Correct Answer - Option 4 : 10%
Concept:
Percentage change in the length = \(\frac{L_f\;-\;L_i}{L_i}\times100\)
Where,
Lf = Final length of work-piece, Li = Initial length of work-piece.
Calculation:
Given:
Lf = 220 mm, Li = 200 mm
Then,
Percentage change in the length = \(\frac{L_f\;-\;L_i}{L_i}\times100\)
= \(\frac{220\;-\;200}{200}\times100\)
= 10%
Elongation in the bar under compressive and tensile load (P)
Elongation (δ) = \(\frac{PL}{AE}\)
Where,
P = Tensile or Compressive load, L = Initial length of bar, A = Cross-section area of bar, E = Modulus of elasticity.