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A steel wire of $40\;m$ in length is stretched through $2.0\; mm.$ The cross sectional area of the wire is $2.0 \;mm^2$ . If young's modulus of steel is $2.0 \times 10^{11}\;N/m^2$ the energy density of wire is

\[\begin {array} {1 1} (a)\;2.5 \times 10^4 J/m^3 \\ (b)\;5.0 \times 10^4 J/m^3 \\ (c)\;7.5 \times 10^4 J/m^3 \\ (d)\;10.0 \times 10^4 J/m^3 \end {array}\]

1 Answer

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$U= \large\frac{1}{2} $$ \times stress \times strain $
$\quad= \large\frac{1}{2}$$ \times y \times (stain)^2$ $\qquad [strain =\large\frac{\Delta L}{L}]$
$\quad= \large\frac{1}{2}$$ \times 2.0 \times 10^{11} \times \bigg(\large\frac{2 \times 10^{-3}}{4}\bigg)^2$
$\quad= 2.5 \times 10^{4} J/m^3$
Hence a is the correct answer.


answered Oct 2, 2013 by meena.p
edited Feb 18, 2014 by meena.p

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