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.. 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 × 10–3I know my post is very long..but plz help me understanding the concept well.. thanks in advance

Priyanka , 10 Years ago

Grade 12th pass