A mild steel wire of Length 2L and cross sectional Area A is stretched well within elastic limit, horizontally between two pillars. A man m is suspended from the mid point 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)`

Explanation

Increase in length = Bo + OC –BC
or `DeltaL=2B0-2L`
`Delta L=2(L^(2)+x^(2))^(1//2)-2L`
`DeltaL=2L[1+(x^(2))/(L^(2))]^(1//2)-2L`
`DeltaL=(x^(2))/(L)`
So strain = `(Delta L)/(2L) = (x^(2))/(Lxx2L)=(x^(2))/(2L^(2))`