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Chapter 1: Q27E (page 359)

A metal rod that is 4.00 m long and 0.50cm2in cross-sectional area is found to stretch 0.20 cm under a tension of 5000 N. What is Young’s modulus for this metal?

Short Answer

Expert verified

2×1011Pa

Step by step solution

01

Given information

Length: l0=4.00m,

Area: A=0.50cm2,

Elongation: ∆l=0.20cm,

Tension Force: F = 4000 N .

02

Concept/Formula used

Y=l0F∆l

Where, Y is Young’s modulus, l0 is length of muscle, F is muscle force, A is cross-sectional area and ∆l is elongation.

03

Young’s modulus Calculation

Y=(4.00m)(5000N)(0.50×10-4m2)(0.20×10-2m)=2×1011Pa

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