Question

A metallic rod of length $l$ and cross sectional area $A$ is made of a material of Young's modulus $Y$. If the rod is elongated by an amount $y$, then the work done is proportional to

• A. $y$
• B. $1/y$
• C. $y^2$
• D. $1/y^2$

Answer 1

Answer: (C)

$y^2$

strain=ElongationOriginallength=ylstrain=ElongationOriginallength=yl
Young's modulus Y=StressStrainY=StressStrain
work done, W=12×stress×strain×volumeW=12×stress×strain×volume
W=12×Y×(Strain)2×Al=12(YAl)y2W=12×Y×(Strain)2×Al=12(YAl)y2
W∝y2