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A steel wire of 8m in length is stretched by 1mm of cross sectional area $2mmm^2$ .If young modulus of steel is $2\times 10^{11}N/m^2$.What is elastic potential energy?

$\begin{array}{1 1}(A)\;.025J\\(B)\;.25J\\(C)\;.035J\\(D)\;0.045J\end{array} $

1 Answer

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$\Delta l$=1mm=$10^{-3}$m
$a=2mm^2=2\times 10^{-6}m^2$
$V=l\times a=8\times 2\times 10^{-6}=16\times 10^{-6}m^3$
Elastic energy =$\large\frac{1}{2}$$\times (Y)\times (strain)^2\times V$
$\Rightarrow \large\frac{1}{2}$$\times 2\times 10^{11}\times (\large\frac{10^{-3}}{8})^2$$\times 16\times 10^{-6}$
$\Rightarrow \large\frac{16}{64}$$\times 10^{-1}J$
$\Rightarrow .025J$
Hence (A) is the correct answer.
answered May 7, 2014 by sreemathi.v

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