Textbook Question
A circular steel wire 2.00 m long must stretch no more than 0.25 cm when a tensile force of 700 N is applied to each end of the wire. What minimum diameter is required for the wire?
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Ch 12: Fluid Mechanics
Chapter 12, Problem 11
A solid gold bar is pulled up from the hold of the sunken RMS Titanic. (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.
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Identify the relationship between bulk modulus (B), pressure change (ΔP), and volume change (ΔV). The formula to use is B = -V * (ΔP/ΔV), where V is the original volume.
Rearrange the formula to solve for ΔV, the change in volume. This gives ΔV = -V * (ΔP/B).
Note that the bulk modulus of lead is one-fourth that of gold, i.e., B_lead = 1/4 * B_gold.
Substitute the bulk modulus values into the volume change formula for both lead and gold. This results in ΔV_lead = -V * (ΔP/(1/4 * B_gold)) and ΔV_gold = -V * (ΔP/B_gold).
Calculate the ratio of the volume changes for lead and gold. This is done by dividing the volume change formula of lead by that of gold: (ΔV_lead / ΔV_gold) = ((-V * (ΔP/(1/4 * B_gold))) / (-V * (ΔP/B_gold))). Simplify this expression to find the ratio.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Bulk Modulus
The bulk modulus is a measure of a material's resistance to uniform compression. It quantifies how much a material's volume decreases under pressure, defined as the ratio of the change in pressure to the relative change in volume. A higher bulk modulus indicates that a material is less compressible. In this context, the bulk modulus of gold is compared to that of lead to determine how each material responds to the same pressure change.
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Example 1
Volume Change
Volume change refers to the alteration in the volume of a material when subjected to external forces, such as pressure. For solids, this change can be calculated using the formula ΔV = -V0 * (ΔP / B), where ΔV is the change in volume, V0 is the original volume, ΔP is the change in pressure, and B is the bulk modulus. Understanding how volume change relates to pressure and material properties is essential for solving the problem.
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Ratio of Volume Changes
The ratio of volume changes between two materials under the same pressure change is determined by their respective bulk moduli. If the bulk modulus of lead is one-fourth that of gold, it implies that lead will experience a greater volume change than gold for the same pressure increase. This concept is crucial for comparing the compressibility of different materials and is key to solving the problem presented.
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Related Practice
Textbook Question
Stress on a Mountaineer's Rope. A nylon rope used by mountaineers elongates 1.10 m under the weight of a 65.0-kg climber. If the rope is 45.0 m in length and 7.0 mm in diameter, what is Young's modulus for nylon?
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Textbook Question
A specimen of oil having an initial volume of 600 cm3 is subjected to a pressure increase of 3.6×10^6 Pa, and the volume is found to decrease by 0.45 cm^3. What is the bulk modulus of the material and the compressibility?
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Textbook Question
In the Challenger Deep of the Marianas Trench, the depth of seawater is 10.9 km and the pressure is 1.16×10^8 Pa (about 1.15×10^3 atm). (a) If a cubic meter of water is taken from the surface to this depth, what is the change in its volume? (Normal atmospheric pressure is about 1.0×10^5 Pa. Assume that k for seawater is the same as the freshwater value given in Table 11.2.)
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Textbook Question
A square steel plate is 10.0 cm on a side and 0.500 cm thick. (a) Find the shear strain that results if a force of magnitude 9.0×10^5 N is applied to each of the four sides, parallel to the side. (b) Find the displacement x (in centimeters).
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