First slide

Elastic Modulie

Question

When a force is applied on a wire of uniform crosssectional area 3 x $10^{- 6} m^{2}$ and length 4 m, the increase in length is I mm. Energy stored in it will be (Y = 2 x $10^{11} N / m^{2}$ )

Moderate

A

6250 J
B

0.177 J
C

0.075 J
D

0.150 J

Solution

$U = \frac{1}{2} \times \frac{{YAl}^{2}}{L} = \frac{1}{2} \times \frac{2 \times 10^{11} \times 3 \times 10^{- 6} \times {(1 \times 10^{- 3})}^{2}}{4}$
= 0.075 J

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