A mild steel wire of length 2L and crosssectional area A is stretched, well within elastic limit, horizontally between two pillars as shown in figure. A mass m is suspended from the mid point of the wire, the mid point lower by distance x. Then 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

Increase length in wire,

`therefore Delta L = 2 BO - 2 BD [ because BO = OC, BD= DC]=2 (BO -BD)`
`=2 [ (x ^(2) + L ^(2)) ^(1//2) - L]`
`=2L [ ((x ^(2))/( L ^(2)) + 1 ) ^(1//2) -1 ]`
`=2L [ (1 + (x ^(2))/( L ^(2))) ^(1//2) -1 ]`
`=2L [(1 + (1)/(2) (x ^(2))/( L ^(2)) + ...)-1 ] [because x lt lt L] `
By binomial theorem upto two terms,
`= 2L xx 1/2 (x^(2))/( L ^(2))`
`Delta L = (x ^(2))/( L )`
`therefore` Strain `= (Delta L )/( 2L ) = (x ^(2) // L )/( 2 L ) = (x ^(2))/( 2 L ^(2))`