When a wire of uniform cross-sectional area, length l and density ρ is suspended from one end, its elongation is found to be x. Then Young's modulus of its material is
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a
ρgl2/x
b
2ρgl2x
c
ρgl2/4x
d
ρgl2/2x
answer is D.
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Detailed Solution
Mass of wire, m=SlpElongation, x=mg.(l2)S.Y=Slp.g.l2SY⇒Y=ρgl22x