Solution:

`(i) text(Work done)/(volume) = (1F)/(2L) . (Deltal)/(l)`

`:.` Work done 1`/2 F xx A (Deltal)/(l)`
`= 1/2` Tension x extension
(ii) The wire has length l, area of crosssection A made of material constant Y. Let a force F be applied and at any instance, x be the extension associated `(x < L),` where L is the maximum extension . At this instant , `F = (AY.x)/(l)`

Since force is a variable with `x`, Work done to stretch is,

`W = int_(0)^(L) Fdx`

`W= 1/2 (AY)/l . L^2`

`W = 1/2 (Al)((Y.L))/(l) (L/l)`
`=1/2` x Volume x Stress x Strain
`:.` Work done per unit volume
`= 1/2` Stress `xx` Strain

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Solution:

Young's modulus of elasticity is the ratio of the longitudinal stress to the longitudinal strain.

`Y = ((F/a)/(Deltal)/(l))` Elastic fatigue is the delay in getting back to the original status when the
deforming force is removed.

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Solution:

Similarities :
(i) Electrical by nature.
(ii) Active over short distances.

Differences :
(i) Force between molecules or atoms.
(ii) Molecular forces are weaker than interatomic forces.

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Solution:

Elastic limit.

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Solution:

When the deforming force is increased, a limit is reached beyond which the solid does not come back to its original shape or size but remains deformed on removal of applied force. This limit is called the elastic limit.

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Solution:

It is a material constant. So all the physical quantities that might alter the nature like density can alter it.

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Solution:

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