Calculate the % increase in the length of a wire of diameter 2.5 mm st

1.	Calculate the % increase in the length of a wire of diameter 2.5 mm stretched by a force of 100 kgf. (Y for the wire = 125 × 1011 dyne cm-2)
Answer» Given : Y = 12.5 × 10¹¹ dyne cm^-2 = 12.5 × 10¹⁰ Nm^-2 Diameter, D = 2.5 mm = 2.5 × 10^-3 m. ∴ radius, r = \(\frac{D}{2}\) = 1.25 × 10^-3 m. F = 100 kgf = 100 × 9.8 N = 980 N. \(\frac{∆L}{L}\) × 100 = ? A = πr² = π(1.25 × 10^-3)² m². From the relation, Y = \(\frac{FL}{A∆L},\) We get % increase in length = \(\frac{∆L}{L}\) × 100 = \(\frac{F}{AY}\) × 100 = \(\frac{F}{πr^2Y}\) × 100 = \(\frac{980}{3.142×(1.25×10^−3)×12.5×10^{10}}\) × 100 = 0.16 %

Calculate the % increase in the length of a wire of diameter 2.5 mm stretched by a force of 100 kgf. (Y for the wire = 125 × 1011 dyne cm-2)

Answer»

Given :

Y = 12.5 × 10¹¹ dyne cm^-2

= 12.5 × 10¹⁰ Nm^-2

Diameter, D = 2.5 mm = 2.5 × 10^-3 m.

∴ radius, r = \(\frac{D}{2}\) = 1.25 × 10^-3 m.

F = 100 kgf = 100 × 9.8 N = 980 N. \(\frac{∆L}{L}\) × 100 = ?

A = πr² = π(1.25 × 10^-3)² m².

From the relation, Y = \(\frac{FL}{A∆L},\)

We get % increase in length = \(\frac{∆L}{L}\) × 100

= \(\frac{F}{AY}\) × 100 = \(\frac{F}{πr^2Y}\) × 100

= \(\frac{980}{3.142×(1.25×10^−3)×12.5×10^{10}}\) × 100

= 0.16 %