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Find the elastic potential energy per unit volume of water (in \(\left.\times 10^{3} \mathrm{Jm}^{-3}\right)\) at a depth of \(1 \mathrm{~km}\). Given, compressibility of water \(=5 \times 10^{10} \mathrm{Sl}\) units and density of water \(=1000 \mathrm{~kg} \mathrm{~m}^{-3}\).

Short Answer

Expert verified
The elastic potential energy per unit volume at 1 km depth is approximately \( 24 \times 10^3 \, \text{Jm}^{-3} \).

Step by step solution

01

Understand the Formula of Elastic Potential Energy

Elastic potential energy per unit volume, also known as strain energy density, can be calculated using the formula:\[ U = \frac{1}{2} P \Delta V \]Where \( U \) is the elastic potential energy per unit volume, \( P \) is the pressure exerted on the fluid and \( \Delta V \) is the change in volume per unit volume of the fluid due to pressure.
02

Calculate Pressure at Depth

The pressure exerted by a column of fluid can be calculated using the formula:\[ P = \rho g h \]Where \( \rho = 1000 \, \text{kg/m}^3 \) (density of water), \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity), and \( h = 1000 \, \text{m} \) (depth). Substituting the values:\[ P = 1000 \times 9.8 \times 1000 = 9.8 \times 10^6 \, \text{Pa} \]
03

Determine Volume Change Per Unit Volume

Compressibility \( (\beta) \) of a fluid relates to volume change under pressure:\[ \beta = \frac{\Delta V}{V \times P} \]Given that \( \beta = 5 \times 10^{-10} \), therefore:\[ \Delta V = \beta \times V \times P \]Since it is volume change per unit volume, \( V = 1 \):\[ \Delta V = 5 \times 10^{-10} \times 9.8 \times 10^6 \approx 4.9 \times 10^{-3} \]
04

Calculate Elastic Potential Energy per Unit Volume

Using the formula from step 1, substitute the pressure \( P \) and the change in volume per unit volume \( \Delta V \):\[ U = \frac{1}{2} \times 9.8 \times 10^6 \times 4.9 \times 10^{-3} \]\[ U = 24.01 \, \text{kJm}^{-3} \]\( \approx 24 \times 10^3 \, \text{Jm}^{-3} \) (in units of \( 10^3 \text{ Jm}^{-3} \))

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Key Concepts

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

Strain Energy Density
Strain energy density refers to the amount of elastic potential energy stored per unit volume of a material under deformation. This concept is essential to understand when considering how materials like fluids or solids respond to stress or pressure.

In the context of fluids, the energy is stored when the fluid is compressed. This energy can be calculated using the formula:
  • \( U = \frac{1}{2} P \Delta V \)
  • Where \( U \) is the elastic potential energy per unit volume, \( P \) is pressure and \( \Delta V \) is the change in volume per unit volume.
Here, strain energy density helps us quantify how much energy a fluid can store and can help in analyzing scenarios where fluids are subjected to varying pressures, such as in deep-water environments.
Compressibility of Water
Compressibility is a measure of how much the volume of a substance decreases under pressure. For fluids like water, compressibility plays a crucial role in understanding how they behave under different conditions.

In simpler terms, it tells us how "squishable" the fluid is. The compressibility of water is given as \( \beta = 5 \times 10^{-10} \), indicating that water is relatively incompressible compared to gases.

This value is used to calculate the change in volume of the water under pressure using:
  • \( \beta = \frac{\Delta V}{V \times P} \)
  • where \( \Delta V \) is the change in volume and \( P \) is the pressure exerted on the water.
Understanding compressibility is vital in engineering applications where pressure conditions change, such as designing underwater structures or assessing the impacts of natural phenomena like tides and currents.
Pressure at Depth
Pressure at depth is a fundamental concept in fluid mechanics. It describes how pressure increases with depth due to the weight of the fluid above.

