A mild steel wire of length `1.0m` and cross sectional area `2L` is strethched, within its elastic limit horizontally between two pillars(figure). A mass of `m` is suspended form the midpont 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

Consider the diagram below

Hence, change in length
`DeltaL = BO + OC - (BD+DC)`
`=2BO - 2BD " "(because BO = OC, BD = DC)`
`=2[BO-BD]`
`=2[(x^(2) + L^(2))^(1//2)-L]`
`=2L [(1+ (x^(2))/(L^(2)))^(1//2)-1]`
`~~2L[1+ (1)/(2)(x^(2))/(L^(2))-1]=(x^(2))/(L)" "[because x ltltL]`
`therefore " Strain" = (DeltaL)/(2L)= (x^(2)//L)/(2L)=(x^(2))/(2L^(2))`