Due to extension produced in the cord, energy is stored in it which is converted into kinetic energy when the stone flies away. Assuming that there is no loss of energy in this process, the kinetic energy of the stone is given by
W = \(\frac{1}{2}\)mv² = 4J
This must be equal to the work done in stretching the cord.
Using the equation.
W = \(\frac{1}{2}\)FΔl = 4J
Where F is the stretching force. Since
Δl = 20 cm = 0.2 m
F= \(\frac { 4\times 2 }{ 0.2 } \)= 40 N
Stress = \(\frac { F }{ A } =\frac { 40 }{ { \pi r }^{ 2 } } \)
Now, r = 3 mm = 3 x 10^-3 m
Hence, Stress = \(\frac { 40 }{ \pi (3\times 10^{ - })^{ 2 } } \)=1.415 x 10⁶ Nm^-2
But Strain = \(\frac { 20 }{ 42 } \)=0.476
Young's modulus = \(\Upsilon =\frac { 1.415\times { 10 }^{ 6 } }{ 0.476 } \) 
= 2.97 x 10⁶ Nm^-2.

Numerical: Let I be the length of the wire at 20°C and ley the length at O°C. Then
l-l₀ = ∝l₀ ΔT = 20 ∝.l₀
Compressive strain =\(\frac { l-{ l }_{ 0 } }{ { l }_{ 0 } } \) =20∝ = 20 x 12 x 10^-6 = 2.4 x 10^-4
\(\Upsilon \)= \(\frac { Stress }{ Strain } \)
Stress = \(\Upsilon \) x strain = \(\frac{F}{A}\)
Hence, tension T=\(\Upsilon \)A x Strain
= 21 x 10¹¹ x 0.5 x 10^-2 x 2.4 x10^-4
= 2.52 x 10⁶ dyne
= 25.2 N
This is the tension in the wire when the temperature falls to O°C.

For Hooke's law and Young's modulus of elasticity, see fact that matter on pages 386-387 of this text book.
Let L be the original length of the wire, A be its .area of cross-section and W be the load attached to the wire. Then, Young's modulus of the wire is given by
Y = \(\frac { W\times L }{ A\times \triangle L } \)
Since ΔL = 90-L
∴ \(\Upsilon =\frac { W\times L }{ A(90-L) } \)      ......(1)
Volume of weight attached = \(\frac { W }{ density\quad of\quad weight } \)
= \(\frac { W }{ 7.6 } \)cm³
Weight of water displaced = \(\frac { W }{ 7.6 } \) x density of water
\(\frac { W }{ 7.6 } \times 1=\frac { W }{ 7.6 } \)
∴ Net weight after immersing in water is
W'=W-\(\frac { W }{ 7.6 } =\frac { 6.6W }{ 7.6 } \)
Length of wire after immersing in water
= (90 - 0.18) = 89.82 cm
∴ Change in length on immersing in water,
ΔL'=(89.82-L)cm
∴ \(\Upsilon \)=\(\frac { W'L }{ A\triangle L' } \)
= \(\frac { 6.6W\times L }{ 7.6\times A(89.82-L) } \)     ....(2)
Comparing eqns. (1) and (2), we get
\(\frac { W\times L }{ A(90-L) } =\frac { 6.6W\times L }{ 7.6\times A\times (89.82-L) } \)
7.6 x (89.82-L) = 6.6 (90-L)
682.632-7.6L = 594-6.6L
∴ L = 88.632 cm

For Poisson's ratio, see text.
Poisson's ratio
where r is the radius of the wire and l its length.
Volume of the wire before expansion V₁
Volume of the wire after expansion
V₂ = \(\pi\)(r-Δr)²(l+Δl)
If the volume is to remain unchanged during expansion, we require that
V₁ = V₂, i.e.,
\(\pi\)r²l = \(\pi\)[r²-2r Δr + (Δr)²](l+Δl)
= \(\pi\)r²(l+Δl)-2\(\pi\)r Δrl-2r Δr Δl \(\pi\) + (Δr)²(l+Δl)\(\pi\)
or 2\(\pi\)r Δrl =\(\pi\)r² Δl + terms containing the product of Δr and Δl, which can be neglected.
Hence \(\frac { \triangle r/r }{ \triangle l/l } =\frac { 1 }{ 2 } \)
or σ = 0.5
Thus the volume of the wire does not change if the Poisson's ratio of the material of the wire is 0.5.

Consider a wire of length L and area of crosssection A. Let a force F be applied to stretch the wire (Fig. a). If Ibe the length through which the wire is stretched, then
Longitudinal strain = \(\frac{l}{L}\)
and tensile stress = \(\frac{F}{A}\)
∴ Young's modulus of elasticity,
\(\Upsilon =\frac { stress }{ strain } =\frac { F/A }{ l/L } =\frac { FL }{ Al } \)
⇒ F= \(\frac { \Upsilon Al }{ L } \)               ......(i)

