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

Explanation

Hooke's law states that for a small deformation, when the stress and strain are proportional to each other.
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 spring is suspended from a fixed point O. A pan and a pointer are attached at the free end of the wire as shown in Figure 7.2. Using a vernier scale arrangement the extension produced on the wire is measured. From experiment it is known that for a given load, the corresponding stretching force is F and the elongation produced on the wire is `DeltaL`.

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. This graph is a straight line passing through the origin as shown in Figure 7.3.

`thereforeDeltaL=(slope)F`
Multiplying and dividing by volume,
`V=AL`,
`F(slope)=(AL)/(AL)DeltaL`
Rearranging, we get
`(F)/(A)=((L)/(A(slope)))(DeltaL)/(L)`
Therefore, `(F)/(A)prop((DeltaL)/(L))`
From the equations, we get
Stress, `sigma=("Force")/("Area")=(F)/(A)`
Strain, `epsilon=("Change in size")/("Original size")=(Deltal)/(l)`
`sigmapropepsilon`
i.e., the stress is proportional to the strain in the elastic limit.