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Chapter 7: Problem 899

Two similar wires under the same load yield elongation of \(0.1 \mathrm{~mm}\) and \(0.05 \mathrm{~mm}\) respectively. If the area of Cross-section of the first wire is \(4 \mathrm{~mm}^{2}\). Then what is the area of cross - section of the second wire? (A) \(6 \mathrm{~mm}^{2}\) (B) \(8 \mathrm{~mm}^{2}\) (C) \(10 \mathrm{~mm}^{2}\) (D) \(12 \mathrm{~mm}^{2}\)

Short Answer

Expert verified
The area of cross-section of the second wire is \(8 \mathrm{~mm}^{2}\) (Option B).

Step by step solution

01

Write down the relationship between elongation, length and area of cross-section

According to the relationship between elongation, length, and area of cross-section, we have: \[\frac{\text{Elongation}}{\text{Area of cross-section}} \propto \text{Length}\]
02

Write down the given information

We are given the following information: - Elongation of the first wire = 0.1 mm - Elongation of the second wire = 0.05 mm - Area of cross-section of the first wire = 4 mm虏 - We need to find the area of cross-section of the second wire.
03

Write equations for both wires

Using the relationship mentioned in Step 1, we can write equations for both wires as: \[\frac{0.1}{4} = kL \Rightarrow kL = 0.025 \,\,\,...(1)\] \[\frac{0.05}{A_2} = kL \Rightarrow kL = 0.05 \times A_2^{-1}\,\,\,...(2)\] Here, \(k\) is a proportionality constant and \(L\) is the length of the wire (which is the same for both the wires since they are similar). \(A_2\) is the area of cross-section of the second wire that we need to find.
04

Set up the equation to solve for the area of cross-section of the second wire

Since both equations (1) and (2) are equal to \(kL\), we can set them equal to each other: \[0.025 = 0.05 \times A_2^{-1}\]
05

Solve for the area of cross-section of the second wire

Now, isolate \(A_2\) by dividing both sides of the equation by 0.05: \[A_2^{-1} = \frac{0.025}{0.05} \Rightarrow A_2^{-1} = 0.5\] Invert both sides of the equation to find \(A_2\): \[A_2 = \frac{1}{0.5} \Rightarrow A_2 = 2\] Since the area of cross-section of the first wire is 4 mm虏 and the ratio of the areas of cross-section is 2, the area of cross-section of the second wire is: \[A_2 = 4 \times 2 = 8 \mathrm{~mm}^2\] So, the correct answer is (B) \(8 \mathrm{~mm}^{2}\).

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Key Concepts

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

Elasticity
Elasticity is a material property that describes its ability to return to its original shape after deformation. Imagine a rubber band. When you stretch it, it elongates but immediately snaps back once released. This behavior is a classic example of elasticity.

In engineering and physics, elasticity is quantified by Young's Modulus, a constant that measures the stiffness of a material. A higher Young's Modulus means the material is stiffer and requires more force to stretch it.
  • Key Idea: Materials with high elasticity can deform under stress but will return to their original shape.
  • Young鈥檚 Modulus: Determines how much a material will stretch under a given load.
Understanding elasticity is crucial when dealing with stresses and strains in materials, as it helps predict how materials will behave under various forces.
Cross-sectional Area
The cross-sectional area of a wire or any object is essentially the area of its "slice" or section perpendicular to its length. For example, if you cut a wire, the shape of the cut is its cross-section. This area plays a critical role in determining how much the material will stretch.

A larger cross-sectional area means more material to bear the load, thus less elongation for the same force compared to a material with a smaller area.
  • Relation: The larger the area, the less it stretches under the same force.
  • Impact: Engineering designs often optimize the cross-sectional area to balance strength and material use.
In the exercise example, by understanding the relationship between load, elongation, and area, you can find unknown quantities like the area of a second wire.
Elongation
Elongation refers to the extent to which a material stretches under a load. It's like measuring how far a rubber band extends when you pull it. Typically, it's expressed in units of length, such as millimeters or inches.

The amount of elongation depends on several factors such as the material's elasticity (Young's Modulus), the applied force, and the cross-sectional area.
  • Factors: More specific elasticity and lesser cross-sectional area lead to greater elongation.
  • Description: Elongation helps in understanding the limits of material use for various applications.
Engineering calculations often require understanding how much a material will elongate under specific conditions to ensure safety and functionality.
Proportionality Constant
In physics, a proportionality constant relates two variables that are directly proportional. In the given exercise, the proportionality constant, represented by \(k\), links elongation with cross-sectional area and length for the same material.

This constant helps simplify complex relationships into more manageable forms by converting them into equations.
  • Significance: Useful in forming equations that help solve for unknowns in problems.
  • Example: In our exercise, we've used \(k\) to relate elongation and cross-sectional area.
Understanding this concept allows you to recognize patterns and predict material behavior under different conditions. Addressing these connections assists in engineering and design applications, simplifying calculations.

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Most popular questions from this chapter

In capillary pressure below the curved surface at water will be (A) Equal to atmospheric (B) Equal to upper side pressure (C) More than upper side pressure (D) Lesser than upper side pressure

8000 identical water drops are combined to form a bigdrop. Then the ratio of the final surface energy to the initial surface energy of all the drops together is (A) \(1: 10\) (B) \(1: 15\) (C) \(1: 20\) (D) \(1: 25\)

To what depth below the surface of sea should a rubber ball be taken as to decrease its volume by \(0.1 \%\) [Take : density of sea water \(=1000\left(\mathrm{~kg} / \mathrm{m}^{3}\right)\) Bulk modulus of rubber \(=9 \times 10^{8}\left(\mathrm{~N} / \mathrm{m}^{2}\right)\), acceleration due to gravity \(\left.=10\left(\mathrm{~m} / \mathrm{s}^{2}\right)\right]\) (A) \(9 \mathrm{~m}\) (B) \(18 \mathrm{~m}\) (C) \(180 \mathrm{~m}\) (D) \(90 \mathrm{~m}\)

When two soap bubbles of radius \(\mathrm{r}_{1}\) and \(\mathrm{r}_{2}\left(\mathrm{r}_{2}>\mathrm{r}_{1}\right)\) coalesce, the radius of curvature of common surface is........... (A) \(r_{2}-r_{1}\) (B) \(\left[\left(r_{2}-r_{1}\right) /\left(r_{1} r_{2}\right)\right]\) (C) \(\left[\left(\mathrm{r}_{1} \mathrm{r}_{2}\right) /\left(\mathrm{r}_{2}-\mathrm{r}_{1}\right)\right]\) (D) \(\mathrm{r}_{2}+\mathrm{r}_{1}\)

Shearing stress causes change in (A) length (B) breadth (C) shape (D) volume
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