First slide

Elastic Modulie

Question

A uniform heavy rod of weight W, cross-sectional area A and initial length I is suspended from a rigid support. f is the Young's modulus of the material of the wire. If no Iateral contraction is taken into account, the elongation of the rod will be

Difficult

A

$\frac{WL}{2 AY}$
B

$\frac{WL}{AY}$
C

$\frac{2 AY}{WL}$
D

$\frac{WL}{4 AY}$

Solution

Let the whole length of the rod be divided into a large number of small length elements. Consider a small length dx of the rod at a distance x from the fixed rod. The length of the rod below this part will be (L -x). The tension in the rod at the element will be equal to the weight of the rod below it, i.e.
, $T = (\frac{W}{L}) (L - X)$
Elongation in the element
$= \frac{original length \times stress}{Young's modulus} = \frac{Tdx}{AY}$
$= \{\frac{W}{L} (L - x)\} \times \frac{dx}{AY}$
Total elongation $= \int_{0}^{L} \frac{W (L - x) dx}{LAY}$
$= \frac{W}{LAY} {[LX - \frac{X^{2}}{2}]}_{0}^{L} = \frac{WL}{2 AY}$

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