Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 7: Problem 859

An area of a cross-section of rubber string is \(2 \mathrm{~cm}^{3}\). Its length is doubled when stretched with a linear force of \(2 \times 10^{5}\) dynes. What will be young's modulus of the rubber in dynes? (A) \(4 \times 10^{5}\) (B) \(1 \times 10^{5}\) (C) \(2 \times 10^{5}\) (D) \(1 \times 10^{4}\)

Short Answer

Expert verified
The short answer based on the step-by-step solution is: The Young's modulus of the rubber is \(1 \times 10^5 \, \text{dynes/cm}^2\). The correct answer is (B) \(1 \times 10^{5}\).

Step by step solution

01

Write down the given values

The area of the cross-section is given as 2 cm². The force applied is given as \(2 \times 10^5\) dynes. When the rubber is stretched, its length doubles.
02

Calculate the stress

Stress is defined as force per unit area. In this case, we have: Stress = Force / Area of cross-section Stress = \(2 \times 10^5 \, \text{dynes} / 2 \, \text{cm}^2\) Stress = \(10^5 \, \text{dynes/cm}^2\)
03

Calculate the strain

Strain is defined as the change in length divided by the original length. Since the length of the rubber doubles when it is stretched, the change in length is equal to the initial length: Strain = Change in length / Original length Strain = 1 (since the change and original lengths are equal)
04

Determine Young's modulus

Now we can determine the Young's modulus, which is defined as the ratio of stress to strain: Young's modulus (Y) = Stress / Strain Y = \(10^5 \, \text{dynes/cm}^2\ / 1\) Y = \(1 \times 10^5 \, \text{dynes/cm}^2\) The Young's modulus of the rubber is \(1 \times 10^5 \, \text{dynes/cm}^2\). The correct answer is (B) \(1 \times 10^{5}\).

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

Read the assertion and reason carefully and mark the correct option given below. (a) If both assertion and reason are true and the reason is the correct explanation of the assertion. (b) If both assertion and reason are true but reason is not the correct explanation of the assertion. (c) If assertion is true but reason is false. (d) If the assertion and reason both are false. Assertion: Tiny drops of liquid resist deforming forces better than bigger drops. Reason: Excess pressure inside a drop is directly proportional to surface tension. (A) a (B) \(b\) (C) (D) d

Which is the dimensional formula for modulus of rigidity? (A) \(\mathrm{M}_{1} \mathrm{~L}^{1} \mathrm{~T}^{-2}\) (B) \(\mathrm{M}^{1} \mathrm{~L}^{-1} \mathrm{~T}^{-2}\) (C) \(\mathrm{M}^{1} \mathrm{~L}^{-2} \mathrm{~T}^{-1}\) (D) \(\mathrm{M}^{1} \mathrm{~L}^{-2} \mathrm{~T}^{-2}\)

A beaker is completely filled with water at \(4^{\circ} \mathrm{C}\) It will overflow if (A) Heated above \(4^{\circ} \mathrm{C}\) (B) Cooled below \(4^{\circ} \mathrm{C}\) (C) Both heated and cooled above and below \(4^{\circ} \mathrm{C}\) respectively (D) None of these

The bulk modulus of an ideal gas at constant temperature..... (A) is equal to its volume \(\mathrm{V}\) (B) is equal to \(\mathrm{P} / 2\) (C) is equal to its pressure \(\mathrm{P}\) (D) cannot be determined

When the room temperature becomes equal to the dew point the relative humidity of the room is (A) \(100 \%\) (B) \(0 \%\) (C) \(70 \%\) (D) \(85 \%\)
See all solutions

Recommended explanations on English Textbooks

Summary Text

Read Explanation

Rhetorical Analysis Essay

Read Explanation

English Language Study

Read Explanation

Linguistic Terms

Read Explanation

Blog

Read Explanation

Listening and Speaking

Read Explanation
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.

Sign-up for free