Compute the percentage increase in length of a wire of diameter 2.2 mm

1.	Compute the percentage increase in length of a wire of diameter 2.2 mm stretched by a load of 100 kg. Young's modulus of wire is 12.5 x 1010 Nm-2.
Answer» F = 100 x 9.8 N Y = 12.5 x 10¹⁰ Nm^-2 A = (π/4)d² = (π/4) x (2.2 x 10^-3)² = 3.8 x 10^-6 m³ Y = FL/Al ⇒ l/L = F/AY = {100 x 9.8}/{(3.8 x 10^-6) x (12.5 x 10¹⁰)} = 0.0021 % increase in length = (l/L) x 100 = 0.021 x 100 = 0.21%

Compute the percentage increase in length of a wire of diameter 2.2 mm stretched by a load of 100 kg. Young's modulus of wire is 12.5 x 1010 Nm-2.

Answer»

F = 100 x 9.8 N

Y = 12.5 x 10¹⁰ Nm^-2

A = (π/4)d² = (π/4) x (2.2 x 10^-3)² = 3.8 x 10^-6 m³

Y = FL/Al

⇒ l/L = F/AY = {100 x 9.8}/{(3.8 x 10^-6) x (12.5 x 10¹⁰)}

= 0.0021

% increase in length = (l/L) x 100 = 0.021 x 100 = 0.21%