First slide

Elastic Modulie

Question

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

Difficult

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 $\frac{3 x}{4}$
D

elongation of the lower half is x/8

Solution

$x = \frac{mg (l / 2)}{AY} = \frac{mgl}{2 AY}$
$x_{1} = Elongation of AC= \frac{(mg / 2) (l / 4)}{AY} + \frac{(mg / 2) (l / 2)}{AY}$
$\Rightarrow x_{1} = \frac{3 mgl}{8 AY} = \frac{3 x}{4}$
$x_{2} =$ Elongation of the lower half $CB = \frac{(\frac{mg}{2}) \frac{x}{4}}{AY} = \frac{mgl}{8 AY} = \frac{x}{4}$

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