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A steel wire of l=15m and r=1mm is stretched by a unknown force F.If young modulus of steel wire is $2.4\times 10^{11}N/m^2$ and change in length is $5\times 10^{-2}mm$

$\begin{array}{1 1}(A)\;7.48\times 10^5N\\(B)\;8.53\times 10^5N\\(C)\;7.53\times 10^5N\\(D)\;7.63\times 10^5N\end{array} $

1 Answer

$l$=15m
$r=$1mm
Area(a)=$\pi(1\times 10^{-3})^2m^2$
$\qquad\;\;=10^{-6}\pi m^2$
$Y=2.4\times 10^{11}N/m^2$
$\Delta l=5\times 10^{-2}$mm
$\quad\;=5\times 10^{-5}$m
As we know
Y=$\large\frac{\text{Stress}}{\text{Strain}}$
$\Rightarrow $Stress=Y.Strain
$\Rightarrow F=\large\frac{Y\Delta lA}{l}$
$\Rightarrow \large\frac{2.4\times 10^{11}\times 5\times 10^{-5}\times 10^{-6}\pi}{5\times 10^{-5}}$
$\Rightarrow 7.53\times 10^5N$
Hence (C) is the correct answer.
answered May 7, 2014 by sreemathi.v
 

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