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A member of length 200 mm and diameter 25 mm is subjected to a tensile load of 20 kN. The final length of the member is found to be 220 mm. The percentage increase in the length of the member is


1. 2%
2. 1%
3. 5%
4. 10%

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Best answer
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.

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