First slide

Elastic behaviour of solids

Question

A brass rod of cross-sectional area 1 cm² and length 0.2 m is compressed lengthwise by a weight of 5 kg. If Yourg's modulus of elasticity of brass is $1 \times 10^{11} N / m^{2}$ and g = 10 m $s^{- 2}$ then increase in the energy of the rod will be

Moderate

A

$10^{- 5} J$
B

$2.5 \times 10^{- 5} J$
C

$5 \times 10^{- 5} J$
D

$2.5 \times 10^{- 4} J$

Solution

$\begin{matrix} U = \frac{1}{2} \times \frac{(Stress)^{2}}{Y} \times Volume \\ = \frac{1}{2} \times \frac{F^{2} \times A \times L}{A^{2} \times Y} \\ = \frac{1}{2} \times \frac{F^{2} L}{AY} = \frac{1}{2} \times \frac{(50)^{2} \times 0 .2}{1 \times 10^{- 4} \times 1 \times 10^{11}} = 2 .5 \times 10^{- 5} J \end{matrix}$

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