The relationship is defined by the equation:
  • \( P = \rho g h \)
  • where \( \rho \) is the density of the fluid, \( g \) is the acceleration due to gravity, and \( h \) is the depth of the fluid column.
For water, density is typically \( 1000 \, \text{kg/m}^3 \) and gravity is \( 9.8 \, \text{m/s}^2 \). So, at a depth of \( 1 \, \text{km} \), the pressure is extremely high, approaching \( 9.8 \times 10^6 \, \text{Pa} \).

This increase in pressure with depth is critical, for instance, when calculating how submarines and deep-sea vehicles withstand underwater forces or when studying the earth's crust activities.
Volume Change in Fluids
Volume change in fluids under pressure is a cornerstone for understanding fluid dynamics, especially in non-atmospheric environments. When a fluid is under pressure, it undergoes a slight reduction in volume, which relates directly to its compressibility.

The formula for volume change per unit volume is given by:
  • \( \Delta V = \beta \times V \times P \)
  • where \( \Delta V \) is the change in volume, \( \beta \) is the compressibility, and \( P \) is the applied pressure.
Even though water is typically incompressible, under large pressures like those at ocean depths, there are still measurable changes in volume. This understanding is critical in scenarios like:
  • Calculating the buoyancy force which influences flotation and submersion of objects.
  • Designing resilient structures to withstand changes in fluid volume.
Recognizing these minimal, but significant changes ensures safety and efficiency in fluid-related engineering solutions.

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Most popular questions from this chapter

Choose the correct statements from the following: (1) Steel is more elastic than rubber. (2) The stretching of a coil spring is determined by the Young's modulus of the wire of the spring. (3) The frequency of a tuning fork is determined by the shear modulus of the material of the fork. (4) When a material is subjected to a tensile (stretching) stress the restoring forces are caused by interatomic attraction.

A rod is made of uniform material and has non-uniform cross section. It is fixed at both the ends as shown and heated at mid-section. Which of the following statements are not correct? (1) Force of compression in the rod will be maximum at mid-section. (2) Compressive stress in the rod will be maximum at left end. (3) Since rod in fixed at both the ends, its length will remain unchanged. Hence, no strain will be induced in it. (4) None of the above.

A rubber rope of length \(8 \mathrm{~m}\) is hung from the ceiling of a room. What is the increase in length of rope due to its own weight? (Given: Young's modulus of elasticity of rubber \(=5 \times 10^{6} \mathrm{~N} / \mathrm{m}\) and density of rubber \(=1.5 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}\). Take \(\left.g=10 \mathrm{~ms}^{-2}\right)\) (1) \(1.5 \mathrm{~mm}\) (2) \(6 \mathrm{~mm}\) (3) \(24 \mathrm{~mm}\) (4) \(96 \mathrm{~mm}\)

A nylon rope \(2 \mathrm{~cm}\) in diameter has a breaking strength of \(1.5 \times 10^{5} \mathrm{~N}\). The breaking strength of a similar rope \(1 \mathrm{~cm}\) in diameter is (1) \(0.375 \times 10^{5} \mathrm{~N}\) (2) \(2 \times 10^{5} \mathrm{~N}\) (3) \(6 \times 10^{5} \mathrm{~N}\) (4) \(9 \times 10^{4} \mathrm{~N}\)

A copper wire of negligible mass, \(1 \mathrm{~m}\) length and cross-sectional area \(10^{-6} \mathrm{~m}^{2}\) is kept on a smooth horizontal table with one end fixed. A ball of mass \(1 \mathrm{~kg}\) is attached to the other end. The wire and the ball are rotating with an angular velocity of \(20 \mathrm{rad} / \mathrm{s}\). If the elongation in the wire is \(10^{-3} \mathrm{~m}\). If for the same wire as stated above the angular velocity is increase to \(100 \mathrm{rad} / \mathrm{s}\) and the wire breaks down, the breaking stress is (1) \(1.5 \times 10^{10} \mathrm{~N} / \mathrm{m}^{2}\) (2) \(10^{10} \mathrm{~N} / \mathrm{m}^{2}\) (3) \(2.0 \times 10^{10} \mathrm{~N} / \mathrm{m}^{2}\) (4) \(3.0 \times 10^{10} \mathrm{~N} / \mathrm{m}^{2}\)

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