State Hooke's law and verify it with the help of an experiment.

Explanation

(i) Hooke's law is for a small deformation. When the stress and strain are proportional to each other.
(ii) It can be verified in a simple way by stretching a thin straight wire (stretches like spring) of length L and uniform cross-sectional area A suspended from a fixed point O.
(iii) A pan and a pointer are attached at the free end of the wire as shown in Figure (a).
(iv) The extension produced on the wire is measured using a vernier scale arrangement. The experiment shows that for a given load, the corresponding stretching force is F and the elongation produced on the wire is `DeltaL`.
(v) It is directly proportional to the original length L and inversely proportional to the area of cross section A. A graph is plotted using F on the X-axis and `DeltaL` on the Y-axis Figure (b).
(vi) This graph is a straight line passing through the origin.

Therefore,
Multiplying and dividing by volume,
V = A L,
F(slope) `= (AL)/(AL) Delta L`
Rearranging, we get
`(F)/(A) = [(L)/(A("slope"))](Delta L)/(L)`.
Therefore, `(F)/(A) alpha[(Delta L)/(L)]`
`[because sigma = (F)/(A) and epsilon = [(Delta L)/(L)]]`
So we `sigma alpha epsilon` i.e., the stress is proportional to the strain in the elastic limit.