Solveeit Logo

Question

Physics Question on elastic moduli

The increase in energy of a metal bar of length 'L' and cross-sectional area 'A' when compressed with a load 'M' along its length is (Y = Young�s modulus of the material of metal bar)

A

FL2AY\frac{FL}{2AY}

B

F2L2AY\frac{F^2 L}{2AY}

C

FLAY\frac{FL}{AY}

D

F2L22AY\frac{F^2 L^2}{2AY}

Answer

F2L2AY\frac{F^2 L}{2AY}

Explanation

Solution

Energy =12×=\frac{1}{2} \times stress ×\times strain ×\times volume
=12( stress )2Y× volume =\frac{1}{2} \frac{(\text { stress })^{2}}{Y} \times \text { volume }
=12(FA)2y×L.A[ as stress =FA, volume =L×A]=\frac{1}{2} \frac{\left(\frac{F}{A}\right)^{2}}{y} \times L. A\left[\text { as stress }=\frac{F}{A}, \text { volume }=L \times A\right]
or energy (E)=F2L2AY(E)=\frac{F^{2} L}{2 A Y}