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Longitudinal Strain

A steel wire ...

Question

A steel wire if $1 m$ long and $1 m m^{2}$ cross section area is hanged from rigid end when weight of $1 k g$ is hang from it, then change in length will be
(Young's coefficient for wire $Y = 2 \times 10^{11} N / m^{2})$

0.5 m m

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0.25 m m

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0.05 m m

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5 m m

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Solution

The correct option is C $0.05 m m$
Young's coefficient $Y$ of a wire of length $l$ and cross section $A$ , having weight $m g$ is given by ,
$Y = \frac{m g l}{A Δ l}$
or $Δ l = \frac{m g l}{A Y}$
where $Δ l =$ change in length
given - $Y = 2 \times 10^{11} N / m^{2}, l = 1 m, A = 1 m m^{2} = 10^{- 6} m^{2}, m = 1 k g$
Hence , $Δ l = \frac{1 \times 10 \times 1}{10^{- 6} \times 2 \times 10^{11}} = 5 \times 10^{- 5} m = 0.05 m m$

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