/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 2 If \(B\) is the bulk modulus of ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

If \(B\) is the bulk modulus of a metal and a pressure \(P\) is applied uniformly on all sides of the metal with density \(D\), then the fractional increase in density is given by (a) \(\frac{B}{P}\) (b) \(\frac{P}{B}\) (c) \(\frac{P D}{B}\) (d) \(\frac{B D}{P}\)

Short Answer

Expert verified
The fractional increase in density is \(\frac{P}{B}\), answer (b).

Step by step solution

01

Understanding the Problem

The exercise asks for the fractional increase in density when a pressure \(P\) is applied uniformly to a metal with bulk modulus \(B\) and density \(D\). The bulk modulus relates to how compressible a material is.
02

Formula for Bulk Modulus

The bulk modulus \(B\) is defined as the ratio of the applied pressure \(P\) to the fractional decrease in volume \(\frac{\Delta V}{V}\), given by \(B = -P \frac{V}{\Delta V}\). This can be rearranged to express \(\frac{\Delta V}{V} = -\frac{P}{B}\).
03

Relation Between Volume and Density

Density \(D\) is the mass \(M\) divided by volume \(V\). Therefore, \(D = \frac{M}{V}\). If \(\Delta V\) represents the change in volume, then the new density \(D'\) would be \(D' = \frac{M}{V - \Delta V}\).
04

Expressing Change in Density

Use the identity \(\frac{1}{V - \Delta V} \approx \frac{1}{V} + \frac{\Delta V}{V^2}\), which is valid when \(\Delta V\) is small. Thus, \(D' \approx \frac{M}{V}(1 + \frac{\Delta V}{V}) = D(1 + \frac{\Delta V}{V})\).
05

Calculating Fractional Increase in Density

The fractional increase in density \(\frac{D' - D}{D}\) is \(\frac{D(1 + \frac{\Delta V}{V}) - D}{D} = \frac{\Delta V}{V}\). Substitute \(\frac{\Delta V}{V}\) with \(-\frac{P}{B}\) from our earlier equation: The fractional increase is \(-(-\frac{P}{B}) = \frac{P}{B}\).
06

Final Answer Selection

Based on the calculated fractional increase in density, the correct answer is (b) \(\frac{P}{B}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Fractional Increase in Density
When pressure is applied to a material, its density changes proportionally. But how can we calculate this change? We do this by finding the fractional increase in density, which tells us by what percentage the density of a material increases.
This fractional increase helps us understand how much denser a material becomes when it is compressed. For instance, imagine pushing a sponge; its size seems to shrink, but its mass remains the same, making it denser.
To calculate the fractional increase in density, we need to use the equation \(\frac{P}{B}\). Here, \(P\) represents the applied pressure and \(B\) is the bulk modulus, a measure of the material's resistance to being compressed. So, when we say the fractional increase in density is \(\frac{P}{B}\), it tells us how much the density will increase in response to the applied pressure, specific to the material's compressibility.
Pressure and Volume Relationship
Pressure and volume have a unique relationship characterized by the bulk modulus. When we apply pressure to a material, its volume changes, typically decreasing to accommodate the pressure.
The bulk modulus \(B\) is key in understanding this relationship. It's defined by the ratio of the pressure \(P\) applied to a material to the relative change in its volume. Mathematically, we express this as \(B = -P \frac{V}{\Delta V}\), where \(\Delta V\) represents the decrease in volume and \(V\) the original volume.
This formula highlights how a material's volume responds to external pressure. Understanding this helps us calculate how much a material can compress when a specific pressure is applied. By understanding this relationship, you better explain why certain materials like metals compress less than others, such as gases, under similar pressures.
Compressibility of Materials
The bulk modulus \(B\) is a fundamental concept in understanding how compressible a material is. It tells us about a material's ability to withstand changes in volume under pressure.
A higher bulk modulus means a material is less compressible. It can withstand higher pressures without significantly changing its volume. For instance, metals typically have a high bulk modulus, making them stiffer and less susceptible to compression.
On the other hand, materials like rubber or gases have a lower bulk modulus. They compress more easily under pressure because their structure allows for more movement and adjustment.
Knowing a material's compressibility is crucial in applications like engineering and construction, where understanding how materials respond to pressure ensures safety and efficiency. By comprehending a material's bulk modulus, you predict how it will behave in different scenarios.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Water rises to a height of \(10 \mathrm{~cm}\) in a capillary tube and mercury falls to a depth of \(3.42 \mathrm{~cm}\) in the same capillary tube. If the density of mercury is \(13.6\) and angle of contact is \(135^{\circ}\), the ratio of surface tension for water and mercury is (Angle of contact for water and glass is \(8^{\circ}\) ) (a) \(1: 0.5\) (b) \(1: 3\) (c) \(1: 6.5\) (d) \(1.5: 1\)

A particle of mass \(m\) under the influence of a force \(F\) which varies with the displacement \(x\) according to the relation \(F=-k x+F_{o}\), in which \(k\) and \(F_{\text {ore constants. }}\) The particle when disturbed will oscillate (a) about \(x=0\) with \(\omega \neq \sqrt{\frac{k}{m}}\) (b) about \(x=0\) with \(\omega=\sqrt{\frac{k}{m}}\) (c) about \(x=\frac{F_{0}}{k}\) with \(\omega=\sqrt{\frac{k}{m}}\) (d) about \(x=\frac{F_{0}}{k}\) with \(\omega \neq \sqrt{\frac{k}{m}}\)

The pressure applied from all directions on a cube is \(P\). How much its temperature should be raised to maintain the original volume? (The volume elasticity of the cube is \(\beta\) and the coefficient of volume expansion is \(\alpha\) ) (a) \(\frac{P}{\alpha \beta}\) (b) \(\frac{P \alpha}{\beta}\) (c) \(\frac{P \beta}{\alpha}\) (d) \(\frac{\alpha \beta}{P}\)

A uniform solid cylinder made of steel, is compressed along the axis. Which of the following statements is correct? (a) Decrease in volume of the cylinder is independent of its area of cross- section but directly proportional to its length. (b) Decrease in volume of the cylinder is independent of its area of cross- section but inversely proportional to its length. (c) Decrease in volume of the cylinder is directly proportional to its cross- sectional area but is independent of its length (d) Decrease in volume of the cylinder is inversely proportional to its volume as well as length.

When a spring is stretched by a distance \(x\), it exerts a force, given by \(F=\left(-5 x-16 x^{3}\right) \mathrm{N}\). The work done, when the spring is stretched from \(0.1 \mathrm{~m}\) to \(0.2 \mathrm{~m}\) is (a) \(8.7 \times 10^{-2} \mathrm{~J}\) (b) \(12.2 \times 10^{-2} \mathrm{~J}\) (c) \(8.7 \times 10^{-1} \mathrm{~J}\) (d) \(12.2 \times 10^{-1} \mathrm{~J}\)

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.