Question

What would be the percent increase in length of wire of diameter 2.5 x I0^-3m stretched by a force of 980 N. Young’s modulus of elasticity of wire is 8 x 10^10 N/m^2.

(a) 0.129

(b) 0.364

(c) 0.249

(d) 0.098

I have been asked this question at a job interview.

My enquiry is from Strength of Materials in division Strength of Materials and Construction Materials of Farm Machinery

Answer 1

Correct option is (c) 0.249

Explanation: Cross-sectional area of wire = πr^2 = π * (1.25 x 10^-3)^2 m^2

From the relation

Y = \(\frac{F.L.}{A.l} \)

Percent increase in length = \(\frac{l}{L} = \frac{F}{A.Y} \)

\(\frac{l}{L}\)*100=\(\frac{F}{A.Y}\)*100=\(\frac{980*7*100}{[22(1.25*10^{-3})^2*8*10^{10}]} \)

= 0.249