If the wire is stretched through a length dl, then work done is given by
dW = Fdl = \(\frac { \Upsilon Al }{ L } \)dl              .....(ii)
∴ Total work done to stretch the wire through length l can be calculated by integrating
eqn. (ii) between the limits l=0 to ml=l
i.e., \(\int { dW } =\int _{ 0 }^{ 1 }{ \frac { \Upsilon A }{ L } } \)
⇒ W= \(\frac { \Upsilon A }{ L } \frac { { l }^{ 2 } }{ 2 } =\frac { 1 }{ 2 } \left( \frac { \Upsilon Al }{ L } \right) \times l\)
or W = \(\frac{1}{2}\)F x l
[F = \(\frac { \Upsilon Al }{ L } \) , from eqn. (i)]
Numerical:
Hence, work done =\(\frac{1}{2}\) stretching  force x extension
W₁ = \(\frac{1}{2}\)F₁(l₁-l) and W₂=\(\frac{1}{2}\)F₂(l₂-l)
U = W₂-W₁=\(\frac{1}{2}\)F₂(l₂-l)-\(\frac{1}{2}\)F₁(l₁-l)
= \(\frac{1}{2}\)[F₂l₂-F₁l₁+(F₁-F₂)l]            .....(i)
Now, \(\frac { { F }_{ 1 } }{ { F }_{ 2 } } =\frac { { l }_{ 1 }-l }{ { l }_{ 2 }-l } \) or F₁l₁-F₁l=F₂l₁-F₂l
or (F₂-F₁)=F₂l₁-F₁l₂
⇒ l = \(\frac { { F }_{ 2 }{ l }_{ 1 }-{ F }_{ 1 }{ l }_{ 2 } }{ { F }_{ 2 }-{ F }_{ 1 } } \)
From eqn. (i)
U=\(\frac{1}{2}\)[F₂l₂-F₁l₁+(F₁-F₂)\(\frac { { F }_{ 2 }{ l }_{ 1 }-{ F }_{ 1 }{ l }_{ 2 } }{ { F }_{ 2 }-{ F }_{ 1 } } \)]
= \(\frac{1}{2}\)[F₂l₂-F₁l₁-F₂l₁+F₁l₂]
= \(\frac{1}{2}\)[(F₂+F₁)l₂-(F₂+F₁)l₁]
U = \(\frac{1}{2}\)(F₂+F₁)(l₂-l₁).

Given, side of a cube (l) = 10 cm = 0.1 m
Area of its each face (A) = l² = (0.1)² = 0.01 m²
Load(m) = 100 kg
Tangential force acting on one face of the cube, F = mg = 100\(\times \) 9.8 =980 N
Shear stress acting on this face = \(\frac { F }{ A } \) = \(\frac { 980 }{ 0.01 } \) N/m²
= 9.8 \(\times \)10⁴ N/m²
Shear modulus of aluminium (\(\eta \)) = 25 GPa
= 25\(\times \) 10⁹ N/m²
Shear modulus (\(\eta \))=\(\frac { Shearing\ stress }{ Shearing\ strain } \)
or    shearing strain \(\left( \frac { \Delta L }{ L } \right) \) = \(\frac { Shearing\ stress }{ Shear\ modulus } \)
or      \(\Delta L\)   = \(\frac { Shearing\ stress }{ Shear\ modulus } \) \(\times \) L = \(\frac { 9.8\times { 10 }^{ 4 } }{ 25\times { 10 }^{ 9 } } \)\(\times \) 0.1
= 0.0392 \(\times \)10^-5m
= 3.92 \(\times \) 10^-7m

Bulk modulus(B)\(=\frac { 1 }{ Compressibility } =\frac { 1 }{ k } \)     
\(=0.25\times { 10 }^{ 5 }\times 1.013\times { 10 }^{ 5 }\ N/m^{ 2 }\)
\(\\=2.533\times { 10 }^{ -4 }{ m }^{ 3 }\)
\(\\ Volume,V\ =100cm^{ 3 }=10^{ -4 }m^{ 3 }\)
\(\\ Pressure,P=100\ atm=100\times 1.013\times { 10 }^{ 5 }\ N/m^{ 2 }\)
\(\\ =1.013\times { 10 }^{ 5 }N/m^{ 2 }\)
\(\\ Now\quad apply\frac { 1 }{ B } =k=\frac { \Delta V }{ pV } \)
\(\\ \therefore \Delta V=\frac { pV }{ B } =\frac { 1.013\times { 10 }^{ 7 }\times { 10 }^{ -4 } }{ 2.533\times { 10 }^{ 9 } } \)
\( \Delta V=0.4\times { 10 }^{ -6 }{ m }^{ 3 }=0.4{ cm }^{ 3 }\) 

Since, Young's modulus depends only on the nature of material and not on ita dimensions. So, the value of Young's modulud of the copper wire is not changed when its radius decreases. Thus , Young's modulus of the wire remains the same.

Given area of cross section A = 2m²
Force  F = 3kN = 3 x 10³ N
Length   L = 4m
Young's modulus Y = 110 x 10⁹ N/m²
Change in length, \(\Delta L=?\)
Apply,    \(Y=\frac { FL }{ A\Delta L } \)
\(\Rightarrow \Delta L=\frac { Fl }{ AY } =\frac { 3\times { 10 }^{ 3 }\times 4 }{ 2\times 110\times { 10 }^{ 9 } } =0.0545\times { 10 }^{ -6 }m\)
\( \Delta L=54.5\times 10^{ -3 }mm\)

Given, angle of shear = 30\(^0\)  and change in length  \(\Delta L\) = 250 cm = 2.5 m
\(\therefore\)   Shear strain, tan \(\theta \) = \(\frac { \Delta L }{ L } \) \(\Longrightarrow \)  tan 30\(^0\) = \(\frac { 2.5 }{ L } \) 
L = \(\frac { 2.5 }{ tan30^{ 0 } } \) = \(\frac { 2.5 }{ 0.577 } \) = 4.332 m
Volume, V = L³ = 81.309 m³

\(\eta \) of material is smaller than its Y. As,we know that it is easier to slide layers of atome of solids over one another than to pull them apart or to squeeze them to close together.