A mild steel wire of length 2L and cross-sectional area A is stretched, well within elastic limit, horizontally between two pillars,

A mass m is suspended from the midpoint of the wire. Strain in the wire is

Options

A. `x^2/(2L^2)`
B. `x/L`
C. `x^2/L`
D. `x^2/(2L)`

Answer

A

Explanation

Change in length , `triangleL=AC-AO` (From `triangleACO` )
`=[L^2+x^2]^(1//2) -L=L[1+1/2 x^2/L^2]-L=x^2/(2L)`
`therefore` Longitudinal strain = `(DeltaL)/L = (x^2//2L)/L=x^2/(2L^2)`