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Q.

A thin rod AB of uniform cross sectional area is suspended from end A. If x be the elongation of the rod, then

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a

elongation of the lower half is x/2

b

elongation of the upper half is x/2

c

elongation of the upper half is 3x4

d

elongation of the lower half is x/8

answer is C.

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Detailed Solution

x=mgl/2AY=mgl2AYx1=Elongation of AC=mg/2l/4AY+mg/2l/2AY⇒x1=3mgl8AY=3x4x2=Elongation of the lower half CB=mg2x4AY=mgl8AY=x4
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