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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.. 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

Shobhit Varshney IIT Roorkee
6 years ago
Dear student,

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

Here, Young’s Modulus should be used as the force is tensile (longitudinal)
I hope it helps you.

Thanks.