Question

When a wire of length 5m and diameter 1mm was stretched by a load of 5 kg, elongation produced in the wire was 1mm. Find the energy stored in per unit volume of the wire.​

Answer 1

Answer:

energy=

2

1

×stress×strain×volume

volume

energy

=

2

1

×stress×strain

=

2

1

×γ×strain×strain

=

2

1

×1.9×10

4

×

5000

1

×

5000

1

=3.8×10

−4

J/m

3