/*! 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 7 The ratio of coefficients of cub... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The ratio of coefficients of cubical expansion and linear expansion is (A) \(1: 1\) (B) \(3: 1\) (C) \(2: 1\) (D) None of these

Short Answer

Expert verified
The ratio of coefficients of cubical expansion and linear expansion is not constant and depends on the change in temperature and the cubical coefficient of expansion. Therefore, the correct answer is (D) None of these.

Step by step solution

01

Recall the definitions of coefficients of cubical and linear expansion

The coefficient of linear expansion (\(\alpha\)) is the change in length per unit length for a temperature change of 1°C. The coefficient of cubical expansion (\(\beta\)) gives the change in volume per unit volume for a temperature change of 1°C.
02

Coefficient of cubical expansion

Given a substance with a linear, area, and volume increase of ΔL, ΔA, and ΔV, the increase in volume due to temperature change can be obtained by: \[ \beta = \frac{\Delta V}{V_0\Delta T} \] Where \(V_0\) is the initial volume, and \(\Delta T\) is the change in temperature.
03

Relate cubical expansion to linear expansion

When the length, width, and depth of a substance all increase with a linear expansion, the product of these increases is the cubical increase. Let's denote the increase in length, width, and depth as \(\Delta L_1, \Delta L_2,\) and \(\Delta L_3\), respectively. Then the cubical increase is given by: \[ \Delta V = \Delta L_1 \cdot \Delta L_2 \cdot \Delta L_3 \]
04

Using the coefficients of linear expansion

Using the coefficient of linear expansion \(\alpha\), we can express the increases in length, width, and depth as follows: \[ \Delta L_1 = \alpha L_1\Delta T \] \[ \Delta L_2 = \alpha L_2\Delta T \] \[ \Delta L_3 = \alpha L_3\Delta T \] Where \(L_1, L_2, L_3\) are the initial length, width, and depth, respectively.
05

Calculate the cubical increase using the coefficients of linear expansion

Now substituting these linear increases into the cubical increase expression, we get: \[ \Delta V = (\alpha L_1\Delta T) \cdot (\alpha L_2\Delta T) \cdot (\alpha L_3\Delta T) \] \[ \Delta V = \alpha^3 L_1L_2L_3 (\Delta T)^3 \] Now, \(L_1L_2L_3 = V_0\), the initial volume of the substance, so we can rewrite the equation: \[ \Delta V = \alpha^3 V_0 (\Delta T)^3 \]
06

Relate the cubical increase and the coefficients of linear and cubical expansion

Using the equation for the coefficient of cubical expansion, we can rewrite the equation: \[ \beta = \frac{\Delta V}{V_0\Delta T} \] Substitute the expression for \(\Delta V\): \[ \beta = \frac{\alpha^3 V_0 (\Delta T)^3}{V_0\Delta T} \] This simplifies to: \[ \beta = \alpha^3 (\Delta T)^2 \]
07

Ratio of coefficients of cubical expansion to linear expansion

Now we want to find the ratio of \(\beta\) to \(\alpha\). Dividing both sides of the equation by \(\alpha^3\), we get: \[ \frac{\beta}{\alpha^3} = (\Delta T)^2 \] Divide both sides by \((\Delta T)^2\): \[ \frac{\beta}{(\Delta T)^2}= \alpha^3 \] Now take the cube root of both sides of the equation: \[ \sqrt[3]{\frac{\beta}{(\Delta T)^2}}= \alpha \] Now to find the ratio of \(\alpha : \beta\): \[ \frac{\alpha}{\beta} = \frac{\sqrt[3]{\frac{\beta}{(\Delta T)^2}}}{\beta} \] \[ \frac{\alpha}{\beta} = \frac{1}{\beta^{2/3} (\Delta T)^{4/3}} \] The ratio of coefficients of cubical expansion and linear expansion is not constant and depends on the change in temperature and the cubical coefficient of expansion. Therefore, the correct answer is: (D) None of these

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.

