We know that some materials take appreciable time to recover their original condition completely. In other words, the strain persists even when the stress is removed. This lagging behind of strain is called elastic hysteresis.

Compressibility of the material of a body is defined as the reciprocal of its bulk modulus. It is, thus, defined as the fractional change in volume per unit increase in pressure.
Compressibility, K=\(\frac { 1 }{ B } =-\left( \frac { \triangle V }{ V } \right) \times \frac { 1 }{ P } \)
The solids are least compressible whereas gases are most compressible. It is on account of the fact that in solids neighbouring atoms are tightly coupled but molecules in gases are very poorly coupled to their neighbours.

Elastomers are those substances which can be stretched to cause large strains. Substances like tissue of aorta, rubber etc., are elastomers.
The stress-strain curve for an elastomer is as shown in figure below. Although elastic region is very large but the material does not obey Hooke's law over most of the region Moreover, there is no well defined plastic region.

The shearing stress is exerted on the rectangular surface (2.0 cm x 2.0 cm x 0.2 cm) that is the boundary of the hole. The area of this surface is (see Fig)
A = 2.0 x 10^-2 x 4 x 0.2 x 10^-2 m²
= 1.6 x 10^-4 m²
Since the minimum shearing stress to rupture the steel is
\(\left( \frac { F }{ A } \right) _{ min }\)=.5 x 10⁸ Nm^-2
The force required is given by
F=3.5 x 10⁸ x 1.6 x 10^-4N=5.6 x 10⁴ N

l = 4m, r = 1.5 mm = 1.5 x 10^-3 m,
M = 50 kg,
 \(\Upsilon \)=7 x 10¹⁰ Nm^-2, F = 50 x 9.8 N = 490 N, Δl = ?
∴ \(\Upsilon =\frac { F\times l }{ A\times \triangle l } \)
or \(\triangle l=\frac { F\times l }{ { \pi r }^{ 2 }\times \Upsilon } =\frac { 490\times 4\times 7 }{ 22\times (1.5\times { 10 }^{ -3 })^{ 2 }\times 7\times 10^{ 10 } } \)m
= 39.6 x 10^-4 = 39.6 x 10^-4 x 10³ mm = 3.96 mm

Here, the pulling force F (= mg) is same on both the wires. Let Δl₁ , Δl₂ be the elongations in the two wires.

As, \(\Upsilon =\frac { Fl }{ { \pi r }^{ 2 }\triangle l } \) or \(\triangle l=\frac { Fl }{ \Upsilon { \pi r }^{ 2 } } \)
For wire 'A', \({ \triangle l }_{ 1 }=\frac { mgl }{ \Upsilon \pi { r }^{ 2 } } \)
For wire 'B', \({ \triangle l }_{ 2 }=\frac { mg(2l) }{ \Upsilon \pi (2{ r }^{ 2 }) } =\frac { mgl }{ 2\Upsilon \pi { r }^{ 2 } } \)
Total elongation = Δl₁ + Δl₂
=\(\frac { mgl }{ \Upsilon \pi { r }^{ 2 } } +\frac { 1 }{ 2 } \frac { mgl }{ \Upsilon \pi { r }^{ 2 } } =\frac { 3 }{ 2 } \frac { mgl }{ \Upsilon \pi { r }^{ 2 } } \).

When a deforming force acts on a body, the atoms of the substance get displaced from their original positions. Due to this, the configuration of the body (substance) changes. The moment, the deforming force is removed, the atoms return to their original positions and hence the substance or body regains its original configuration. That is why, elasticity is said to be internal property of matter.

Let l and A be the original length and area of cross-section of the metal wire.
Change in length in the first case = (l₁ - l)
Change in length in the second case = (l₂ - l)
\(\Upsilon =\frac { { T }_{ 1 } }{ A } \times \frac { l }{ ({ l }_{ 1 }-l) } =\frac { { T }_{ 2 } }{ A } \times \frac { l }{ ({ l }_{ 2 }-l) } \)
or T₂l₂ - T₁l = T₂l₁-T₂l
or l(T₂-T₁) = T₂l₁-T₁l₂ ⇒ l = \(\frac { { T }_{ 2 }{ l }_{ 1 }-{ T }_{ 1 }{ l }_{ 2 } }{ ({ T }_{ 2 }-{ T }_{ 1 }) } \).

We know, K = \(\frac { PV }{ \triangle V } \)
∴ ΔV =  \(\frac { PV }{ K } \)
Now P = 5 x 10⁸ Nm^-2;
V = \(\frac { 4 }{ 3 } \pi \)r³ = \(\frac { 4 }{ 3 } \pi \)(0.1)³ m³
= 4.19 x 10^-3 m³
[∵ r = 10 cm = 0.1 m]
K = 3.14 x 10¹¹ Nm^-2
ΔV = \(\frac { 5\times 10^{ 8 }\times 4.19\times 10^{ -3 } }{ 3.14\times 10^{ 11 } } \)
= 6.67 x 10^-6 m³.

