Question

A copper wire 4 in long has diameter of 1 mm. if a load of 10 kg wt is attached at other end. What extension is produced, if Poisson's ratio is 0.26? How much lateral compression is produced in it ? ($Y_{cu}$ = 12.5 x 10¹⁰ N / m²) (Ans : 3.992 mm, 2.595 x 10⁻⁴ mm)

Answer 1

length of copper wire, l = 4m

diameter of copper wire, d = 1mm = 0.001m

Load , F = 10kgwt = 98N

use formula, Y = Fl/A∆l

Where ∆l is extension produced in wire.
So, ∆l = Fl/AY

= 98 × 4/(πd²/4 × 12.5 × 10^10)

= (98 × 16)/{22/7 × (0.001)² × 12.5 × 10^10}

= 3.992 mm

we know, Poisson's ratio = lateral strain/longitudinal strain

longitudinal strain = 3.992 mm

So, lateral strain = longitudinal strain × Poisson's ratio

= 3.992 × 0.26

= 1.03792 mm

so, lateral compression = lateral strain/original length
= 1.03792/4 mm

= 2.595 × 10^-4 m

Answer 2

Answer:

iiiiiiiiiiiiiiiiiiiiiiiiii