Coefficient of Cubical Expansion
In thermal physics, understanding how substances expand or contract with temperature changes is crucial. The coefficient of cubical expansion, denoted as \(\beta\), describes how a material's volume changes with temperature. It is defined as the change in volume per unit volume for every degree change in temperature.
To mathematically express this, the formula is:
  • \[\beta = \frac{\Delta V}{V_0 \Delta T}\]
Here, \(\Delta V\) represents the change in volume, \(V_0\) is the initial volume, and \(\Delta T\) is the temperature change. By incorporating the principle of cubical expansion, engineers can predict how materials will behave in different thermal environments. This is particularly useful for applications involving large structures or components exposed to variable temperatures, like buildings and bridges.
Coefficient of Linear Expansion
The coefficient of linear expansion, represented by \(\alpha\), measures how a material's length changes with temperature. This coefficient is defined as the change in length per unit length for each degree of temperature change.
The equation for linear expansion is:
  • \[\Delta L = \alpha L_0 \Delta T\]
In this equation:
  • \(\Delta L\) is the change in length due to temperature change,
  • \(L_0\) is the original length of the material, and
  • \(\Delta T\) is the change in temperature.
Materials expand to different extents based on their atomic structure. This concept is widely used in designing devices such as thermal sensors, where precise length changes can signal a specific temperature adjustment.
Thermal Physics
Thermal physics is the field of study that deals with the relationship between heat and other forms of energy. A key aspect of thermal physics is understanding how substances react to changes in temperature, particularly in terms of expansion and contraction.
The concepts of linear and cubical expansions are part of the broader study of thermal physics. They reflect how materials respond to thermal energy, influencing both macroscopic physical changes in materials and microscopic particle dynamics.
  • Temperature affects particle vibrations in solids, leading to expansion.
  • Expansion coefficients (linear and cubical) help quantify these changes.
Applications span from everyday objects that need to withstand heat fluctuations to understanding the behavior of materials in extreme environments. By leveraging these principles, scientists and engineers can manipulate and use materials efficiently in various technologies and industries.

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

On the Celsius scale, the absolute zero of temperature is at (A) \(0^{\circ} \mathrm{C}\) (B) \(-32^{\circ} \mathrm{C}\) (C) \(100^{\circ} \mathrm{C}\) (D) \(-273.15^{\circ} \mathrm{C}\)

Water flows at the rate of \(0.1500 \mathrm{~kg} / \mathrm{min}\) through a tube and is heated by a heater dissipating \(25.2 \mathrm{~W}\). The inflow and outflow water temperatures are \(15.2^{\circ} \mathrm{C}\) and \(17.4^{\circ} \mathrm{C}\), respectively. When the rate of flow is increased to \(0.2318 \mathrm{~kg} /\) min and the rate of heating to \(37.8 \mathrm{~W}\), the inflow and outflow temperatures are unaltered. Find the rate of loss of heat from the tube

1000 drops of a liquid of surface tension \(\sigma\) and radius \(r\) join together to form a big single drop. The energy released raises the temperature of the drop. If \(\rho\) be the density of the liquid and \(S\) be the specific heat, the rise in temperature of the drop would be \((J=\) Joule's equivalent of heat) (A) \(\frac{\sigma}{J r S \rho}\) (B) \(\frac{10 \sigma}{J r S \rho}\) (C) \(\frac{100 \sigma}{J r S \rho}\) (D) \(\frac{27 \sigma}{10 J r S \rho}\)

Heat is associated with, (A) Kinetic energy of random motion of molecules. (B) Kinetic energy of orderly motion of molecules. (C) Total kinetic energy of random and orderly motion of molecules. (D) Kinetic energy of random motion in some cases and kinetic energy of orderly motion in other.

If mass-energy equivalence is taken into account, when water is cooled to form ice, then the mass of water should (A) increase. (B) remain unchanged. (C) decrease. (D) first increase then decrease.

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.