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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...& plz tell me where to use Young’s modulous & sheer modulous... 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...& plz tell me where to use Young’s modulous & sheer modulous...
 
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

shashi K Sharma
46 Points
6 years ago
Dear Student
 
Physics is depthless and of undefined depth. At your level whenever you encounter a pulling or a pushing force which is normal to the cross section, go for young’s modulus. When ever the force or any component of it is parallel to the area of cross section go for shear modulus. If the force appears to be hydrostatic i.e. acting equally from all sides and normal to surface everywhere go for bulk modulus. Need not confuse atleast at this stage.

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