Volume of wire
V = \(\pi\)r²L
∴ Density = \(\frac { mass }{ V } \)
9 x 10³ = \(\frac { 31.4 }{ \pi r^{ 2 }\times L } \)
L = \(\frac { 3.14 }{ { \pi r }^{ 2 }\times 9\times 10^{ 3 } } \)
L = \(\frac { 31.4 }{ 3.14\times 10^{ -6 }\times 9\times 10^{ 3 } } \)
∴ L = \(\frac { 10 }{ 9 } \) x 10³m
=\(\frac { { 10 }^{ 4 } }{ 9 } \)m
Now \(\Upsilon =\frac { mg\quad L }{ { \pi r }^{ 2 }l } \)
l = \(\frac { mg\quad L }{ { \pi r }^{ 2 }.\Upsilon } \)
=\(\frac { 3.14\times 9.8\times { 10 }^{ 4 } }{ 3.14\times 10^{ -6 }\times 9\times 9.8\times 10^{ 10 } } \)
= \(\frac { 10 }{ 9 } \)m
Now the work done =\(\frac { 1 }{ 2 } \)F.l
=\(\frac { 1 }{ 2 } \) x 3.14 x 9.8 x \(\frac { 10 }{ 9 } \)
75%of the work done is converted into heat energy
∴ Heat energy =\(\frac { 1 }{ 2 } \) x 31.4 x 9.8 x \(\frac { 10 }{ 9 } \times \frac { 75 }{ 100 } \)
But heat energy  =mass x S.P. heat x temp difference
= 1.4 x 490 x t
∴ 31.4 x 490 x t = \(\frac { 1 }{ 2 } \) x31.4 x 9.8 x \(\frac { 10 }{ 9 } \times \frac { 75 }{ 100 } \)
∴ t = \(\frac { \frac { 1 }{ 2 } \times 3.14\times 9.8\times \frac { 10 }{ 9 } \times \frac { 75 }{ 100 } }{ 3.14\times 490 } \)
t = \(\frac { 1 }{ 120 } \)K or 0.0083⁰C

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

If the rod stretches along its length, then the stress produced is the tensile stress whereas the strain produced is longitudinal strain.
Radius, r = 8 mm = 8 x 10^-3 m,
Length, L = 2 m
Force, F = 150 kN = 15 x 10⁴ N
Area, A = \(\pi \) r² = \(\pi \) x ( 8 x 10^-3)² = 201 x 10^-6 m²
\(\Delta \) L = 7.46 mm = 7.46 x 10^-3 m, Percentage strain = ?
Stress = \(\frac { F }{ A } \) = \(\frac { 15\times { 10 }^{ 4 } }{ 201\times { 10 }^{ -6 } } \) = 0.0746 x 10¹⁰ N/m²
= 7.46 x 10⁸ N/m²
Longitudinal strain = \(\frac { \Delta L }{ L } \) = \(\frac { 7.46\times { 10 }^{ -3 } }{ 2 } \) = 3.73 x 10^-3
Percentage strain = 3.73 x 10^-3 x 100 = 0.37 %

Here B = 2.3 x 10⁹ Nm^-2 = \(\frac { 2.3\times { 10 }^{ 9 } }{ 1.01\times { 10 }^{ 5 } } \) atmosphere = 2.27 x 10⁴ atmosphere
(a) ∴ Compressibility k =\(\frac { 1 }{ B } =\frac { 1 }{ 2.27\times 10^{ 4 } } \) = 4.4 x 10^-4 atm^-1
(b) As \(\frac { \triangle V }{ V } \) = -0.1% = -\(\frac { 0.1 }{ 100 } \) = -0.001
∴ Increase in pressure p=B\(\left( -\frac { \triangle V }{ V } \right) \) = 2.27 x 10⁴ x 0.001 = 22.7 atm.

Original length, 1 = 2 \(\pi\) r; Extension, Δl = 2 \(\pi\)r = 2\(\pi\)(R - r)
Strain = \(\frac { \triangle l }{ l } =\frac { \pi (R-r) }{ 2\pi r } =\frac { R-r }{ r } \)
Young's modulus, E= \(\frac { F/A }{ (R-r)/r } \) or F = EA\(\left( \frac { R-r }{ r } \right) \).

The material whose plastic range is comparatively large.

