When a body of mass 5kg is hung from a wire of length 1 m and radius 2mm,the length increases by 0.1 mm, IF the poisson's ratio is 0.4 , what will be the change in the radius of the wire? If the load is reduced to 2kg,how will the radius change?

Explanation

Poisson. s ratio , `sigma=d/D.L/l`
[where D,L are the initial diameter and the length of the wire and d,l are the changes in the diameter and in the length of the wire respectively]
`therefored=(sigmaDl)/L=(0.4times0.004times0.0001)/1m[becausesigma=0.4,D=2times2=4mm=0.004m,l=0.1mm=0.0001m,L=1m]`
or, `d=16 times 10^-8 m`
`therefore` Change in radius `=(16times10^-8)/2=8times10^-8m`
Now,`Y=(mgL)/(pir^2l)or,l=(mgL)/(Ypir^2)`
So, for the same wire , `l prop m`. Then we get,
`m_1/m_2=l_1/l_2or,5/2=0.0001/l_2or,l_2=4times10^-5m`
`therefore` Change in diameter,
`d_2=(sigmaDl_2)/L=(0.4times0.004times4times10^-5)/1=64times10^-9m`
`therefore` Change in radius `(64times10^-9)/2=3.2times10^-8m`