Question

A steel wire of length 10.0 m is stretched through
1.0 mm. The cross-sectional area of the wire is
4 mm2. The energy density of the wire is (Young's
modulus of wire = 2.0 * 1011 N/m2)​

Answer 1

Answer:

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.

Answer 2

Answer:

The energy density of the wire is  $10^{7}$ J/m^3 .

Explanation:

As per the data given in the question
We have to calculate the energy density of the wire.

As per the questions

It is given that

A steel wire of length 10.0 m is stretched through 1.0 mm. The cross-sectional area of the wire is 4 mm2.Young's modulus of wire = 2.0 * 10^11 N/m2.

Here l=10m

Δl=1×$10^{-3}$ m

Area = 4×$10^{-6}$ m^2

Y= 2×$10^{11}$ N/m^2

The energy density of stretched wire

u= $\frac{1}{2}$×stress×strain^2

u= $\frac{1}{2}$ ×2×$10^{11}$ ×$(\frac{1*10^{-3} }{10} )^2$

u= 1×$10^{11}$ ×$(10^{-2}) ^{2}$

u=$\frac{10^{11} }{10^{4} }$

⇒u=$10^{7}$ J/m^3

Hence, the energy density of the wire is  $10^{7}$ J/m^3 .

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