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A wire whose cross-sectional area is 4mm 2 is stretched by 0.1 mm by a certain load. if a...

Question:

A wire whose cross-sectional area is 4mm{eq}^2 {/eq} is stretched by 0.1 mm by a certain load. if a similar wire double the area of cross-section is under the same load, then find the longation.

Elongation of the wire

According to Hooke's law, the stress is directly proportional to the strain. The strain is defined as the ratio of the change in the length to the original length of the object or wire. And the proportionality constant is known as Young Modulus.

Answer and Explanation:

Given

Area of the wire {eq}A = 4 \ mm^{2} {/eq}

Length of the elongation {eq}\delta = 0.1 \ mm {/eq}

Now, the elongation of the wire is given by

{eq}\delta = \dfrac{PL}{AE} {/eq}

Where

  • P is the applied load
  • L is the original length of the wire
  • E is the young modulus of the wire

Now, if the area has been doubled, then

{eq}\delta_{2} = \dfrac{PL}{A_{2}E} \\ \delta_{2} = \dfrac{PL}{2AE} \\ \delta_{2} = \dfrac{\delta}{2} \\ \delta_{2} = 0.05 \ mm {/eq}


Learn more about this topic:

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Young's Modulus: Definition & Equations

from National Eligibility Test (AIPMT): Study Guide

Chapter 8 / Lesson 3
22K

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