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Ncert question..plz i’m a bit confused..plz helpQuestion 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 mBreadth of the piece of copper, b = 15.2 mm = 15.2 × 10–3 mArea of the copper piece:A = l × b= 19.1 × 10–3 × 15.2 × 10–3= 2.9 × 10–4 m2Tension force applied on the piece of copper, F = 44500 NSheer Modulus of elasticity of copper, η = 42 × 109 N/m2Modulus of elasticity, η = stress/strain = F/A/strainTherefore, 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

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.