First slide

Elastic behaviour of solids

Question

When a force is applied on a wire of uniform crosssectional area $3 \times 10^{- 6} m^{2}$ and length 4 m, the increase in length is 1 mm. Energy stored in it will be $(Y = 2 \times 10^{11} N / m^{2})$

Moderate

A

6250 J
B

0.177 J
C

0.075 J
D

0.150 J

Solution

$\begin{array}{l} U = \frac{1}{2} \times \frac{Y A l^{2}}{L} = \frac{1}{2} \times \frac{2 \times 10^{11} \times 3 \times 10^{- 6} \times {(1 \times 10^{- 3})}^{2}}{4} \\ = 0.075 J \end{array}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App