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.

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.

Let volume of the wire before the expansion
V₁ = \({ \pi r }^{ 2 }l\)
Volume of the wire after expansion
V₂ = \(\pi \)(r-Δr)² x (l+Δl)
But volume remains same during the expansion
∴ \({ \pi r }^{ 2 }l\) = \(\pi \)(r-Δr)² x (l+Δl)
r²l = (r²+Δr²-2r x Δr) x (l x Δl)
=r²(l+Δl) + Δr²(l+Δl)-2r x Δr(l+Δl)
=r²l + r² x Δl + Δr²l + Δr2 x Δl - 2rΔr x Δl
⇒ 2rΔrl = r²Δl [Rejecting the higher power of Δr and Δr x Δl]
⇒ \(\frac { r.\triangle r.l }{ r^{ 2 }.\triangle l } =\frac { 1 }{ 2 } \)
⇒ \(\frac { \frac { \triangle r }{ r } }{ \frac { \triangle l }{ l } } =\frac { 1 }{ 2 } \)
\(\sigma =\frac { 1 }{ 2 } \)=0.5.
Therefore, if the volume of the wire does not change, the value of the poission ratio o is maximum equal to 0.5.

For steel wire A, l₁ =I; A_z = 1 mm²: \(\Upsilon \)₁ = 2 x 10¹¹ Nm^-2
For aluminium wire B, l₂= I; A₂ = 2mm²; \(\Upsilon \)₂ = 7 x 10¹⁰ Nm^-2
(a) Let mass m be suspended from the rod at distance x from the end where wire A is connected. Let F₁ and F₂ be the tensions in two wires and there is equal stress in two wires, then
\(\frac { { F }_{ 1 } }{ { A }_{ 1 } } =\frac { { F }_{ 2 } }{ { A }_{ 2 } } \Rightarrow \frac { { F }_{ 1 } }{ { F }_{ 2 } } =\frac { { A }_{ 1 } }{ { A }_{ 2 } } =\frac { 1 }{ 2 } \)           ...(i)
Taking moment of forces about the point of suspension of mass from the rod, we have
F₁x = F₂(1.05 - x) or \(\frac { 1.05-x }{ x } =\frac { { F }_{ 1 } }{ F_{ 2 } } =\frac { 1 }{ 2 } \)
or 2.10-2x = x ⇒ x = 0.70 m = 70 cm
(b) Let mass m be suspended from the rod at distance x from the end where wire A is connected. Let F₁ and F₂ be the tension in the wires and there is equal strain in the two wires i.e.,
\(\frac { { F }_{ 1 } }{ { A }_{ 1 }{ \Upsilon }_{ 1 } } =\frac { F_{ 2 } }{ { A }_{ 2 }{ \Upsilon }_{ 2 } } \Rightarrow \frac { { F }_{ 1 } }{ { F }_{ 2 } } =\frac { { A }_{ 1 }{ \Upsilon }_{ 1 } }{ { A }_{ 2 }{ \Upsilon }_{ 2 } } =\frac { 1 }{ 2 } \times \frac { 2\times 10^{ 11 } }{ 7\times 10^{ 10 } } =\frac { 10 }{ 7 } \)
As the rod is stationary, so F1x = F2 (1.05 - x) or \(\frac { 1.05-x }{ x } =\frac { { F }_{ 1 } }{ { F }_{ 2 } } =\frac { 10 }{ 7 } \)
⇒ 10x = 7.35 - 7x or x = 0.4324 m
= 43.2 cm.

Compressibility of water,
k =\(\frac{1}{B}\) = 45.8 x 10^-11 Pa^-1
Change in pressure,
Δp = 80 atm - 1 atm
= 79 atm = 79 x 1.013 x 10⁵ Pa.
\(\rho \)=1.03 x 10³ kg m^-3
As B = \(\frac { \triangle p.V }{ \triangle V } \) or \(\frac { \triangle V }{ V } =\frac { \triangle p }{ B } =\triangle p\times \frac { 1 }{ B } \)=Δp x k
or \(\\ \frac { \triangle V }{ V } \)=\(\frac { \triangle V }{ V } =\frac { (M/\rho )-(M/\rho ') }{ (M/\rho ) } =1-\frac { \rho }{ { \rho }^{ ' } } \)
or \(\frac { \rho }{ { \rho }^{ ' } } =1-\frac { \triangle V }{ V } \)
or \({ \rho }^{ ' }=\frac { \rho }{ 1-(\triangle V/V) } \)
or \({ \rho }^{ ' }=\frac { 1.03\times 10^{ 3 } }{ 1-3.665\times 10^{ -3 } } =\frac { 1.03\times 10^{ 3 } }{ 0.996 } \)
= 1.034 x 10³ kg/m³.

