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Chapter 7: Problem 18

Two wires of the same material have lengths in the ratio \(1: 2\) and their radii are in the ratio \(1: \sqrt{2}\). If they are stretched by applying equal forces, the increase in their lengths will be in the ratio (1) \(\sqrt{2}: 2\) (2) \(2: \sqrt{2}\) (3) \(1: 1\) (4) \(1: 2\)

Short Answer

Expert verified
The ratio of the increase in lengths is 1:1, which corresponds to option (3).

Step by step solution

01

Understand the Problem

We need to determine the ratio of the increase in lengths of two wires with different initial lengths and radii, when stretched by equal forces. The wires are made of the same material, so we can apply Hooke's Law: \( \Delta L = \frac{FL}{AY} \), where \( F \) is the force, \( L \) is the length, \( A \) is the cross-sectional area, and \( Y \) is the Young's modulus.
02

Determine Cross-Sectional Areas

The cross-sectional area \( A \) of a wire with radius \( r \) is given by \( A = \pi r^2 \). Given the radii are in the ratio \( 1 : \sqrt{2} \), let the radii of the wires be \( r \) and \( r\sqrt{2} \). Hence, the areas are \( A_1 = \pi r^2 \) and \( A_2 = \pi (r\sqrt{2})^2 = 2\pi r^2 \).
03

Apply the Formula for Length Increase

Using Hooke鈥檚 Law, the increase in length for each wire can be expressed as: \( \Delta L_1 = \frac{FL_1}{A_1Y} = \frac{FL_1}{\pi r^2 Y} \) for the first wire and \( \Delta L_2 = \frac{FL_2}{A_2Y} = \frac{FL_2}{2\pi r^2 Y} \) for the second wire.
04

Substitute Length Ratios

Given their lengths are in the ratio \(1:2\), we set \(L_1 = L\) and \(L_2 = 2L\). Substituting these into the expressions, we get: \( \Delta L_1 = \frac{FL}{\pi r^2 Y} \) and \( \Delta L_2 = \frac{F(2L)}{2\pi r^2 Y} = \frac{FL}{\pi r^2 Y} \).
05

Calculate the Ratio of Length Increases

The ratio of the increases in length is \[ \frac{\Delta L_1}{\Delta L_2} = \frac{\frac{FL}{\pi r^2 Y}}{\frac{FL}{\pi r^2 Y}} = 1:1 \]. Therefore, the increases in length are in the ratio 1:1.

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Key Concepts

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

Young's modulus
Young's modulus is a fundamental property that describes the stiffness of a material. It's denoted by the symbol \( Y \) and is a measure of a material's ability to resist deformation under stress. Young's modulus is defined as the ratio of tensile stress to tensile strain:\[ Y = \frac{\text{Tensile Stress}}{\text{Tensile Strain}} \]In simpler terms, it helps us understand how much a material will stretch or compress under a given force. A higher Young's modulus indicates a stiffer material, meaning it will deform less when subjected to stress.
  • It is expressed in units of pressure, such as Pascals (Pa).
  • Young's modulus is intrinsic to the material and does not depend on the size or shape of the object.
  • Materials with a high Young's modulus, like steel, are useful for construction and applications where little deformation is necessary.
Cross-sectional area
The cross-sectional area \( A \) is crucial in determining how a material will deform under a given force. It represents the area of a cut made perpendicular to the length of the object.For a wire with radius \( r \), the cross-sectional area is given by the formula:\[ A = \pi r^2 \]This equation reflects that the amount of material present in the cross-section directly affects the wire's ability to stretch.
  • Increased cross-sectional area means more material is resisting the deforming force.
  • It plays a significant role in applications of Hooke's Law, as seen in the exercise where the cross-sectional areas are different due to differing radii.
  • In the example, the cross-sectional areas lead to the same force having different effects on wires of different sizes.
Length increase ratio
The length increase ratio is an important concept when dealing with deformation, especially in problems dealing with materials under stress. It compares the change in length of different objects, helping us understand how different factors affect deformation.In the exercise, Hooke's Law, \( \Delta L = \frac{FL}{AY} \), guides this analysis:
  • \( \Delta L \) is the change in length.
  • The formula shows how force \( F \), initial length \( L \), cross-sectional area \( A \), and Young's modulus \( Y \) interact to determine the length change.
Using the known ratios of lengths and radii alongside Hooke's Law allows us to find that the length increase ratio is 1:1. This reveals both wires stretch equally despite different initial conditions.

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Most popular questions from this chapter

When a body of mass \(M\) is attached to lower end of a wire (of length \(L\) ) whose upper end is fixed, then the elongation of the wire is \(l\). In this situation, mark out the correct statement(s). (1) Loss in gravitational potential energy of \(M\) is \(M g l .\) (2) Elastic potential energy stored in the wire is \(\frac{M g l}{2}\). (3) Elastic potential energy stored in the wire is \(\mathrm{Mgl}\). (4) Elastic potential energy stored in the wire is \(\frac{M g l}{3}\).

A uniform plank is resting over a smooth horizontal floor and is pulled by applying a horizontal force at its one end. Which of the following statements are not correct? (1) Stress developed in plank material is maximum at the end at which force is applied and decrease linearly to zero at the other end. (2) A uniform tensile stress is developed in the plank material. (3) Since plank is pulled at one end only, plank starts to accelerate along direction of the force. Hence, no stress is developed in the plank material. (4) None of the above.

A wire of length \(L\) and radius \(r\) is fixed at one end. When a stretching force \(F\) is applied at free end, the elongation in the wire is \(l\). When another wire of same material but of length \(2 L\) and radius \(2 r\), also fixed at one end is stretched by a force \(2 F\) applied at free end, then elongation in the second wire will be (1) \(l / 2\) (2) \(l\) (3) \(2 l\) (4) \(l / 4\)

According to Hooke's law, within the elastic limit stress/strain = constant. This constant depends on the type of strain or the type of force acting. Tensile stress might result in compressional or elongative strain; however, a tangential stress can only cause a shearing strain. After crossing the elastic limit, the material undergoes elongation and beyond a stage breaks. All modulus of elasticity are basically constants for the materials under stress. If stress/strain is \(x\) in elastic region and \(y\) in the region of yield, then (1) \(x=y\) (2) \(x>y\) (3) \(x

Compressibility of water is \(5 \times 10^{-10} \mathrm{~m}^{2} / \mathrm{N}\). Find the decrease in volume (in \(\mathrm{mL}\) ) of \(100 \mathrm{~mL}\) of water when subjected to a pressure of \(15 \mathrm{MPa}\).
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