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

Here change in length is
`Delta l = [AC + BC] - 2l = 2(l^(2 + x^(2)))^(1//2) - 2l`
`= 2l (1 + (x^(2))/(l^(2))) - 2l = 2l(1 + (1)/(2) (x^(2))/(l^(2))) - 2l = (x^(2))/(l)`

`therefore "strain" = (Delta l)/(2l) = (x^(2))/(2l^(2))`