A spring will be better one if a large restoring force is set up in it on being deformed, which in turn depends upon the elasticity of the material of the spring. Since the Young's modulus of elasticity of steel is more than that of copper, hence steel is preferred in making the springs.

Consider two pieces of wires, one of steel and the other of rubber. Suppose both are of equal length (L) and of equal area of cross-section (a). Let each be stretched by equal forces, each being equal to F. We find that the change in length of the rubber wire (l_r) is more than that of the steel (l_s) i.e. l_r < I_s If \(\Upsilon \)_s and \(\Upsilon \)_r, the Young's moduli of steel and rubber respectively, then from the definition of Young's modulus,
\({ \Upsilon }_{ s }=\frac { F.L }{ a.{ l }_{ s } } \) and \({ \Upsilon }_{ r }=\frac { F.L }{ a.{ l }_{ r } } \)
∴ \(\frac { { \Upsilon }_{ s } }{ { \Upsilon }_{ r } } =\frac { { l }_{ r } }{ { l }_{ s } } \)
As l_r > l_s ∴ \(\frac { { \Upsilon }_{ s } }{ { \Upsilon }_{ r } } \)>1 or \({ \Upsilon }_{ s }>{ \Upsilon }_{ r }\).
i.e., the Young's modulus of steel is more than that of rubber. Hence steel is more elastic than rubber.
(or ) Any material which offers more opposition to the deforming force to change its configuration is more elastic.

(i) Energy density = \(\frac { 1 }{ 2 } \left( \frac { \Upsilon l }{ L } \right) .\frac { 1 }{ L } \)
=\(\frac { 1 }{ 2 } \left[ \frac { 2\times 10^{ 11 }\times 2\times { 10 }^{ -3 } }{ 4 } \right] \times \left[ \frac { 2\times 10^{ -3 } }{ 4 } \right] \)
=\(\frac { 1 }{ 2 } \) x 10⁸ x \(\frac { 1 }{ 2 } \) x 10^-3
= 0.25 x 10⁵
= 2.5 x 10⁴ J/m³.
(ii) Potential energy stored in the wire
U = \(\frac { 1 }{ 2 } \left( \frac { \Upsilon Al }{ L } \right) .l\)
= \(\frac { 1 }{ 2 } \left[ \frac { 2\times { 10 }^{ 11 }\times 2\times 10^{ -6 }\times 2\times 10^{ -3 } }{ 4 } \right] \) x 2 x 10^-3
= 10² x 2 x 10^-3
= 0.2 J.

(a) False. The Young's modulus is defined as the ratio of stress to the strain within elastic limit. For a given stretching force elongation is more in rubber and quite less in steel. Hence, rubber is less elastic than steel.
(b) True. Stretching of a coil is determined by its shear modulus. When equal and opposite forces are applied at opposite ends of a coil, the distance as well as shape of helicals of the coil change and it involves shear modulus.

Given, each side of cube()=10cm=0.1m
Hydralic pressure (p)=7×10⁶ Pa
Bulk modulus for copper(B)=140×10⁹ Pa
Volume contraction(△V)=?|
Volume of the cube(V)=I³=(0.1)³=1×10⁻³m³
Bulk modulus for copper(B) \(=\frac { p }{ \triangle V/V } \)
\(\\ \triangle V=\frac { pV }{ B }\)
\( \\ \triangle V=\frac { 7\times 10^{ 6 }\times 1\times 10^{ -3 } }{ 140\times 10^{ 9 } } =\frac { 1 }{ 20 } \times 10^{ -6 }m^{ 3 }\)
\(\\ =0.05\times 10^{ -6 }m^{ 3 }=5\times 10^{ -8 }m^{ 3 }\)

For wire A, \(F_1=F, A_1=\pi(A r)^2, l_1=4 l, \mathrm{Y} _1=\mathrm{Y}, \Delta I_1 =1 \mathrm{~mm}\).
For wire B, \(F_2=F, A_2=\pi(3 r)^2, l_2=5 l\),
\( Y_2=Y, \Delta l_2=? \)
\( Y=\frac{F_1}{A_1} \times \frac{1_1}{\Delta_1}=\frac{F_2}{A_2} \times \frac{l_2}{\Delta l_2} \)
\( \Delta l_2=\frac{F_2}{F_1} \times \frac{A_1}{A_2} \times \frac{l_2}{l_1} \times \Delta l_1 \)
\( =1 \times\left(\frac{4}{3}\right) \times\left(\frac{5}{4}\right) \times 1=\frac{20}{9} \mathrm{~mm}=2.22 \mathrm{~mm}\)