Here total mass to be supported, M = 50,000 kg
∴ Total weight of the structure to be supported = Mg
= 50,000 x 9.8 N
Since this weight is to be supported by 4 columns,
∴ Compressional force on each column (F) is given by
F = \(\frac { Mg }{ 4 } =\frac { 50,000\times 9.8 }{ 4 } \)N
Inner radius of a column, r₁= 30 cm = 0.3 m
Outer radius of a column, r₂ = 60 cm = 0.6 m.
∴ Area of cross-section of each column is given by
A = \(\pi ({ r }_{ 2 }^{ 2 }-{ r }_{ 1 }^{ 2 })\)
= \(\pi \)[(0.6)²-(0.3)²] = 0.27\(\pi \) m².
Young's modulus, \(\Upsilon \) = 2 x 10¹¹ Pa.
Compressional strain of each column =?
\(\Upsilon \)= \(\\ \frac { Compressional\ force \ area }{ Compressional\ Strain } \)
= \(\frac { F/A }{ Compressional\ Strain } \)
or Compressional strain of each column
= \(\frac { F }{ A\gamma } =\frac { 50,000\times 9.8\times 7 }{ 4\times 0.27\times 22\times 2\times 10^{ 11 } } \)
= 0.722 x 10^-6
∴ Compressional strain of all columns is given by
= 0.722 x 10^-6 x 4 = 2.88 x 10^-6
= 2.88 x 10^-6.

Given, mass(m) = 14 .5kg
Length of wire (l ) = 1 m
Angular frequency (v) = 2 revls
Angular velocity (\(\omega \)) = 2\(\pi \)v
= 2\(\pi \)\(\times \)2 rad/s = 4\(\pi \) rad/s
 
Area of cross-section of wire (A) = 0.065 cm²
= 6.5 \(\times \)10^-6 m²
Young's modulus for steel (Y) = 2 \(\times \) 10¹¹ N/m².
At lowest point of the vertical circle, T - mg = ml\({ \omega }^{ 2 }\)
or  T= mg +  m\({ \omega }^{ 2 }\)
= (14.5\(\times \)9.8)+14.5\(\times \)1\(\times \)\(​​{ (4\pi ) }^{ 2 }\)
= 14.5(9.8+16\({ \pi }^{ 2 }\))
= 14.5(9.8\(\times \)16\(\times \)9.87)          [\(\because \) \({ \pi }^{ 2 }\)=9.87]
= 14.5\(\times \) 167.72N=2431.94 N
Young's modulus (Y) =\(\frac { Stress }{ Strain } =\frac { (T/A) }{ \Delta l/l } =\frac { Tl }{ A.\Delta l } \)
\(\therefore \Delta l=\frac { T.l }{ A.Y } =\frac { 2431.94\times 1 }{ 6.5\times { 10 }^{ -6 }\times 2\times { 10 }^{ 11 } } \)
= 1.87\(\times \)10^-3 m =1.87 mm

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

As the cross sections of the wires are equal and same tension exists in both the stresses developed are equal. Lelt the origin/Al lengths of the steel wire and the copper wire be Ls and Lc respectively and the elongation in each wire be I.
\(\frac{l}{L_s}=\text { stres } \frac{s}{2.0 \times 10^{11} \mathrm{~nm}^{-2}} \)....(i)
and \(\frac{l}{L_c}=\) stres \(\frac{s}{1.1 \times 10^1 \mathrm{Nm}^{-2}} \)..ii
Dividing ii by i
\(\frac{L_s}{L_c}=\frac{2.0}{1.1}=20: 11\)

1.67 km