Explanation:
Here,
l
=
4.0
m
;
Δ
l
=
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
=
1
2
×
stress
×
strain
=
1
2
×
Y
×
(
s
t
r
a
i
n
)
2
=
1
2
×
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
)
×
4.0
J
=
20
×
10
−
2
=
0.20
J.
