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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

`(y^2)/(l)`

B

`(y^2)/(3l)`

C

`(y^2)/(4l)`

D

`(2y^2)/(l)`

Text Solution

Verified by Experts

The correct Answer is:
A
`Delta l = 2(y^(2) + l^(2))^(1//2) - 2l = 2l (1 + (y^(2))/(l^(2)))^(1//2) - 2l`
`= 2l (1 + (1)/(2) (y^(2))/(l^(2)))-2l = (y^(2))/(2l^(2)) xx 2l = (y^(2))/(l)`
energy `(1)/(2) xx (Y A Delta l^(2))/(2l) - 2l = (1)/(2) "stress" xx "strain" xx "volime"`
`= (1)/(2)y ("strain")^(2) A(2l) = (1)/(2) y((Delta l)/(2l))^(2) A.2l = (yA(Delta l)^(2))/(4l)`
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