(i) All three states of matter (i.e., solid, liquids and gases) have volume elasticity.
(ii) When we stretch a wire, the work has been done against interatomic forces. This work is stored in the wire in the form of elastic potential energy.
(iii) Elastic limit is the upper limit of deforming force upto which, if deforming force is removed, the body regains its original form completely and beyond which if deforming force is increased, the body loses its property of elasticity and gets permanently deformed.
(iv) Modulus of elasticity ofa body is defined as the ratio of the stress to the corresponding strain produced, within the elastic limit.
It depends upon the nature of material of the body and the manner in which the body is deformed.
(v) It is because for a given stress applied the gases are more compressible than that of solids.
(vi) r₁ : r₂ = 2: 1, F₁ = F₂ = F
\( \text { Stress }=\frac{\text { Force }}{\text { area }}=\frac{F}{\pi r^{2}}, \text { therefore } S \alpha \frac{1}{r^{2}} \)
Hence \( \frac{S_{1}}{S_{2}}=\frac{r_{2}^{2}}{r_{1}^{2}}=\left(\frac{1}{2}\right)^{2}=\frac{1}{4} \) 

(i) Portion OA of the graph is a straight line, showing that stress is directly proportional to strain. Hook's law is fully obeyed in the region OA. Point A is called proportional limit.
(ii) Point B represents the elastic limit. At this point if the wire is unloaded at B then wire will acquire its original configuration. Portion OB is called elastic region.
(iii) If the wire is completely unloaded then strain is equal to 00,. This region is called permanent set and region beyond C is known as plastic region.
(iv) The point B at which the wire yields to the applied stress and begins to flow down. The stress corresponding to yield point B is called yield strength of the wire.
(v) The stress corresponding to point E is called breaking stress or breaking point.
(vi) Elastomers do not obey Hook's law over most of the region. These have no plastic range.

(i) T cos \(\theta\) = mg
T sin  \(\theta\)= mr \(\omega\)² (centripetal force)
Dividing \( \tan \theta=\frac{r \omega^{2}}{g} \text { or } \omega=\sqrt{\frac{g \tan \theta}{r}} \) 
(ii) \( T^{2}=(m g)^{2}+\left(m r \omega^{2}\right)^{2} \) 
or \( T=m\left(g^{2}+r^{2} \omega^{4}\right)^{1 / 2} \)
(iii) \( \text { Stress }=\frac{T}{A}, \text { and strain }=\frac{\Delta L}{L}=\frac{\text { Stress }}{Y} \) 
\( \therefore \ \Delta L=\frac{T L}{A Y}=\frac{m g}{\cos \theta} \times \frac{L}{A Y} \)
(iv) \( \text { Stress }=\frac{T}{A}=\frac{m g}{\cos \theta \cdot A} \)
(v) Elastic potential energy = \( \frac{1}{2} \times \text { stress } \times \text { strain } \times \text { volume } \)
\( \text { Energy density }=\frac{\text { Elastic potential energy }}{\text { volume of wire }} \)
\( \therefore \frac{\text { Elastic potential energy }}{\text { Energy density }}=\text { volume of the wire. } \) 

(i) The temporary delay in regaining the original configuration by an elastic body after the removal of deforming force is called elastic after effect.
(ii) Elastic fatigue is the property of an elastic body by virtue of which its behaviour becomes less elastic under the action of repeated alternating deforming forces.
(iii) Bridges and girders are given I shape to protect it against bending because angle of depression is inversely proportional to cube of depth of the beam.
(iv) Torque \( \tau \) required to produce a given twist in a solid shaft is much smaller than that in a hollow shaft of same length and material.
(v) It is ratio of lateral strain to longitudinal strain
\( \sigma=\frac{-\Delta R / R}{\Delta l / l} \)
(vi) Since breaking load = breaking stress x area, is free from the length of elastic wire, therefore if cable is cut to half of its original length, there is no change in its area of cross-section and breaking stress, Hence there is no effect on the maximum load that the cable can support,
(vii) Since the youngs modulus of elasticity of steel is more than that of copper, hence steel is preferred in making the springs,
(viii) When a bridge is used for long time, it loses its elastic strength, therefore the amount of strain in the bridges for a given stress will become large and ultimately the bridge may collapse,

(i) The decrease in the length of the rod on cooling is
\( \Delta L=L \alpha \Delta t=0.5 \times 10^{-} \times 100=5 \times 10-{ }^{4} \mathrm{~m} \)
(ii) If the rod is to be prevented from contracting, the mass m attached at the lower end must increase its length by an amount \(\triangle\)L = 5 x 10^-4 m. Now
\( Y=\frac{m g L}{A \Delta L} \)
\( \Rightarrow \quad m=\frac{Y A \Delta L}{g L}=\frac{10^{11} \times 4 \times 10^{-6} \times 5 \times 10^{-4}}{10 \times 0.5} \)
= 40 kg
(iii) Energy stored in the rod is
\( U=\frac{1}{2} F \times \Delta L=\frac{1}{2} m g \Delta L \)
\(=\frac{1}{2} \times 40 \times 10 \times 5 \times 10^{-4}=0.1 \mathrm{~J}\) 
(iv) When a wire is stretched, some work is done against the internal restoring forces acting between particles of the wire, This work done appears as elastic potential energy in the wire,
Elastic potential energy per unit volume
\(=\frac{1}{2} \times \text { Young's modulus } \times(\text { strain })^{2}\)