Question

Determine the elongation of the steel bar 1m long and 1.5 cm^(2) cross-sectional area when subjected to a pull of 1.5xx10^(4) N. (Take Y=2.0xx10^(11 )N//m^(2)).

Answer 1

Elongation of wire is 5×10^-² cm

•On application of stress , strain is produced which leads to change in length of wire.

•Force applied (F) = 1.5 × 10⁴ N

•Length of steel bar = 1 m

•Cross-sectional area of wire = 1.5 cm²

•Young's modulus (Y) = 2.0 × 10¹¹ N/m²

Also , Relation between stress and strain is such that , stress is Young's modulus times strain

i.e. Stress = Y . Strain

where , stress is Force per unit Area

strain is change in length per unit length

We get,

F/A = (Y∆L)/L

where ∆L is elongation of wire

Now putting the given values in equation .

(1.5 × 10⁴ )/(1.5 ×10^-⁴) =∆L.(2.0 ×

10¹¹)/(1)

10^8/(2×10¹¹) = ∆L

∆L = 10^-³/2

∆L = 5 × 10^-⁴ m

∆L = 5×10^-² cm

So , Elongation of wire is 5×10^-² cm

by application of Force (F)