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Ncert question..plz i’m a bit confused..plz help Question 9.8: A piece of copper having a rectangular cross-section of 15.2 mm × 19.1 mm is pulled in tension with 44,500 N force, producing only elastic deformation. Calculate the resulting strain? which modulous should i use here , young’s modulous or sheer modulous of copper....i’m really very confused with the answer written below..i found it in a website...But this answer doesn’t match the book answer...According to ncert book answer ,Young’s modulous is used in this case...plz help.. answer;; (This answer is found in a website) Length of the piece of copper, l = 19.1 mm = 19.1 × 10 –3 m Breadth of the piece of copper, b = 15.2 mm = 15.2 × 10 –3 m Area of the copper piece: A = l × b = 19.1 × 10 –3 × 15.2 × 10 –3 = 2.9 × 10 –4 m 2 Tension force applied on the piece of copper, F = 44500 N Sheer Modulus of elasticity of copper, η = 42 × 10 9 N/m 2 Modulus of elasticity, η = stress/strain = F/A/strain Therefore, strain = F/An = 44500/2.9* 10^-4 * 42 * 10^9 = 3.65 × 10 –3 I know my post is very long..but plz help me understanding the concept well.. thanks in advance

 
Ncert question..plz i’m a bit confused..plz help
Question 9.8:
A piece of copper having a rectangular cross-section of 15.2 mm × 19.1 mm is pulled in tension with 44,500 N force, producing only elastic deformation. Calculate the resulting strain?
which modulous should i use here , young’s modulous or sheer modulous of copper....i’m really very confused with the answer written below..i found it in a website...But this answer doesn’t match the book answer...According to ncert book answer ,Young’s modulous is used in this case...plz help..
 
answer;; (This answer is found in a website)
Length of the piece of copper, l = 19.1 mm = 19.1 × 10–3 m
Breadth of the piece of copper, b = 15.2 mm = 15.2 × 10–3 m
Area of the copper piece:
A = l × b
= 19.1 × 10–3 × 15.2 × 10–3
= 2.9 × 10–4 m2
Tension force applied on the piece of copper, F = 44500 N
Sheer Modulus of elasticity of copper, η = 42 × 109 N/m2
Modulus of elasticity, η = stress/strain = F/A/strain
Therefore, strain = F/An
= 44500/2.9* 10^-4  * 42 * 10^9
 
= 3.65 × 10–3
   
 
I know my post is very long..but plz help me understanding the concept well..
 
thanks in advance

Grade:12th pass

1 Answers

Shobhit Varshney IIT Roorkee
askIITians Faculty 33 Points
6 years ago
Dear student,

Thanks for your post.
Before answering this question, you have to first understand the meaning of young’s modulus and shear modulus.

256-1429_shear modulus.jpg
Here, Young’s Modulus should be used as the force is tensile (longitudinal)
I hope it helps you.

Thanks.

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