Question

A piece of steel wire, 200.0cm long and having cross-sectional area of 0.50mm2, is stretched by a force of 50N. Its new length is found to be 200.1cm. Calculate the stress and strain, and the Young’s modulus of steel.

Answer 1

Here, l=4.0m;Δl=2×10

−3

m;a=2.0×10

−6

m

2

,Y=2.0×10

11

N/m

2

(i) The energy density of stretched wire

u=

2

1

× stress × strain

=

2

1

×Y×(strain)

2

=

2

1

×2.0×10

11

×(2×10

−3

)/4)

2

=0.25×10

5

=2.5×10

4

J/m

3

.

(ii) Elastic potential energy = energy density × volume

=2.5×10

4

×(2.0×10

−6

)×4.0J=20×10

−2

=0.20J.