Elastic Modulie

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A long steel rod of uniform cross sectional area is suspended from one end. If $x$ be the elongation of lower half of the rod, the elongation of upper half is

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$Let m be the mass of the rod, l be its length and S be its cross sectional area and Y be the Young' s modulus of steel, then elongation of lower half BC, x = \frac{(\frac{m g}{2}) \frac{l}{4}}{S Y} = \frac{m g l}{8 S Y}$
Elongation of upper half AC, $y = \frac{(\frac{m g}{2}) \frac{l}{4}}{S . Y} + \frac{(\frac{m g}{2}) \frac{l}{2}}{S Y}$
$\Rightarrow y = \frac{3}{8} \frac{m g l}{S Y} \Rightarrow y = 3 x$

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Elastic Modulie

Remember concepts with our Masterclasses.

A long steel rod of uniform cross sectional area is suspended from one end. If x be the elongation of lower half of the rod, the elongation of upper half is

Let m be the mass of the rod, l be its length and S be its cross sectional area and Y be the Young's modulus of steel, then elongation of lower half BC, x=(mg2)l4SY=mgl8SYElongation of upper half AC, y=(mg2)l4S.Y+(mg2)l2SY⇒y=38mglSY⇒y=3x

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A long steel rod of uniform cross sectional area is suspended from one end. If $x$ be the elongation of lower half of the rod, the elongation of upper half is