The following four wires are made of the same material. If same tension is applied to each, the wire having largest extension is

Options

A. length ` 0. 5 ` m, diameter `0. 5 ` mm.
B. length 1 m, diameter 1 mm.
C. length 2 m, diameter 2 mm.
D. length 3 m , dimeter 3 mm,

Answer

A

Explanation

Given, that material is same, so Young.s modulus - constant and tension - constant.
Hence, the extension,
`Delta L = (FL)/(YA)`
So, ` DeltaL = (4F)/(Ypi)xx L/D^(2)" " (because A = (piD^(2))/4)`
`rArr Delta L prop L/D^(2)`
So, now checking the options,
(a) `L = 0.5 m, D = 0.5 mm = 0.5 xx 10^(-3) m `
`Delta L = (4F)/(ypi) xx (0.5)/((0.5xx10^(-3))^(2))=2xx 10^(6)xx (4F)/(Ypi)`
(b) `L = 1 m , D = 1 mm`
`Delta L + (4F)/(Ypi)xx 1.0 xx10^(6)`
(c) `L = 2m, D = 2 mm `
`Delta L = (4F)/(Y4pi)xx0.5xx10^(6)`
(d) `L = 3 m, D = 3 mm `
`Delta L = (4F)/(Ypi)xx0.33xx10^(6)`
Hence, `(4F)/(Ypi)` is a constant in all quantities. Now after
checking the multiplier of `(4F)/(Ypi)`. The highest value of
`Delta L ` is in option (a).