A one meter long $50mm$ diameter rod shortens by $2\mathrm{.}4$

$mm$ under a load. The strain in the lateral direction is

equal to the strain along the axis of the rod multiplied by

the Poisson's ratio, $.$ Using $=0\mathrm{.}33,$ find the diameter of

the rod under the load.

Ans. $50\mathrm{.}0396mm$