According to Hooke's law,
Modulus of elasticity, E = \(\frac { F }{ \pi { r }^{ 2 } } \times \frac { l }{ \Delta l } \) 
or      \(\Delta l=\frac { Fl }{ \pi { r }^{ 2 }E } \)
or    \(\Delta l\propto \frac { Fl }{ { r }^{ 2 } } \)         [ \(\because\)  E is same for two wires]
\(\therefore\)      \(\frac { \Delta { l }_{ 1 } }{ \Delta { l }_{ 2 } } =\frac { { F }_{ 1 } }{ { F }_{ 2 } } \times \frac { { l }_{ 1 } }{ { l }_{ 2 } } \times \frac { { r }_{ 2 }^{ 2 } }{ { r }_{ 1 }^{ 2 } } \) 
= \(\frac { 1 }{ 4 } \times \frac { 2 }{ 1 } \times \left( \frac { 3 }{ 1 } \right) ^{ 2 }=\frac { 9 }{ 2 } \) 
So,           \(\Delta { l }_{ 1 }\quad :\Delta { l }_{ 2 }\) 
= 9 : 2

Given,
L=50 cm=50×10⁻² m
t=10 cm=10×10⁻² m
G=5.6×10⁹ Pa
F=9.0×10⁴ N
Area of the face on which force is applied,
A=50×10=500 cm²=0.05 m²
Let ΔL be the displacement of the upper edge of the slab, due to tangential force F applied.
\( \text { Then, } G=\frac{\left(\frac{F}{A}\right)}{\left(\frac{\Delta L}{L}\right)} \)
\( \Rightarrow \Delta L=\frac{F L}{G A}=\frac{9 \times 10^4 \times 50 \times 10^{-2}}{5.6 \times 10^9 \times 0.05} \)
\( \therefore \Delta L=1.6 \times 10^{-4} \mathrm{~m} \)

Let \(\gamma \) = coefficient of cubical expansion of the cube.
Let K be bulk modulus of elasticity of its material.
V = initial volume,
P = pressure applied
ΔV = Decrease in its volume
∴ By definiton, K = \(\frac { P }{ \frac { \triangle V }{ V } } \)
or ΔV =  \(\frac { PV }{ K } \)
Also ΔV ∝ V
∝ t
or ΔV = \(\gamma \) V x t = \(\gamma \)Vt
where t = rise in its temperature so as to increase the volume by ΔV s.t. it is brought back to its initial volume.
∴ From (i) and (ii), we get
\(\frac { PV }{ K } \)= \(\gamma \)Vt
or t = \(\frac { PV }{ K\gamma V } =\frac { P }{ \gamma K } \) .

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.

As shown in figure, consider a small element of thickness dx at distance x from the fixed support. Force acting on the element dx is                                      
F = Weight of length (l- x) of the rod
= \(\frac { W }{ l } \)(l-x)
Elongation of the element = Original length \(\times \)\(\frac { Stress }{ Y } \)
= \(dx\times \frac { F/A }{ Y } =\frac { W }{ l\quad Ay } (l-x)dx\)               
Total elongation produced in the rod =  \(\frac { W }{ l\quad Ay } \int _{ 0 }^{ l }{ (l-x)dx } =\frac { W }{ l\quad Ay } { \left[ lx-\frac { { x }^{ 2 } }{ 2 } \right] }_{ 0 }^{ l }\)
= \(\frac { W }{ l\quad Ay } \left[ { l }^{ 2 }-\frac { { l }^{ 2 } }{ 2 } \right] =\frac { W }{ 2Ay } \)

Given, compressional force, F = 50000 N
Diameter, D = 0.5 mm = 5\(\times \)10^-4 m
Radius, r = \(\frac { D }{ 2 } \) = 2.5 \(\times \)10^-4m
Pressure at the tip of the anvil (p) = \(\frac { Force }{ Area } \)
\( p=\frac { F }{ \pi { r }^{ 2 } } =\frac { 50000 }{ { \left( 2.5\times { 10 }^{ -4 } \right) }^{ 2 } } \) 
=  2.5 \(\times \)10¹¹ Pa.

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) \(=\frac { p }{ \triangle V/V } \)
\(\\ \triangle V=\frac { pV }{ B }\)
\( \\ \triangle V=\frac { 7\times 10^{ 6 }\times 1\times 10^{ -3 } }{ 140\times 10^{ 9 } } =\frac { 1 }{ 20 } \times 10^{ -6 }m^{ 3 }\)
\(\\ =0.05\times 10^{ -6 }m^{ 3 }=5\times 10^{ -8 }m^{ 3 }\)