A mild steel wire of length 2L and cross-sectional area A is stretched, well within elastic limit, horizontally between two pillars Figure mass m is suspended from the mid point of the wire. Strain in the wire is

(a)\frac{x^{2}}{2L^{2}}
(b)\frac{x}{L}
(c)\frac{x^{2}}{L}
(d) \frac{x^{2}}{2L}

Answers (1)

\Delta L = (AO + BO - AB)

\Delta L = 2[AO - L]

\Delta L = 2 (L^{2} + x^{2})^{\frac{1}{2}}- 1)

\Delta L = 2L [1 +\frac{x^{2}}{2L^{2}} - 1] = \frac{x^{2}}{L}

Strain = \frac{\Delta L}{2L} = \frac{x^{2}}{2L^{2}}. Hence option a is correct.

 

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