Question

A wire 4m long and 2mm diameter is extended by 0.5mm .Find the force applied and energy stored in the stretched wire.

Answer 1

Explanation:

Given,

length = 4 m = 400 cm

diameter = 5 mm = 0.5 cm

=> radius, r = 0.5/2 = 0.25 cm

=> Area = r² = 3.14 x 0.25 x 0.25 = 0.2

cm²

Force = 5 kg wt = 5000 g x 980 cm/s² = 49 x 105

young's modulus = 2.4 x 10¹2 dyne cm²

We know that young's modulus

by the relation,

Y = (F/A)/(AL/L) = FL/AAL

=> AL = FL/YA

= (49 x 105 x 400)/(2.4 x 10¹² x 0.2)

= 0.0041 cm

Hence the increase in length will be 0.0041 cm