A 5 kg mass is attached to one end of a copper wire 2m long and 2mm in diameter. Calculate the leteral compression produced in it. Posisson's ration is 0.3 and Young's modulus of the meterial of the wire is `12.5 xx 10^(10)Nm^(-2)`

Answer

`7.5 xx 10^(-8)m`

Explanation

Here, `F = Mg = 5xx9.8 N, l = 2 m, D = 2xx10^(-3)m, sigma = 0.3, Y = 12.5 xx 10^(10)Nm^(-2)`
`Y =(4F)/(pi D^2) xx (l)/(Deltal)` or `(Delta l)/(l) = (4F)/(pi D^2Y)`
Poisson's ratio, `sigma = (-Delta D//D)/(Delta l//l)`
or `Delta D = - sigma (D Delta l)/(l)` Lateral compression ` =- Delta D`
` = sigma D (Delta l)/(l) = sigma D = (4)F/(pi D^2Y)`
` = (4 sigmaF)/(pi D Y) = (4xx0.3 xx 5xx9.8)/(3.14 xx(2xx10^(-3))xx12.5 xx 10^(10))`
`= 7.5 xx 10^(-8)m`