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)
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)
Updated On: Mar 28, 2026
- $\frac{FL}{2AY}$
- $\frac{F^2 L}{2AY}$
- $\frac{FL}{AY}$
- $\frac{F^2 L^2}{2AY}$
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The Correct Option is B
Solution and Explanation
Energy $=\frac{1}{2} \times$ stress $\times$ strain $\times$ volume
$=\frac{1}{2} \frac{(\text { stress })^{2}}{Y} \times \text { volume }$
$=\frac{1}{2} \frac{\left(\frac{F}{A}\right)^{2}}{y} \times L. A\left[\text { as stress }=\frac{F}{A}, \text { volume }=L \times A\right]$
or energy $(E)=\frac{F^{2} L}{2 A Y}$
$=\frac{1}{2} \frac{(\text { stress })^{2}}{Y} \times \text { volume }$
$=\frac{1}{2} \frac{\left(\frac{F}{A}\right)^{2}}{y} \times L. A\left[\text { as stress }=\frac{F}{A}, \text { volume }=L \times A\right]$
or energy $(E)=\frac{F^{2} L}{2 A Y}$
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