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Chapter 11: Problem 32

A solid gold bar is pulled up from the hold of the sunken RMS Titanic. (a) What happens to its volume as it goes from the pressure at the ship to the lower pressure at the ocean's surface? (b) The pressure difference is proportional to the depth. How many times greater would the volume change have been had the ship been twice as deep? (c) The bulk modulus of lead is one- fourth that of gold. Find the ratio of the volume change of a solid lead bar to that of a gold bar of equal volume for the same pressure change.

Short Answer

Expert verified
(a) Volume increases. (b) Twice as great. (c) Ratio of 4:1.

Step by step solution

01

Understanding Volume Change Due to Pressure

When an object is submerged in a fluid, it experiences pressure from the fluid surrounding it. As the gold bar is moved from the pressure at the depth of the RMS Titanic to the lower pressure at the ocean's surface, the pressure exerted on the bar decreases. According to the principles of fluid mechanics, a decrease in pressure leads to an increase in volume.
02

Calculating the Effect of Depth on Volume Change

The change in pressure is proportional to the depth. Therefore, if the depth is doubled, the change in pressure will also be doubled. According to the formula for volume change \( \Delta V = \frac{1}{B} P \Delta V_0 \), where \( B \) is the bulk modulus and \( P \) is the pressure change, doubling the pressure change would double the volume change. Thus, the volume change would be twice as great if the ship were twice as deep.
03

Comparing Volume Changes Between Gold and Lead

The bulk modulus \( B \) is a measure of a material's resistance to compression. Gold has a higher bulk modulus compared to lead. Given that the bulk modulus of lead is one-fourth that of gold, it implies that lead's volume changes more than gold under the same pressure change. The volume change for lead would be four times that of gold because \( B_{ ext{lead}} = \frac{1}{4} B_{ ext{gold}} \). Consequently, the ratio of the volume change of lead to gold is 4:1.

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Key Concepts

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

Pressure Change
In the intriguing field of fluid mechanics, understanding how pressure impacts objects is fundamental. As the solid gold bar ascends from the depths of the RMS Titanic, it transitions from a high-pressure environment to a significantly lower pressure environment at the ocean鈥檚 surface. This contrast in pressure is what we refer to as a 'pressure change.'
When an object is submerged deeply in a fluid, it experiences more pressure due to the weight of the fluid above it. The deeper the object is, the greater the pressure it faces. Therefore, moving the gold bar from deep underwater to the surface results in a reduction of this pressure. The concept of pressure change is pivotal because it leads to various physical phenomena, one of which is the alteration in an object's volume, as we'll discuss next.
Volume Change
The volume of an object is not a permanent, fixed characteristic but can shift in response to external factors like pressure. In the example of the gold bar, as it is lifted to the surface, the decrease in pressure allows its volume to increase. This principle is rooted in fluid mechanics: when pressure on an object decreases, the volume of the object tends to increase.
  • The direct relationship between pressure change and volume change means the bigger the pressure change, the larger the volume change.
  • Depth influences this change significantly. When the bar is at twice the original depth, it experiences twice the pressure change as it rises, doubling the potential change in volume.
Understanding volume change helps us predict how objects will react as they move through environments with different pressures.
Bulk Modulus
Bulk modulus is a key concept in determining how a material reacts under pressure. It measures the material's resistance to uniform compression. The bulk modulus helps us understand why different materials expand or compress differently under the same conditions.
In comparing materials like gold and lead, their differing bulk moduli have significant implications. Gold, with a higher bulk modulus, compresses less than lead under identical pressure changes. Conversely, lead's lower bulk modulus indicates it expands or compresses more easily. Given that lead鈥檚 bulk modulus is one-fourth of gold鈥檚, the volume change experienced by lead is notably more substantial.
To put it simply, if exposed to the same pressure change, lead's volume would alter four times more than that of gold. This information is crucial for engineers and scientists as they design materials to withstand specific pressure conditions, ensuring their applications meet safety and performance standards.

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Most popular questions from this chapter

Bulk Modulus of an Ideal Gas. The equation of state (the equation relating pressure, volume, and temperature) for an ideal gas is \(p V=n R T,\) where \(n\) and \(R\) are constants. (a) Show that if the gas is compressed while the temperature \(T\) is held constant, the bulk modulus is equal to the pressure. (b) When an ideal gas is compressed without the transfer of any heat into or out of it, the pressure and volume are related by \(p V^{\gamma}=\) constant, where \(\gamma\) is a constant having different values for different gases. Show that, in this case, the bulk modulus is given by \(B=\gamma p\)

\(\mathrm{A} 3.00 \mathrm{-m}-\mathrm{long}, 240-\mathrm{N},\) uniform rod at the zoo is held in a horizontal position by two ropes at its ends (Fig. Ell. 19). The left rope makes an angle of \(150^{\circ}\) with the rod and the right rope makes an angle \(\theta\) with the horizontal. \(\mathrm{A} 90\) -N howler monkey (Alouatta seniculus) hangs motionless 0.50 \(\mathrm{m}\) from the right end of the rod as he carefully studies you. Calculate the tensions in the two ropes and the angle \(\theta\) . First make a free-body diagram of the rod.

An engineer is designing a conveyor system for loading hay bales into a wagon (Fig. P11.81). Each bale is 0.25 \(\mathrm{m}\) wide, 0.50 \(\mathrm{m}\) high, and 0.80 \(\mathrm{m}\) long (the dimension perpendicular to the plane of the figure), with mass 30.0 kg. The center of gravity of each bale is at its geometrical center. The coefficient of static friction between a bale and the conveyor belt is 0.60 , and the belt moves with constant speed. (a) The angle \(\beta\) of the conveyor is slowly increased. At some critical angle a bale will tip (if it doesn't slip first), and at some different critical angle it will slip (if it doesn't tip first). Find the two critical angles and determine which happens at the smaller angle. (b) Would the outcome of part (a) be different if the coefficient of friction were 0.40\(?\)

A uniform rod is 2.00 \(\mathrm{m}\) long and has 1.80 \(\mathrm{kg} . \mathrm{A}\) \(2.40-\) -kg clamp is attached to the rod. How far should the center of gravity of the clamp be from the left-hand end of the rod in order for the center of gravity of the composite object to be 1.20 \(\mathrm{m}\) from the left-hand end of the rod?

Two people carry a heavy electric motor by placing it on a light board 2.00 \(\mathrm{m}\) long. One person lifts at one end with a force of \(400 \mathrm{N},\) and the other lifts the opposite end with a force of 600 \(\mathrm{N}\) . (a) What is the weight of the motor, and where along the board is its center of gravity located? (b) Suppose the board is not light but weighs \(200 \mathrm{N},\) with its center of gravity at its center, and the two people each exert the same forces as before. What is the weight of the motor in this case, and where is its center of gravity located?
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