Question

4) A weight exerts force of 200 N on steel wire of cross-section area of A m2. If the extension produced is 4
mm, for 5 m length of wire then find the diameter of wire. (Given, Y= 2*104) N/m​

Answer 1

Explanation:

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

=0.20