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A steel rod of r=30mm and length 3m.A 50 KN force is applied along its length.Calculate the change in length is $Y=2\times 10^{11}N/m^2$

$\begin{array}{1 1}(A)\;0.52mm\\(B)\;0.23mm\\(C)\;0.26mm\\(D)\;0.50mm\end{array} $

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$r=30\times 10^{-3}$m
$a=\pi r^2$
$\Rightarrow \pi(30\times 10^{-3})^2m^2$
$\Rightarrow (9\pi\times 10^{-4})m^2$
$l=3m$
As $Y=\large\frac{Fl}{a\Delta l}$
$\Rightarrow \Delta l=(\large\frac{F}{a})\frac{l}{r}$
$\Rightarrow \large\frac{50\times 10^3}{9\times \pi\times 10^{-4}}\times \frac{3}{2\times 10^{11}}$m
$\Rightarrow 0.26mm$
Hence (C) is the correct answer.
answered May 7, 2014 by sreemathi.v
 

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