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A mild steel wire of length 1.0m and cro...

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

A

`(x^(2))/(2L^(2))`

B

`(x)/(L)`

C

`x^(2)//L`

D

`x^(2)//2L`

Text Solution

Verified by Experts

The correct Answer is:
A
Consider the diagram below

Hence, change in length
`Delta L = BO + OC- (BD+ DC)`
`= 2BO- 2BD (therefore 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 lt lt L]`
`therefore` Strain `= (Delta L)/(2L)= (x^(2)//L)/(2L)= (x^(2))/(2L^(2))`
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