2}\), estimate the difference in magnitude between the resultant hydrostatic force on the front and back of the tank.

If a body of specific gravity \(\mathrm{SG}_{1}\) is placed in a liquid of specific gravity \(\mathrm{SG}_{2}\), what fraction of the total volume of the body will be above the surface of the liquid? If an iceberg has a specific gravity of 0.95 and is floating in seawater with a specific gravity of \(1.25,\) what fraction of the iceberg is above water?

A 1.6-m-diameter cylinder is filled with a liquid to a depth of \(1.1 \mathrm{~m}\) and rotated about its center axis. (a) Assuming that the cylinder is tall enough for the liquid not to spill, at what rotational speed will the liquid surface intersect the bottom of the cylinder? (b) If the cylinder is rotated at 60 rpm, what is the minimum height of the cylinder that prevents spillage?
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