A 15 kg mass is attached to one end of a copper wire 2m ling and 2mm in diameter. Calculate the lateral compression produced in it. Poission's ratio is 0.30 and ypung's modulus of the material of the wire is `12 xx 10^(10) Nm^(-2)`. Use , g=10ms^(-2)`.

Explanation

Here, `F=Mg= 15 xx 10N = 150 N, l=2m, D-2mm=2 xx 10^(3) m, sigma - 0.30`,
`Y=12 xx 10^(10) Nm^(-2)`
`Y=(F)/((pi D^(2)//4) xx (l)/(Delta l)) or Delta l=(4Fl)/(pi D^(2)y)`
`:. Delta l = (4 xx 150 xx 2)/((3.14 xx (2 xx 10^(-3))^(2) xx 12 xx 10^(10))) = 7.35 xx 10^(-4)m`
Poission's ratio , `sigma =(-Delta D//D)/(Delta l//l)`
or `Delta D = (Delta l xx D xx sigma)/(l) = -((7.35 xx 10^(-4)) xx ( xx 10^(-3)) xx 0.30)/(2) = - 2.205 xx 10^(-7) m =-0.22 mu m`
Thus lateral compression produced in wire is `0.22 mu m`.