The poisson' s ratio of a wire of sigma. Show that if e is the longitudinal strain due to an applied forces, the volume strain will be e(1-2sigma).

The poisson' s ratio of a wire of sigma. Show that if e is the longitu

1.	The poisson' s ratio of a wire of sigma. Show that if e is the longitudinal strain due to an applied forces, the volume strain will be e(1-2sigma).
Answer» Solution :SUPPOSE the LENGTH of the wire is l, RADIUS is R and its volume is V. `thereforeV=pir^2l` or,`dV=pir^2dl+2pilrdr=pi r^2dl-2pir^2sigmadl` `[becausesigma=(-dr//r)/(dl//l),dr=-rsigma(dl)/l]` `=pir^2dl(1-2sigma)` `therefore ` Volume strain `=(dV)/V=(pir^2dl(1-2sigma))/(pir^2l)=(dl)/l(1-2sigma)` `=e(1-2sigma)` [given that `(dl)/l=e`].