Question

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. $\frac{FL}{2AY}$
• B. $\frac{F^{2}L}{2AY}$
• C. $\frac{FL}{AY}$
• D. $\frac{F^2{L}^2}{2AY}$

Answer 1

Answer: (B)

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

Energy,$U=\frac{1}{2}\times stress\times strain\times volume$

Now, $stress=\frac{F}{A}$

$strain=\frac{stress}{Y}=\frac{F}{AY}$

$Volume=AL$

Hence, $U=\frac{1}{2}\left(\frac{F}{A}\right)\left(\frac{F}{AY}\right)AL$

$⟹U=\frac{F^{2}L}{2AY}$

Hence, answer is option-(B).