Question

a steel wire of diameter 2m.m is pulled to increase its length by 1% .what is the restoring force developed in it.​

Answer 1

diameter of steel wire, d = 2mm

so, radius of steel wire , r = 1mm = 10^-3 m

and cross sectional area of steel wire, A = πr² = π(10^-3)² = π × 10^-6 m²

steel wire is pulled to increase its length by 1% .

percentage change in length = 1%

so, ∆l/l × 100 = 1%

or, ∆l/l = strain = 1/100 = 0.01

applying, Young's modulus = stress/strain

we know, Young's modulus of steel = 1.9 × 10¹¹ N/m²

so, 1.9 × 10¹¹ N/m² = stress/0.01

or, stress = 1.9 × 10^9 N/m²

now stress = force/cross sectional area

or, 1.9 × 10^9 N/m² = force/π × 10^-6 m²

or, force = 1.9π × 10³ N = 1900π N

hence, restoring force developed in it is 1900π N.