$2\mathrm{.}1.$ For very large strains, a special liquid$-$metal strain gage can be employed, which is simply

a rubber tube filled with liquid metal $[3].$ It is obvious that the volume of the liquid metal,

$V=LA,$ will not change during deformation of the test specimen. Assuming that the

change in the specific resistance of the liquid metal is negligible, show that $\frac{R}{R}=2lon+$

$lo{n}^2,$ where $lon=\frac{L}{L}.$